Quadratic equations are one of the most important topics on the ACT Math test. Whether you’re solving equations by factoring, using the quadratic formula, or identifying the vertex of a parabola, mastering these concepts can help you answer questions quickly and accurately.
In this 2026 ACT Math guide, you’ll learn how to solve quadratic equations step by step, recognize the different forms of quadratic expressions, and apply proven strategies to common ACT questions. You’ll also find worked examples and practice problems to build confidence before test day.
Quadratic equations can be solved by factoring, completing the square, or using the quadratic formula. While factoring is efficient for simple problems, the quadratic formula is a universal method that works for all quadratic equations, including those that cannot be factored over the integers. You should memorize the quadratic formula: For a quadratic equation $ax^2+bx+c=0$ (where $a\neq0$), $$ x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}. $$ This formula is derived by completing the square. Mastering this technique is also essential for solving problems involving circles or conic sections.
For example, to solve $x^2-4x+2=0$ by completing the square, first move the constant to the right side: $x^2-4x=-2$. Then, add the square of half the coefficient of $x$ [which is $(-2)^2=4$] to both sides: \begin{align*} x^2-4x+4 & = -2+4 \\ (x-2)^2 & = 2 \\ x-2 & = \pm\sqrt{2} \\ x & = 2\pm\sqrt{2}. \end{align*} While completing the square works for all equations, it can be complicated for more complex equations. Therefore, relying on the quadratic formula is generally more efficient during the exam.
Many quadratic problems can be solved by graphing the function on a calculator and finding the $x$-intercepts. However, this method yields decimal approximations, which may not be sufficient if the problem requires an exact answer in radical form.
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For a quadratic equation in the form $ax^2+bx+c=0$, the sum of the solutions is $-\dfrac{b}{a}$ and the product of the solutions is $\dfrac{c}{a}$. Using these relationships is significantly faster than finding the solutions individually. It is recommended that you memorize these formulas and understand their derivation from the quadratic formula.
For example, to find the sum of the solutions of $3x^2-8x+100=0$, directly apply the formula rather than solving for $x$ (which would yield complex numbers). $$ \text{Sum} = \dfrac{-b}{a} = \dfrac{-(-8)}{3} = \dfrac{8}{3}. $$
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If the solutions of a quadratic equation are $p$ and $q$, the equation can be written as $(x-p)(x-q)=0$. Expanding this yields $x^2-(p+q)x+pq=0$.
For example, if the solutions are $-2$ and $5$, the equation is: $$ (x+2)(x-5)=x^2-3x-10=0. $$ If the solutions involve fractions, such as $\dfrac{1}{2}$ and $\dfrac{1}{3}$, the equation becomes: $$ \left(x-\dfrac{1}{2}\right)\left(x-\dfrac{1}{3}\right)=x^2-\dfrac{5x}{6}+\dfrac{1}{6}=0. $$ To eliminate denominators, multiply the entire equation by the least common denominator (in this case, 6): $$ 6\left(x^2-\dfrac{5x}{6}+\dfrac{1}{6}\right)=0 \implies 6x^2-5x+1=0. $$ Alternatively, you can clear the denominators initially by writing the factors with integer coefficients: $(2x-1)$ corresponding to $x=\frac{1}{2}$, and $(3x-1)$ corresponding to $x=\frac{1}{3}$.
When rational coefficients are assumed, irrational roots occur in conjugate pairs. If one root is $4+\sqrt{3}$, the other must be $4-\sqrt{3}$. To find the equation, multiply the factors: $$ (x-(4+\sqrt{3}))(x-(4-\sqrt{3})) = 0. $$ Regroup terms to use the difference of squares pattern $((a-b)(a+b)=a^2-b^2)$: $$ ((x-4)-\sqrt{3})((x-4)+\sqrt{3}) = (x-4)^2 - (\sqrt{3})^2. $$ Simplifying this yields: $$ (x^2-8x+16)-3 = x^2-8x+13=0. $$ The irrational terms cancel out. Finding an equation from complex roots follows a similar process, which is discussed later in this book.
