Fundamental Theorem of Algebra for Quadratics - Algebra 2

1. Over the real numbers, a quadratic $ax^2+bx+c=0$ can have 0, 1, or 2 real solutions. Over the complex numbers, it always has exactly 2 solutions counting multiplicity. The discriminant $b^2-4ac$ determines the solution type: $>0$ two distinct real, $=0$ one repeated real (multiplicity 2), $<0$ two complex conjugates. For $x^2+6x+9=0$, which statement is correct?
   Answer: It has one repeated real solution $x=-3$ with multiplicity 2 (so 2 solutions counted with multiplicity) This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b² - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. For x² + 6x + 9 = 0, let's verify: (1) Calculate discriminant: b² - 4ac = 36 - 36 = 0 (zero). (2) This means 1 real solution with multiplicity 2. (3) Notice x² + 6x + 9 = (x + 3)² = 0, giving x = -3 with multiplicity 2. (4) Count: 2 solutions (one value appearing twice). Choice C correctly recognizes that zero discriminant means one repeated real solution x = -3 with multiplicity 2, maintaining the required total of 2 solutions. Choice A incorrectly claims positive discriminant and two distinct real solutions. Choice B incorrectly claims negative discriminant and complex solutions. Choice D makes the critical error of counting the repeated root as only 1 solution total instead of 2—multiplicity must be counted! Perfect square trinomials: (1) When b² - 4ac = 0, the quadratic is a perfect square. (2) General form: a(x + b/(2a))² = 0. (3) The single solution -b/(2a) has multiplicity 2. (4) In factored form (x - r)², the exponent 2 is the multiplicity. Multiplicity counting: Think of (x + 3)² = 0 as (x + 3)(x + 3) = 0. Each factor contributes one zero at x = -3, so we count it twice. This preserves the degree-2 polynomial having exactly 2 zeros!
2. Over the complex numbers, every quadratic $ax^2+bx+c=0$ has exactly 2 solutions counting multiplicity. The discriminant $b^2-4ac$ determines the type of solutions (real distinct, repeated real, or complex conjugates), but not the total count. Which statement best summarizes this fact for quadratics with real coefficients?
   Answer: A quadratic always has exactly 2 complex solutions counting multiplicity; the discriminant only tells whether they are two real, one repeated real, or a conjugate pair This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The discriminant b² - 4ac determines the TYPE of solutions (real distinct, repeated real, or complex conjugates) but never changes the total count of 2. This is a crucial distinction: over the reals, a quadratic might have 0, 1, or 2 real solutions, but over the complex numbers, it always has exactly 2 solutions counting multiplicity. Choice B correctly summarizes this fundamental fact: quadratics always have exactly 2 complex solutions counting multiplicity, with the discriminant only determining whether they're real or complex. Choice A incorrectly suggests the number of solutions can vary from 0 to 3—this violates the Fundamental Theorem which guarantees exactly 2 for degree 2. Choice C incorrectly claims quadratics always have 2 real solutions, ignoring that negative discriminant gives complex solutions. Choice D makes the critical error of claiming negative discriminant means 0 solutions in ℂ—it actually means 2 complex conjugate solutions! Key insight about discriminant: (1) Discriminant > 0: two distinct real solutions. (2) Discriminant = 0: one real solution with multiplicity 2 (counted twice). (3) Discriminant < 0: two complex conjugate solutions. In all three cases, the total is exactly 2! The discriminant is a type indicator, not a count indicator. Complex completeness: The complex numbers ℂ are algebraically closed, meaning every polynomial has all its roots in ℂ. This completeness is why we can guarantee exactly n roots for degree n, unlike the real numbers where roots might "escape" into the complex realm!
3. Factor the quadratic $x^2+25$ over the complex numbers. Recall that every quadratic $ax^2+bx+c=0$ has exactly 2 complex zeros counting multiplicity (2 distinct real, 1 repeated real, or 2 complex conjugates), and the discriminant indicates the type. Which factorization and zero count is correct?
