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 + 5 = 0, discriminant = 16 - 20 = -4 (negative), so two complex: x = (4 ± sqrt(-4))/2 = (4 ± 2i)/2 = 2 ± i. Choice C correctly recognizes that quadratics always have exactly 2 complex solutions when counting multiplicity and properly identifies the conjugate pair. Choice D says no solutions because discriminant <0—this applies to reals, but over complexes, negative discriminant gives two complex solutions. To apply this transferable strategy: (1) Use quadratic formula. (2) For imaginary sqrt, get conjugates. (3) Always 2. Example: x squared + 9 = 0 has ±3i. Keep shining—you're awesome!

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!

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 - 5x + 6 = 0, discriminant = 25 - 24 = 1 (positive), so two distinct real solutions (x=2, x=3), totaling 2. Choice C correctly recognizes that quadratics always have exactly 2 complex solutions when counting multiplicity and properly identifies the distinct real case from positive discriminant. Choice D says it could have 0, 1, or 2 complex solutions depending on discriminant—this is misleading; over complexes, always exactly 2 total (which may be real or complex). To apply this transferable strategy: (1) Compute discriminant. (2) Positive means two distinct real. (3) Solve via factoring or formula. For example, x squared - 3x + 2 = (x-1)(x-2). You're doing amazingly—practice more!

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!

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 a real-coefficient quadratic with one solution 2 + 3i, the other must be 2 - 3i, as non-real roots come in conjugate pairs to ensure real coefficients. Choice A correctly recognizes that quadratics always have exactly 2 complex solutions when counting multiplicity and properly identifies the conjugate pair requirement for real coefficients. Choice D says no second solution because a quadratic can have only one complex root—this fails because the theorem guarantees exactly 2, and for real coefficients, complex roots pair up as conjugates; you can't have just one. To apply this transferable strategy: (1) If given one complex root for real-coefficient quadratic, the other is its conjugate (flip imaginary sign). (2) Verify by plugging in or using Vieta's formulas. For example, if root is 4 + i, other is 4 - i. You're building strong skills—keep going!

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. The core idea is that over complexes, every quadratic has precisely 2 zeros counting multiplicity, with discriminant just specifying the flavor. Choice B correctly recognizes that quadratics always have exactly 2 complex solutions when counting multiplicity and properly identifies how discriminant affects type, not count. Choice A says 0,1,2, or 3 complex solutions depending on discriminant—this overstates variability; always exactly 2 total (real count as complex with zero imaginary part). To apply this transferable strategy: (1) Remember the theorem guarantees n roots for degree n. (2) Use discriminant for type. (3) Count with multiplicity. For cubics, always 3, etc. You've got this—super progress!

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 - 3) squared = 0, which is x squared - 6x + 9 = 0, discriminant = 0, so one real root 3 with multiplicity 2, totaling 2 solutions counting multiplicity. Choice C correctly recognizes that quadratics always have exactly 2 complex solutions when counting multiplicity and properly identifies the repeated root case. Choice B says exactly 1 solution counting it once—this forgets multiplicity; the theorem counts repeats by their multiplicity, so 3 counts twice here. To apply this transferable strategy: (1) Factor or solve. (2) Count distinct roots and add multiplicities (e.g., $(x-3)^2$ means 3 twice). (3) Total 2 for quadratics. For example, $(x+1)^2$=0 has -1 twice. You're progressing wonderfully!

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 + 2x + 5 = 0, discriminant = 4 - 20 = -16 (negative), so two complex conjugates: x = (-2 ± sqrt(-16))/2 = (-2 ± 4i)/2 = -1 ± 2i. Choice C correctly recognizes that quadratics always have exactly 2 complex solutions when counting multiplicity and properly identifies the conjugate pair from the negative discriminant. Choice D says no solutions because discriminant is negative—this is true over reals but not complexes; complexes ensure exactly 2 solutions always, here complex ones. To apply this transferable strategy: (1) Calculate discriminant. (2) If negative, solutions are p ± qi with p = -b/(2a), q = sqrt(|disc|)/(2a). (3) Count 2 total. For example, x squared + 1 = 0 has ±i. Excellent work—stay confident!

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!

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 + 2x + 5 = 0, discriminant = 4 - 20 = -16 < 0, so solutions x = (-2 ± √(-16))/2 = (-2 ± 4i)/2 = -1 ± 2i. Choice C correctly states the negative discriminant and the conjugate pair. Choice D is a tempting distractor saying no solutions for negative discriminant, but that's only over reals—complexes always provide exactly 2! When facing a negative discriminant, use i for the square root and proceed with the formula to find the pair. Always double-check by plugging back into the equation to verify. Fantastic effort— you're mastering this concept!