College Algebra Quiz: Quadratic Formula And The Discriminant

What this quiz covers

This quiz focuses on Quadratic Formula And The Discriminant, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $a = 1$ (correct answer)
2. $a = \frac{1}{5}$
3. $a = 5$
4. $a = 25$

Explanation: If the roots are $x = 2$ and $x = 7$, then $f(x) = a(x-2)(x-7) = a(x^2 - 9x + 14) = ax^2 - 9ax + 14a$. So $b = -9a$ and $c = 14a$. The discriminant is $\Delta = b^2 - 4ac = (-9a)^2 - 4a(14a) = 81a^2 - 56a^2 = 25a^2$. Given $\Delta = 25$, we have $25a^2 = 25$, so $a^2 = 1$, giving $a = \pm 1$. Since we need a specific value and typically take the positive case unless specified otherwise, $a = 1$. Choice B would result from confusing the discriminant formula. Choice C comes from $\sqrt{25} = 5$. Choice D uses the discriminant value directly.

Question 2

1. $k = 8$ (correct answer)
2. $k = 4$
3. $k = 16$
4. $k = -8$

Explanation: For a quadratic equation to have exactly one real solution, the discriminant must equal zero. Using $ax^2 + bx + c = 0$, we have $a = 2$, $b = -8$, and $c = k$. The discriminant is $b^2 - 4ac = (-8)^2 - 4(2)(k) = 64 - 8k$. Setting this equal to zero: $64 - 8k = 0$, so $k = 8$. Choice B uses $c = 2k$ instead of $k$. Choice C incorrectly uses $b^2$ as the discriminant. Choice D uses the wrong sign.

Question 3

1. $p = 8$; root is $x = -2$
2. $p = -8$; root is $x = 2$
3. $p = \pm 8$; root is $x = \mp 2$ (correct answer)
4. $p = 16$; root is $x = -4$

Explanation: For a repeated root, the discriminant must equal zero: $p^2 - 64 = 0$, so $p^2 = 64$, giving $p = \pm 8$. When $p = 8$: the equation is $2x^2 + 8x + 8 = 0$ or $x^2 + 4x + 4 = 0$, which factors as $(x+2)^2 = 0$, so $x = -2$. When $p = -8$: the equation is $2x^2 - 8x + 8 = 0$ or $x^2 - 4x + 4 = 0$, which factors as $(x-2)^2 = 0$, so $x = 2$. Thus $p = \pm 8$ gives roots $x = \mp 2$ respectively. Choice A only considers one case. Choice B only considers the other case. Choice D uses an incorrect value of $p$.

Question 4

1. $k \leq -4$ or $k \geq 4$
2. $-4 \leq k \leq 4$ (correct answer)
3. $k \leq -4$ or $k > 0$
4. $k < 0$ or $k \geq 4$

Explanation: For real solutions, we need $\Delta \geq 0$. Here $a = k$, $b = 8$, $c = k$, so $\Delta = 8^2 - 4(k)(k) = 64 - 4k^2$. Setting $64 - 4k^2 \geq 0$: $64 \geq 4k^2$, so $16 \geq k^2$, which means $-4 \leq k \leq 4$. We also need $k \neq 0$ for this to be quadratic, but the question asks about real solutions to the equation as given. Choice A gives the complement of the correct interval. Choice C incorrectly excludes some valid $k$ values. Choice D has an incorrect boundary.

Question 5

1. $k = -4$
2. $k = 9$
3. $k = 13$
4. $k = 5$ (correct answer)

Explanation: When you encounter a quadratic function and need to find its minimum value, remember that parabolas opening upward (positive leading coefficient) have their minimum at the vertex. The key is finding the vertex and using it to determine the parameter. For $f_k(x) = x^2 - 6x + k$, you can find the vertex using the formula $x = -\frac{b}{2a}$. Here, $a = 1$ and $b = -6$, so the x-coordinate of the vertex is $x = -\frac{(-6)}{2(1)} = 3$. This means the minimum occurs at $x = 3$ regardless of the value of $k$. The minimum value is $f_k(3) = 3^2 - 6(3) + k = 9 - 18 + k = k - 9$. Since we want this minimum to equal $-4$, we set up the equation: $k - 9 = -4$, which gives us $k = 5$. Looking at the wrong answers: Choice A ($k = -4$) gives a minimum of $-4 - 9 = -13$, not $-4$. This might tempt you if you mistakenly think the minimum value equals $k$ directly. Choice B ($k = 9$) produces a minimum of $9 - 9 = 0$, which occurs when the parabola just touches the x-axis. Choice C ($k = 13$) yields a minimum of $13 - 9 = 4$, the positive version of our target. Study tip: For vertex form problems, always find where the minimum occurs first (the x-coordinate), then substitute back to find the actual minimum value. The parameter $k$ shifts the parabola vertically, but the vertex's x-coordinate stays the same.