   Answer: $x^2+25=(x+5i)(x-5i)$, so the zeros are $5i$ and $-5i$ (2 complex zeros). This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b squared - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. But in ALL three cases, counting multiplicity gives exactly 2 total! This is why we can always factor quadratics as a(x - r1)(x - r2) even if r values are complex or repeated. For x squared + 25 = 0, discriminant = 0 - 100 = -100 (negative), so factors as (x + 5i)(x - 5i), zeros 5i and -5i (two complex). Choice B correctly recognizes that quadratics always have exactly 2 complex solutions when counting multiplicity and properly identifies the conjugate factorization. Choice A factors as (x+5)(x-5) but that's for x squared - 25; +25 requires imaginaries. To apply this transferable strategy: (1) To factor over complexes, use roots from quadratic formula. (2) For x squared + $k^2$ = (x + ki)(x - ki). (3) Always 2 zeros. For example, x squared + 4 = (x+2i)(x-2i). Keep up the fantastic effort!
4. A key fact (sometimes called the Fundamental Theorem for quadratics) is that any quadratic $ax^2+bx+c=0$ has exactly 2 complex solutions counting multiplicity: two distinct real, one repeated real (multiplicity 2), or two complex conjugates. Which choice lists one example for each case and matches it with the correct solution type determined by the discriminant $b^2-4ac$?
   Answer: $x^2-5x+6=0$ (two distinct real); $x^2-4x+4=0$ (repeated real); $x^2+4=0$ (complex conjugates) This question tests your understanding of the Fundamental Theorem of Algebra for quadratics: every quadratic has exactly 2 complex solutions counting multiplicity, with three cases based on the discriminant. The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b² - 4ac: (1) if positive, two distinct real solutions (like x = 2 and x = 3 for x² - 5x + 6), (2) if zero, one real solution with multiplicity 2 (like x = 2 for x² - 4x + 4), (3) if negative, two complex conjugate solutions (like ±2i for x² + 4). But in ALL three cases, counting multiplicity gives exactly 2 total! Choice A correctly matches each example to its type. Choice D is a tempting distractor suggesting quadratics can have 0 or 3 solutions, but FTA fixes it at exactly 2 counting multiplicity—no more, no less! Classify by computing discriminants: positive for distinct reals, zero for repeated, negative for conjugates. Use this to pick examples for each case. You're doing wonderfully—keep exploring these!
5. A quadratic polynomial $g(x)$ has integer coefficients and roots $\alpha = \frac{1+i\sqrt{7}}{2}$ and $\beta = \frac{1-i\sqrt{7}}{2}$. What is the minimum possible value of $|g(1)|$?
   Answer: 2 When you encounter a quadratic polynomial with complex conjugate roots, you're working with the relationship between roots and coefficients. Since the polynomial has integer coefficients, any complex roots must come in conjugate pairs—which is exactly what we have here. First, let's find the simplest polynomial with these roots. If $\alpha = \frac{1+i\sqrt{7}}{2}$ and $\beta = \frac{1-i\sqrt{7}}{2}$ are roots, then the polynomial is $g(x) = a(x-\alpha)(x-\beta)$ for some integer $a$. Using the sum and product of roots: $\alpha + \beta = \frac{1+i\sqrt{7}}{2} + \frac{1-i\sqrt{7}}{2} = 1$ and $\alpha \beta = \frac{(1+i\sqrt{7})(1-i\sqrt{7})}{4} = \frac{1-i^2 \cdot 7}{4} = \frac{1+7}{4} = 2$. So $g(x) = a(x^2 - x + 2)$ where $a$ is any nonzero integer. To find the minimum value of $|g(1)|$, substitute $x = 1$: $g(1) = a(1^2 - 1 + 2) = a(2) = 2a$. Therefore, $|g(1)| = |2a| = 2|a|$. Since $a$ must be a nonzero integer, the minimum value of $|a|$ is 1, making the minimum value of $|g(1)|$ equal to 2. Choice A) 1 is impossible since $|g(1)| = 2|a|$ must be even. Choice B) 4 occurs when $|a| = 2$, which is possible but not minimal. Choice C) 3 is impossible since $2|a|$ can never equal 3 for integer $a$. Choice D) 2 is correct, achieved when $a = \pm 1$. Key strategy: For monic polynomials with integer coefficients, remember that scaling by any nonzero integer gives all possible polynomials with those roots.
6. For the quadratic equation $x^2 + 4x + 4 = 0$ (an example of $ax^2+bx+c=0$), recall: every quadratic has exactly 2 complex solutions counting multiplicity, and the discriminant $b^2-4ac$ determines whether they are 2 distinct real, 1 repeated real (multiplicity 2), or 2 complex conjugates. Using the discriminant, which statement is correct?
