Discriminant & Solution Types - Accuplacer Aaf

1. Two quadratic equations $x^2 + px + q = 0$ and $x^2 + rx + s = 0$ both have discriminant equal to $9$. If $p = r + 4$, what is the relationship between $q$ and $s$?
   Answer: $q = s + 2r + 4$ Both discriminants equal 9: $p^2 - 4q = 9$ and $r^2 - 4s = 9$. Since $p = r + 4$, substituting: $(r + 4)^2 - 4q = 9$, so $r^2 + 8r + 16 - 4q = 9$. From the second equation: $r^2 - 4s = 9$, so $r^2 = 9 + 4s$. Substituting: $9 + 4s + 8r + 16 - 4q = 9$, which simplifies to $4s + 8r + 16 = 4q$, so $q = s + 2r + 4$. The other choices reflect sign errors or coefficient errors in the algebra.
2. The discriminant of a quadratic equation with integer coefficients is 20. Which of the following correctly describes the solutions?
   Answer: Two distinct irrational solutions The nature of the solutions is determined by the discriminant, $D$. Since $D = 20$ is positive, the equation has two distinct real solutions. To determine if they are rational or irrational, we check if the discriminant is a perfect square. Since 20 is not a perfect square, the solutions will be irrational.
3. How many distinct real solutions does the equation $\frac{1}{3}x^2 + x + \frac{2}{3} = 0$ have?
   Answer: Two To simplify the calculation, we can clear the fractions by multiplying the entire equation by the least common denominator, 3. This gives $x^2 + 3x + 2 = 0$. Now we find the discriminant, $D = b^2 - 4ac$, with $a=1$, $b=3$, and $c=2$. $D = (3)^2 - 4(1)(2) = 9 - 8 = 1$. Since the discriminant is positive ($D > 0$), the equation has two distinct real solutions.
4. The equation $x^2 - 6x + p = 0$ has discriminant $\Delta_1$, and the equation $x^2 + 6x + p = 0$ has discriminant $\Delta_2$. Which statement is always true?
   Answer: $\Delta_1 = \Delta_2$ for all real values of $p$ For the first equation: $\Delta_1 = (-6)^2 - 4(1)(p) = 36 - 4p$. For the second equation: $\Delta_2 = (6)^2 - 4(1)(p) = 36 - 4p$. Therefore $\Delta_1 = \Delta_2$ for all real values of $p$. The discriminant depends on $b^2$, and $(-6)^2 = 6^2$. Choices A and B incorrectly suggest one is always larger. Choice D incorrectly suggests the relationship varies with $p$.
5. Describe the nature of the roots of the equation $-x^2 + 8x - 16 = 0$.
   Answer: One repeated rational root To find the nature of the roots, we compute the discriminant $D = b^2 - 4ac$. For $-x^2 + 8x - 16 = 0$, we have $a=-1$, $b=8$, and $c=-16$. The discriminant is $D = (8)^2 - 4(-1)(-16) = 64 - 64 = 0$. A discriminant of 0 indicates that there is exactly one real solution, which is a repeated rational root.
6. Which of the following best describes the solutions to the equation $4x^2 - 9 = 0$?
   Answer: Two distinct rational solutions In the equation $4x^2 - 9 = 0$, we have $a=4$, $b=0$, and $c=-9$. The discriminant is $D = b^2 - 4ac = (0)^2 - 4(4)(-9) = 0 + 144 = 144$. Since the discriminant is positive ($D > 0$), there are two distinct real solutions. Because the discriminant (144) is a perfect square ($12^2$), the solutions are rational.
7. A quadratic equation $ax^2 + bx + c = 0$ has discriminant $\Delta = 25$. If the equation is modified by replacing $c$ with $c + 6$, what can be concluded about the number of real solutions of the new equation?
