Quadratic Equation Practice Questions

Factor the left side of the equation. Find two numbers that multiply to 6 and add to −5:

\(x^2 – 5x + 6 = (x – 2)(x – 3) = 0\)

By the zero product property, if the product of two factors equals zero, then at least one factor must equal zero:

\(x – 2 = 0 \implies x = 2\)

\(x – 3 = 0 \implies x = 3\)

Choice C has the correct numbers but the wrong signs. Always check your solutions by substituting them back into the original equation.

To solve an equation of the form \(x^2 = c\), take the square root of both sides. Remember to include both the positive and negative roots:

\(x = \pm\sqrt{49} = \pm 7\)

This gives two solutions: \(x = 7\) and \(x = -7\). A common mistake is to forget the negative root — since both \((7)^2 = 49\) and \((-7)^2 = 49\), both values are valid solutions.

Factor the quadratic. Multiply the leading coefficient by the constant: \(2 \times (-3) = -6\). Find two numbers that multiply to −6 and add to 5:

\(6 \times (-1) = -6\:\) and \(\:6 + (-1) = 5\)

Rewrite and factor by grouping:

\(2x^2 + 6x – x – 3 = 0\)

\(2x(x + 3) – 1(x + 3) = 0\)

\((2x – 1)(x + 3) = 0\)

Set each factor equal to zero:

\(2x – 1 = 0 \implies x = \frac{1}{2}\)

\(x + 3 = 0 \implies x = -3\)

The number of real solutions is determined by the discriminant. For a quadratic equation \(ax^2 + bx + c = 0\), the discriminant is:

\(\Delta = b^2 – 4ac\)

Substitute \(a = 1\), \(b = 4\), and \(c = 7\):

\(\Delta = (4)^2 – 4(1)(7) = 16 – 28 = -12\)

Since the discriminant is negative, the equation has no real solutions. The rules are:

• \(\Delta > 0\): Two distinct real solutions
• \(\Delta = 0\): Exactly one real solution (a repeated root)
• \(\Delta \lt 0\): No real solutions

The quadratic formula states:

\(x = \dfrac{-b \pm \sqrt{b^2 – 4ac}}{2a}\)

Identify the coefficients: \(a = 1\), \(b = -6\), \(c = 2\). Substitute into the formula:

\(x = \dfrac{-(-6) \pm \sqrt{(-6)^2 – 4(1)(2)}}{2(1)}\:\)\(\:= \dfrac{6 \pm \sqrt{36 – 8}}{2}\)\(\:= \dfrac{6 \pm \sqrt{28}}{2}\)

Simplify the radical: \(\sqrt{28} = 2\sqrt{7}\). Then simplify the fraction:

\(= \dfrac{6 \pm 2\sqrt{7}}{2} = 3 \pm \sqrt{7}\)

Factor out the GCF, which is \(x\):

\(x^2 + 8x = x(x + 8) = 0\)

Apply the zero product property and set each factor equal to zero:

\(x = 0\)

\(x + 8 = 0 \implies x = -8\)

A common mistake is to divide both sides by \(x\), which eliminates the solution \(x = 0\). Never divide both sides of an equation by the variable — always factor instead.

Take the square root of both sides. Remember to include \(\pm\):

\(x – 4 = \pm\sqrt{20}\)

Simplify the radical: \(\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}\). Then isolate \(x\):

\(x = 4 \pm 2\sqrt{5}\)

Choice B is not fully simplified because \(\sqrt{20}\) can be reduced to \(2\sqrt{5}\).

Move the constant to the right side:

\(x^2 + 10x = -18\)

To complete the square, take half of the coefficient of \(x\), square it, and add it to both sides. Half of 10 is 5, and \(5^2 = 25\):

\(x^2 + 10x + 25 = -18 + 25\)

\((x + 5)^2 = 7\)

Take the square root of both sides:

\(x + 5 = \pm\sqrt{7}\)

\(x = -5 \pm \sqrt{7}\)

This equation is in vertex form, where \((h, k)\) is the vertex of the parabola:

\(y = a(x – h)^2 + k\)

Comparing with \(y = (x – 3)^2 + 5\), we identify \(h = 3\) and \(k = 5\).

Therefore, the vertex is \((3, 5)\).

Choice B is a common error that comes from forgetting that the formula uses \((x – h)\), not \((x + h)\). Since the equation shows \((x – 3)\), the value of \(h\) is positive 3, not −3.

Isolate \(x^2\) by adding 48 to both sides and then dividing by 3:

\(3x^2 = 48\)

\(x^2 = 16\)

Take the square root of both sides, remembering to include both roots:

\(x = \pm\sqrt{16} = \pm 4\)

Choice A divides 48 by 3 incorrectly, and choice C forgets the negative root.

The maximum height occurs at the vertex of the parabola. For a quadratic in the form \(at^2 + bt + c\), the time at the vertex is:

\(t = \dfrac{-b}{2a}\)

Substitute \(a = -16\) and \(b = 32\):

\(t = \dfrac{-32}{2(-16)} = \dfrac{-32}{-32} = 1\)

The maximum height occurs at \(t = 1\) second. Substitute this back into the height function:

\(h(1) = -16(1)^2 + 32(1) + 5\)\(\:= -16 + 32 + 5 = 21\)

The maximum height is 21 feet.

Factor the trinomial. Find two numbers that multiply to −15 and add to −2:

\(-5 \times 3 = -15\:\) and \(\:-5 + 3 = -2\)

So:

\(x^2 – 2x – 15 = (x + 3)(x – 5) = 0\)

Set each factor equal to zero:

\(x + 3 = 0 \implies x = -3\)

\(x – 5 = 0 \implies x = 5\)

To convert to vertex form, complete the square. Start with the \(x\) terms:

\(y = (x^2 – 8x) + 19\)

Take half of the coefficient of \(x\) and square it: half of −8 is −4, and \((-4)^2 = 16\). Add and subtract 16 inside the expression:

\(y = (x^2 – 8x + 16 – 16) + 19\)

\(= (x^2 – 8x + 16) + 19 – 16\)

\(= (x – 4)^2 + 3\)

Let \(w\) represent the width. The length is \(w + 4\). Since area equals length times width:

\(w(w + 4) = 60\)

Expand and rearrange into standard form:

\(w^2 + 4w – 60 = 0\)

Factor the trinomial. Find two numbers that multiply to −60 and add to 4:

\(10 \times (-6) = -60\:\) and \(\:10 + (-6) = 4\)

\((w + 10)(w – 6) = 0\)

This gives \(w = -10\) or \(w = 6\). Since a width cannot be negative, the width is 6 units.

Using the quadratic formula with \(a = 3\), \(b = 2\), and \(c = -5\):

\(x = \dfrac{-2 \pm \sqrt{(2)^2 – 4(3)(-5)}}{2(3)}\)

\(= \dfrac{-2 \pm \sqrt{4 + 60}}{6}\)

\(= \dfrac{-2 \pm \sqrt{64}}{6} = \dfrac{-2 \pm 8}{6}\)

Now evaluate the two solutions separately:

\(x = \dfrac{-2 + 8}{6} = \dfrac{6}{6} = 1\)

\(x = \dfrac{-2 – 8}{6} = \dfrac{-10}{6} = -\dfrac{5}{3}\)