   Answer: The discriminant is $0$, so there is one real solution $x=-2$ with multiplicity $2$ (total of 2 solutions counting multiplicity). This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b squared - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. But in ALL three cases, counting multiplicity gives exactly 2 total! This is why we can always factor quadratics as a(x - r1)(x - r2) even if r values are complex or repeated. For x squared + 4x + 4 = 0, discriminant = 16 - 16 = 0, so one repeated real root x = -4/2 = -2 with multiplicity 2 (from (x + 2) squared = 0), totaling 2 solutions counting multiplicity. Choice A correctly recognizes that quadratics always have exactly 2 complex solutions when counting multiplicity and properly identifies the repeated real root case from zero discriminant. Choice B says no solutions because discriminant is zero—this is incorrect, as zero discriminant means a repeated real root, not none; over complexes, there are always 2, here both being -2. To apply this transferable strategy: (1) Compute discriminant. (2) If zero, root is -b/(2a) with multiplicity 2. (3) Total always 2. For example, (x - 1) squared = 0 has x=1 twice. Great job—remember multiplicity counts repeats!
7. If $z_1$ and $z_2$ are the roots of $3x^2 - 12x + 15 = 0$, what is the value of $|z_1|^2 + |z_2|^2$?
   Answer: 10 When you encounter a quadratic with complex roots and need to find expressions involving absolute values, think about using Vieta's formulas combined with properties of complex conjugates rather than solving for the individual roots. For the quadratic $3x^2 - 12x + 15 = 0$, first check the discriminant: $b^2 - 4ac = (-12)^2 - 4(3)(15) = 144 - 180 = -36$. Since this is negative, the roots are complex conjugates. Let's call them $z_1 = a + bi$ and $z_2 = a - bi$. From Vieta's formulas: $z_1 + z_2 = \frac{12}{3} = 4$ and $z_1 \cdot z_2 = \frac{15}{3} = 5$. Since the roots are conjugates, $z_1 + z_2 = (a + bi) + (a - bi) = 2a = 4$, so $a = 2$. Also, $z_1 \cdot z_2 = (a + bi)(a - bi) = a^2 + b^2 = 5$, so $4 + b^2 = 5$, giving us $b^2 = 1$. Therefore: $|z_1|^2 = a^2 + b^2 = 4 + 1 = 5$ and $|z_2|^2 = a^2 + b^2 = 4 + 1 = 5$. So $|z_1|^2 + |z_2|^2 = 5 + 5 = 10$. Choice A (8) might come from incorrectly using just $2a = 4$. Choice B (16) could result from squaring the sum of roots: $(z_1 + z_2)^2 = 16$. Choice C (12) might arise from adding the coefficients incorrectly. Remember: for complex conjugate roots of a quadratic, $|z_1|^2 + |z_2|^2$ always equals the product of the roots times 2, since both absolute values are equal.
8. A quadratic polynomial with real coefficients can be written as $ax^2+bx+c$ with $a\ne 0$. In the complex numbers it always has exactly 2 zeros counting multiplicity, which previews the Fundamental Theorem of Algebra (degree-$n$ polynomials have exactly $n$ complex zeros counting multiplicity). Which statement correctly interprets "counting multiplicity" for the quadratic $(x-3)^2=0$?
   Answer: It has one solution value, $x=3$, and it is counted twice (multiplicity 2), giving 2 solutions counting multiplicity. This question tests your understanding of the Fundamental Theorem of Algebra for quadratics: every quadratic has exactly 2 complex zeros counting multiplicity, with a focus on interpreting multiplicity for repeated roots. The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b² - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. But in ALL three cases, counting multiplicity gives exactly 2 total! For (x - 3)² = 0, it's one value x = 3 with multiplicity 2, totaling 2 solutions. Choice B correctly interprets counting multiplicity this way. Choice C is a tempting distractor claiming only one total violates the theorem, but including multiplicity upholds it at exactly 2! To interpret multiplicity, look at the factored form: exponent indicates how many times to count the root. Apply this to any polynomial for FTA compliance. You're fantastic—keep going strong!