   Answer: The number of real solutions cannot be determined without more information The original discriminant is $\Delta_1 = b^2 - 4ac = 25$. The new discriminant is $\Delta_2 = b^2 - 4a(c + 6) = b^2 - 4ac - 24a = 25 - 24a$. Since we don't know the value of $a$, we cannot determine the sign of $\Delta_2$. If $a > \frac{25}{24}$, then $\Delta_2 < 0$ (no real solutions). If $a = \frac{25}{24}$, then $\Delta_2 = 0$ (one real solution). If $0 < a < \frac{25}{24}$, then $\Delta_2 > 0$ (two real solutions).
8. If the discriminant of $ax^2 + bx + c = 0$ is $64$, what is the discriminant of $ax^2 + bx + (c + \frac{b^2}{4a}) = 0$?
   Answer: $0$ The original discriminant is $\Delta_1 = b^2 - 4ac = 64$. For the new equation, the discriminant is $\Delta_2 = b^2 - 4a(c + \frac{b^2}{4a}) = b^2 - 4ac - 4a \cdot \frac{b^2}{4a} = b^2 - 4ac - b^2 = 0$. The term $\frac{b^2}{4a}$ added to $c$ creates a perfect square trinomial, making the discriminant zero regardless of the original discriminant value. Choices A, C, D reflect misconceptions about how the added term affects the discriminant.
9. The graph of a quadratic function $y = ax^2 + bx + c$ is a parabola that is tangent to the x-axis. What is the value of the discriminant $b^2 - 4ac$?
   Answer: It is equal to 0. When the graph of a quadratic function is tangent to the x-axis, it means the parabola touches the x-axis at exactly one point (the vertex). This corresponds to the quadratic equation having exactly one real solution (a repeated root). This occurs precisely when the discriminant, $b^2 - 4ac$, is equal to 0.
10. For which value of $n$ does the system of equations $y = x^2 + 2x + n$ and $y = 5$ have exactly one solution?
    Answer: $n = 6$ Setting the equations equal: $x^2 + 2x + n = 5$, which rearranges to $x^2 + 2x + (n-5) = 0$. For exactly one solution, the discriminant must be zero: $\Delta = 2^2 - 4(1)(n-5) = 4 - 4(n-5) = 4 - 4n + 20 = 24 - 4n = 0$. Solving: $n = 6$. Choice A gives $\Delta = 8 > 0$. Choice C gives $\Delta = 4 > 0$. Choice D gives $\Delta = 12 > 0$.
11. The quadratic equation $mx^2 + 4x + 1 = 0$ has two distinct real solutions when $m$ satisfies which condition?
    Answer: $m < 4$ and $m \neq 0$ For two distinct real solutions, we need $\Delta > 0$ and $m \neq 0$ (to ensure it's quadratic). $\Delta = 4^2 - 4(m)(1) = 16 - 4m > 0$, so $m < 4$. Combined with $m \neq 0$, the answer is $m < 4$ and $m \neq 0$. Choice A reverses the inequality. Choice B excludes negative values of $m$ unnecessarily. Choice D uses the wrong boundary value and wrong inequality direction.
12. Which of the following quadratic equations has no real solutions?
    Answer: $3x^2 - x + 1 = 0$ An equation has no real solutions if its discriminant $D = b^2 - 4ac$ is negative. We check each option: A) $D = (-8)^2 - 4(1)(16) = 64 - 64 = 0$. B) $D = (5)^2 - 4(2)(-3) = 25 + 24 = 49 > 0$. C) $D = (-1)^2 - 4(3)(1) = 1 - 12 = -11 < 0$. D) $D = (0)^2 - 4(1)(-7) = 28 > 0$. Only option C has a negative discriminant, so it is the correct answer.
13. The equation $x^2+bx+1=0$ has no real solutions. Which of the following must be true for $b$?
    Answer: $-2 < b < 2$ For no real solutions, the discriminant must be negative. Here $a=1, c=1$. $D = b^2 - 4(1)(1) < 0$, which means $b^2 - 4 < 0$, or $b^2 < 4$. Taking the square root of both sides, we get $\sqrt{b^2} < \sqrt{4}$, which is $|b| < 2$. This inequality is equivalent to $-2 < b < 2$.