9. Consider the quadratic equation $x^2 - 6x + 13 = 0$ (a special case of $ax^2+bx+c=0$). Over the complex numbers, every quadratic has exactly 2 solutions counting multiplicity (two distinct real, one repeated real with multiplicity 2, or two complex conjugates). Use the discriminant $b^2-4ac$ to determine the solution type and then find all solutions in $\mathbb{C}$. Which option correctly states the solutions (counting multiplicity)?
   Answer: Two complex conjugate solutions: $x=3+2i$ and $x=3-2i$ This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b² - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. For x² - 6x + 13 = 0, let's verify exactly 2 complex solutions: (1) Calculate discriminant: b² - 4ac = 36 - 52 = -16 (negative). (2) This means 2 complex conjugate solutions (not real). (3) Using quadratic formula: x = (6 ± √-16)/2 = (6 ± 4i)/2 = 3 ± 2i. (4) Solutions: 3 + 2i and 3 - 2i (two complex conjugates). Count: 2 solutions. Choice C correctly recognizes that quadratics always have exactly 2 complex solutions when counting multiplicity and properly identifies the complex conjugate pair 3 + 2i and 3 - 2i. Choice A incorrectly states one real solution with multiplicity 1, which would violate the Fundamental Theorem—quadratics must have 2 solutions counting multiplicity! Choice B has the wrong imaginary parts (should be ±2i, not ±i√4 = ±2i written incorrectly). Choice D incorrectly claims real solutions when the negative discriminant guarantees complex ones. Solution counting with multiplicity: (1) Calculate discriminant to determine solution type. (2) Use quadratic formula when discriminant is negative: x = (-b ± √(b² - 4ac))/(2a). (3) For negative discriminant, √(negative) = i√(positive), giving conjugate pair. (4) Always verify you get exactly 2 solutions total. Conjugate pair recognition: for real-coefficient quadratics with negative discriminant, solutions are p + qi and p - qi where p = -b/(2a) and q = √|discriminant|/(2a). They're symmetric about the real axis!
10. For a quadratic $ax^2+bx+c=0$ with real coefficients, the discriminant $b^2-4ac$ determines the type of solutions: $>0$ two distinct real, $=0$ one repeated real (multiplicity 2), $<0$ two complex conjugates. Regardless of type, there are exactly 2 complex solutions counting multiplicity. For $x^2-2x+5=0$, which choice correctly gives the solutions in $\mathbb{C}$?
    Answer: $x=1\pm 2i$ This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b² - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. For x² - 2x + 5 = 0, let's find the solutions: (1) Calculate discriminant: b² - 4ac = 4 - 20 = -16 (negative). (2) This means 2 complex conjugate solutions. (3) Using quadratic formula: x = (2 ± √-16)/2 = (2 ± 4i)/2 = 1 ± 2i. (4) Solutions: 1 + 2i and 1 - 2i. Choice A correctly gives the complex conjugate pair x = 1 ± 2i as the solutions. Choice B incorrectly has ±4i in the solutions—this comes from not dividing √-16 = 4i by 2 in the quadratic formula. Choice C has the wrong real and imaginary parts reversed. Choice D makes the fundamental error of claiming "no solutions" for negative discriminant—negative discriminant means complex solutions, not no solutions! The complex numbers were invented precisely to ensure equations like this have solutions! Quadratic formula with complex results: (1) When discriminant < 0, write √(negative) = i√(positive). (2) x = (-b ± i√|b² - 4ac|)/(2a). (3) This gives x = p ± qi where p = -b/(2a) is the real part and q = √|discriminant|/(2a) is the imaginary part. (4) The ± ensures we get both conjugates. Common error prevention: When using the quadratic formula with negative discriminant, remember to divide BOTH the real part (-b) and the imaginary part (±i√|discriminant|) by 2a. Missing this division is a frequent source of errors!
11. For the quadratic equation $x^2+4x+4=0$ (a case of $ax^2+bx+c=0$ with real coefficients), remember: every quadratic has exactly 2 complex solutions counting multiplicity. The discriminant $b^2-4ac$ determines whether the solutions are 2 distinct real, 1 repeated real (counted twice), or 2 complex conjugates. Which statement correctly describes the solutions (including multiplicity)?
    Answer: One repeated real solution $x=-2$ with multiplicity 2 (so 2 solutions counting multiplicity) This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b² - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. But in all cases, counting multiplicity gives exactly 2 total! For x² + 4x + 4 = 0, the discriminant is 16 - 16 = 0, so one repeated real root x = -2 with multiplicity 2 (from (x + 2)² = 0), totaling 2 solutions. Choice C correctly recognizes this as one repeated real solution with multiplicity 2, ensuring the total count is 2 as per the theorem. Choice B fails by counting it as only 1 complex solution without multiplicity, but the theorem requires counting repeats—multiplicity matters! For a transferable strategy, factor or use the quadratic formula, then count roots with multiplicity: for $(x - r)^k$, count r k times; this always gives 2 for quadratics. You're doing great—keep verifying with examples to solidify this concept!
12. A quadratic with real coefficients (of the form $ax^2+bx+c=0$) always has exactly 2 complex solutions counting multiplicity. If its discriminant is negative, the solutions are non-real and come in complex conjugate pairs. If $2+3i$ is a solution of such a quadratic, what must the other solution be?
    Answer: $2-3i$ This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b² - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. But in all cases, counting multiplicity gives exactly 2 total! For a real-coefficient quadratic with one complex root like 2 + 3i, the other must be its conjugate 2 - 3i to maintain real coefficients, ensuring exactly two solutions. Choice A correctly identifies 2 - 3i as the other solution, leveraging the conjugate pair property. Choice D tempts by suggesting the same root again, but that's incorrect—conjugates are distinct unless imaginary part is zero, and multiplicity would still require proper counting. To master this, remember: for real coefficients, non-real roots come in p + qi and p - qi pairs; if you know one, flip the sign of i-term for the other. Great job exploring this—you're unlocking the beauty of complex numbers!
13. Use the discriminant to analyze $x^2-4x+5=0$. For a quadratic $ax^2+bx+c=0$, the discriminant $b^2-4ac$ determines the type of solutions, but in the complex numbers there are always exactly 2 solutions counting multiplicity (previewing the Fundamental Theorem of Algebra for degree 2). Which option correctly gives the solutions?
    Answer: Two complex conjugate solutions: $x=2\pm i$ This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b² - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. But in ALL three cases, counting multiplicity gives exactly 2 total! For x² - 4x + 5 = 0, discriminant = 16 - 20 = -4 < 0, solutions x = (4 ± √(-4))/2 = (4 ± 2i)/2 = 2 ± i. Choice C correctly identifies the complex conjugates. Choice D is a tempting distractor claiming no solutions for negative discriminant, but over complexes, there are always 2—FTA ensures it! Apply the quadratic formula with i for negative discriminants to find pairs. Verify by expanding (x - (2 + i))(x - (2 - i)) back to the original. Excellent job—you're getting this!
14. For a quadratic $ax^2+bx+c=0$ with real coefficients, the discriminant $b^2-4ac$ determines whether the two complex solutions (counting multiplicity) are 2 distinct real, 1 repeated real, or 2 complex conjugates. If a quadratic has discriminant $b^2-4ac<0$, which must be true?
    Answer: It has exactly two complex solutions that are conjugates of each other. This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b squared - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. But in ALL three cases, counting multiplicity gives exactly 2 total! This is why we can always factor quadratics as a(x - r1)(x - r2) even if r values are complex or repeated. If discriminant <0 for real-coefficient quadratic, there are two complex conjugate solutions (not real, but still exactly two). Choice B correctly recognizes that quadratics always have exactly 2 complex solutions when counting multiplicity and properly identifies conjugates for negative discriminant. Choice A says exactly one complex solution since discriminant negative—this ignores that for real coefficients, complex roots come in pairs; can't have odd number. To apply this transferable strategy: (1) Negative disc means conjugates p ± qi. (2) Real coefficients ensure pair. (3) Total 2. Example: if disc=-16, q=sqrt(16)/(2a)=2/|a| or similar. You're excelling—believe in yourself!
15. The quadratic $h(x) = x^2 + px + q$ satisfies $h(2+i) = 0$. If $p$ and $q$ are real numbers, which equation must also be satisfied?
    Answer: $h(2-i) = 0$ and $p = -4, q = 5$ Since h(x) has real coefficients and 2+i is a root, its complex conjugate 2-i must also be a root by the Fundamental Theorem of Algebra. Using Vieta's formulas: sum of roots = (2+i) + (2-i) = 4 = -p, so p = -4. Product of roots = (2+i)(2-i) = 4 + 1 = 5 = q. Choice B has wrong signs and wrong conjugate. Choice C has wrong sign for q. Choice D suggests a different root structure entirely.