14. Find the set of all values of $m$ for which the equation $x^2 - 5x + m = 0$ has no real solutions.
    Answer: $m > \frac{25}{4}$ An equation has no real solutions if its discriminant is negative ($D < 0$). For the equation $x^2 - 5x + m = 0$, we have $a=1$, $b=-5$, and $c=m$. The discriminant is $D = (-5)^2 - 4(1)(m) = 25 - 4m$. We set this to be less than 0: $25 - 4m < 0$. Adding (4m) to both sides gives $25 < 4m$. Dividing by 4 gives $\frac{25}{4} < m$, or $m > \frac{25}{4}$.
15. A quadratic equation with rational coefficients has two distinct real solutions. Which condition guarantees that these solutions are also rational?
    Answer: The discriminant is a perfect square greater than zero. The solutions of a quadratic equation are given by the quadratic formula, which involves $\sqrt{D}$, where D is the discriminant. For the solutions to be rational, $\sqrt{D}$ must be rational. This occurs when D is a perfect square. Since the solutions must also be distinct and real, D must be positive. Therefore, the discriminant must be a perfect square greater than zero.
16. Which of the following values for the discriminant of a quadratic equation with rational coefficients corresponds to an equation with two distinct, irrational roots?
    Answer: 24 For a quadratic equation to have two distinct, irrational roots, its discriminant must be positive (for two real roots) and not be a perfect square (for irrational roots). A) -1 is negative (no real roots). B) 0 results in one repeated rational root. C) 24 is positive and not a perfect square, so it corresponds to two distinct irrational roots. D) 49 is a perfect square ($7^2$), so it corresponds to two distinct rational roots.
17. For what value of $k$ does the equation $2x^2 - 8x + k = 0$ have exactly one real solution, and what is that solution?
    Answer: $k = 8$; solution is $x = 2$ For exactly one real solution, the discriminant must equal zero: $\Delta = b^2 - 4ac = (-8)^2 - 4(2)(k) = 64 - 8k = 0$. Solving: $k = 8$. When there's exactly one solution, it equals $x = -\frac{b}{2a} = -\frac{(-8)}{2(2)} = 2$. Choice B has the wrong solution value. Choice C has the wrong $k$ value (confusing $4ac$ with $2ac$). Choice D has both wrong values.
18. A quadratic equation has discriminant $\Delta = -12$. If each coefficient of the equation is multiplied by $3$, what is the discriminant of the resulting equation?
    Answer: $-108$ If the original equation is $ax^2 + bx + c = 0$ with discriminant $\Delta = b^2 - 4ac = -12$, then multiplying each coefficient by 3 gives $(3a)x^2 + (3b)x + (3c) = 0$. The new discriminant is $(3b)^2 - 4(3a)(3c) = 9b^2 - 36ac = 9(b^2 - 4ac) = 9(-12) = -108$. Choice A multiplies by 3 instead of $3^2$. Choice C incorrectly assumes the discriminant stays the same. Choice D gets the magnitude right but wrong sign.
19. For what value of $k$ does the quadratic equation $x^2 - 10x + k = 0$ have exactly one real solution?
    Answer: 25 A quadratic equation has exactly one real solution when its discriminant, $D = b^2 - 4ac$, is equal to 0. For the equation $x^2 - 10x + k = 0$, we have $a=1$, $b=-10$, and $c=k$. Setting the discriminant to zero: $D = (-10)^2 - 4(1)(k) = 0$. This simplifies to $100 - 4k = 0$. Solving for $k$ gives $4k = 100$, so $k = 25$.
20. How many distinct real solutions does the equation $2x^2 - 7x + 8 = 0$ have?
    Answer: None To determine the number of distinct real solutions, we use the discriminant, $D = b^2 - 4ac$, for the quadratic equation $ax^2 + bx + c = 0$. Here, $a=2$, $b=-7$, and $c=8$. The discriminant is $D = (-7)^2 - 4(2)(8) = 49 - 64 = -15$. Since the discriminant is negative ($D < 0$), the equation has no real solutions; it has two complex conjugate solutions.

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