16. For the quadratic equation $x^2+4x+4=0$ (with real coefficients), remember: every quadratic $ax^2+bx+c=0$ has exactly 2 complex solutions counting multiplicity. The discriminant $b^2-4ac$ determines whether the solutions are two distinct real ($>0$), one repeated real (=$0$, multiplicity 2), or two complex conjugates ($<0$). Which statement correctly describes the solutions of $x^2+4x+4=0$ in the complex numbers?
    Answer: One repeated real solution: $x=-2$ with multiplicity 2 (so 2 solutions counting multiplicity) This question tests your understanding of the Fundamental Theorem of Algebra for quadratics: every quadratic has exactly 2 complex solutions counting multiplicity, with the discriminant revealing if they're distinct real, repeated real, or complex conjugates. The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b² - 4ac: (1) if positive, two distinct real solutions (like x = 2 and x = 5), (2) if zero, one real solution with multiplicity 2 (like x = 3 counted twice from (x - 3)²), (3) if negative, two complex conjugate solutions (like 2 + 3i and 2 - 3i). But in ALL three cases, counting multiplicity gives exactly 2 total! For x² + 4x + 4 = 0, discriminant = 16 - 16 = 0, so one repeated real root: x = -4/2 = -2 with multiplicity 2, or factored as (x + 2)² = 0. Choice C correctly recognizes this, emphasizing multiplicity to reach exactly 2 solutions. Choice D is a tempting distractor because it says exactly one solution since the parabola touches the x-axis once—this counts distinct values over reals, but we must count multiplicity for the full 2 in complexes. To handle repeated roots, factor the quadratic or use the formula; if discriminant is zero, the single value counts twice. Always verify the total count: for (x - r)² = 0, it's r twice. Great job tackling this—you've got the hang of multiplicity!
17. Solve $x^2+2x+5=0$ in the complex numbers. Recall: every quadratic $ax^2+bx+c=0$ has exactly 2 complex solutions counting multiplicity, and the discriminant $b^2-4ac$ determines whether they are 2 distinct real, 1 repeated real (multiplicity 2), or 2 complex conjugates.
    Answer: Two complex solutions: $x=-1+2i$ and $x=-1-2i$ This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b² - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. But in all cases, counting multiplicity gives exactly 2 total! For x² + 2x + 5 = 0, discriminant 4 - 20 = -16 < 0, so two complex: x = [-2 ± √(-16)] / 2 = [-2 ± 4i] / 2 = -1 ± 2i. Choice C correctly lists these two complex solutions, fulfilling the theorem's count. Choice D wrongly claims no solutions due to negative discriminant, but complexes ensure solvability—always two! Practice by solving with the formula and verifying conjugates for negative discriminants; it always works out to two. Keep up the fantastic work—you've got this!
18. A preview of the Fundamental Theorem of Algebra says a polynomial of degree $n$ has exactly $n$ complex zeros counting multiplicity. For degree $2$, this means every quadratic $ax^2+bx+c=0$ has exactly 2 complex solutions counting multiplicity (two distinct real, one repeated real counted twice, or two complex conjugates). Which quadratic below has exactly one real solution value, but still counts as 2 solutions because of multiplicity?
    Answer: $x^2-4x+4=0$ This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always! We need to find which quadratic has discriminant = 0 (giving one repeated real solution counted twice). Let's check each: (A) x² - 5x + 6: discriminant = 25 - 24 = 1 > 0 (two distinct reals). (B) x² + 4: discriminant = 0 - 16 = -16 < 0 (two complex conjugates). (C) x² - 4x + 4: discriminant = 16 - 16 = 0 (one repeated real). (D) x² + 2x + 5: discriminant = 4 - 20 = -16 < 0 (two complex conjugates). Only C has discriminant 0! Indeed, x² - 4x + 4 = (x - 2)² = 0 gives x = 2 with multiplicity 2. Choice C correctly identifies x² - 4x + 4 = 0 as having one real solution value (x = 2) that counts as 2 solutions due to multiplicity 2. Choice A has two distinct real solutions (x = 2 and x = 3). Choice B has two complex conjugate solutions (x = ±2i). Choice D has two complex conjugate solutions (x = -1 ± 2i). Discriminant quick check: (1) For ax² + bx + c, discriminant = b² - 4ac. (2) If you recognize a perfect square pattern (first and last terms are perfect squares, middle term is ±2 times their square roots), discriminant = 0. (3) Example: x² - 6x + 9 = (x)² - 2(x)(3) + (3)² = (x - 3)². (4) This pattern guarantees one repeated root. Multiplicity interpretation: When we say "one real solution with multiplicity 2," we mean the value appears twice as a root. Graphically, the parabola touches the x-axis at exactly one point (the vertex) rather than crossing it at two points!
19. Factor the quadratic $x^2+25$ over the complex numbers. Recall: every quadratic $ax^2+bx+c=0$ has exactly 2 complex zeros counting multiplicity (two real, one repeated real, or a complex conjugate pair), and this idea previews the Fundamental Theorem of Algebra: a degree-$n$ polynomial has exactly $n$ complex zeros counting multiplicity. Which factorization and zero count is correct for $x^2+25$?
    Answer: $x^2+25=(x+5i)(x-5i)$; zeros $\pm 5i$ (2 complex zeros) This question tests your understanding of the Fundamental Theorem of Algebra for quadratics: every quadratic has exactly 2 complex zeros counting multiplicity, allowing factorization over complexes even if not over reals. The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b² - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. But in ALL three cases, counting multiplicity gives exactly 2 total! For x² + 25, discriminant = 0 - 100 = -100 < 0, so zeros ±5i, factoring as (x + 5i)(x - 5i). Choice B correctly gives this factorization and zero count. Choice C is a tempting distractor claiming irreducibility over complexes with 0 zeros, but FTA guarantees factorization into linears over complexes—always 2 for quadratics! To factor, solve x² + 25 = 0 → x = ±5i, then write (x - r1)(x - r2). This works for any quadratic—try it with positives or zeros too. Keep up the great work—you're building strong skills!
20. Show that $x^2-4x+5=0$ has exactly 2 complex solutions counting multiplicity (as every quadratic $ax^2+bx+c=0$ does in $\mathbb{C}$). Use the discriminant $b^2-4ac$ to determine the type, then find the solutions. Which option is correct?
    Answer: Two complex conjugate solutions: $x=2+i$ and $x=2-i$ This question tests your understanding of the Fundamental Theorem of Algebra as applied to quadratics: every quadratic equation has exactly 2 solutions in the complex number system when counting multiplicity (repeated roots counted by how many times they appear). The Fundamental Theorem of Algebra states that every polynomial of degree n has exactly n complex zeros counting multiplicity. For quadratics (degree 2), this means exactly 2 solutions always—no exceptions! The type of solutions depends on the discriminant b² - 4ac: (1) if positive, two distinct real solutions, (2) if zero, one real solution with multiplicity 2, (3) if negative, two complex conjugate solutions. For x² - 4x + 5 = 0, let's verify exactly 2 complex solutions: (1) Calculate discriminant: b² - 4ac = 16 - 20 = -4 (negative). (2) This means 2 complex conjugate solutions (not real). (3) Using quadratic formula: x = (4 ± √-4)/2 = (4 ± 2i)/2 = 2 ± i. (4) Solutions: 2 + i and 2 - i (two complex conjugates). Count: 2 solutions. Choice A correctly recognizes that the negative discriminant yields two complex conjugate solutions 2 + i and 2 - i. Choice B incorrectly claims positive discriminant and real solutions 2 ± 1 = 3 and 1, but we can verify these don't work: 3² - 4(3) + 5 = 9 - 12 + 5 = 2 ≠ 0. Choice C incorrectly claims zero discriminant and repeated real solution. Choice D violates the Fundamental Theorem by claiming only one complex solution—complex solutions of real-coefficient quadratics must come in conjugate pairs! Verification technique: (1) Always check your solutions by substitution. (2) For x = 2 + i: (2 + i)² - 4(2 + i) + 5 = 4 + 4i - 1 - 8 - 4i + 5 = 0 ✓. (3) Complex arithmetic: (a + bi)² = a² + 2abi + (bi)² = a² - b² + 2abi. (4) Both solutions must check out for the answer to be correct. Complex conjugate theorem: For polynomials with real coefficients, complex roots come in conjugate pairs. This maintains real coefficients when the polynomial is expanded from its factored form!

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