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To find a missing coefficient when a root (or factor) is known, substitute the root into the equation and solve.
For example, if $x-3$ is a factor of $5x^2-4x+k=0$, what is $k$? Since $x-3$ is a factor, $x=3$ is a root. Substitute $3$ for $x$: \begin{align*} 5(3)^2 -4(3)+k & = 0 \\ 45-12 +k & = 0 \\ 33+k & = 0 \\ k & = -33. \end{align*} If the question asks for a coefficient such that the quadratic equation has exactly one solution, the expression must be a perfect square trinomial. The most efficient method is to set the discriminant ($b^2 - 4ac$) equal to zero and solve for the unknown coefficient.
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To find the vertex of the quadratic function $y=ax^2+bx+c$, use the formula $x=\dfrac{-b}{2a}$ to find the $x$-coordinate. Then, substitute this value into the original equation to find the corresponding $y$-coordinate. Alternatively, you can complete the square to rewrite the equation in vertex form, $y=a(x-h)^2+k$, where $(h,k)$ is the vertex.
For example, given $y=x^2+6x +11$, calculate the $x$-coordinate: $$ x=\dfrac{-b}{2a}=\dfrac{-6}{2(1)}=-3. $$ Substitute $-3$ for $x$ to find $y$: $$ y=(-3)^2+6(-3)+11=9-18+11=2. $$ Thus, the vertex is $(-3,2)$.
Using the completing the square method: $$ y=(x^2+6x+9)+2=(x+3)^2+2. $$ Comparing this to the vertex form, the vertex is $(-3,2)$.
Graphing the equation on a calculator and identifying the minimum or maximum point is another valid approach. It is recommended to practice this method, though algebraic methods are essential for problems where graphing is impractical.
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To factor a quadratic expression $ax^2+bx+c$, find two numbers that sum to $b$ and multiply to the product $ac$.
For example, consider $3x^2+14x+8$. Find two numbers that add to $14$ and multiply to $3(8)=24$. These numbers are $2$ and $12$. Split the middle term and factor by grouping: \begin{align*} 3x^2+2x+12x+8 & = x(3x+2)+4(3x+2) \\ & = (x+4)(3x+2). \end{align*} Alternatively, solve the corresponding equation using the quadratic formula and rewrite the result in factored form, ensuring the leading coefficient is maintained.
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29. One square has a side length of $x+3$ cm and another has a side length of $x -2$ cm. What expression represents the difference in the areas of these two squares in square centimeters? 3
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Answer key: 1.D, 2.D, 3.B, 4.A, 5.B, 6.C, 7.D, 8.C, 9.A, 10.B, 11.A, 12.B, 13.D, 14.B, 15.D, 16.A, 17.C, 18.C, 19.A, 20.B, 21.D, 22.D, 23.B, 24.B, 25.D, 26.C, 27.D, 28.A, 29.D, 30.B, 31.D
Method 2 (Reverse Engineering): Find the roots using the quadratic formula: $$ x=\dfrac{-19 \pm \sqrt{361-240}}{20}=\dfrac{-19 \pm 11}{20} \implies x=-\dfrac{3}{2} \text{ or } x=-\dfrac{2}{5}. $$ Convert the roots to integer factors: $$ x=-\dfrac{3}{2} \implies 2x+3=0 \quad \text{and} \quad x=-\dfrac{2}{5} \implies 5x+2=0. $$ Thus, the factors are $(2x+3)(5x+2)$.
Method 3 (Expansion): Expand the answer choices using FOIL to see which one matches the original expression. For choice (A): $$ (2x+3)(5x+2)=10x^2+4x+15x+6=10x^2+19x+6. $$
Method 4 (Substitution): Substitute a specific value for $x$, such as $x=2$. The original expression yields $10(2)^2+19(2)+6=84$. Substituting $x=2$ into choice (A) yields $(4+3)(10+2)=(7)(12)=84$. Since the values match, (A) is correct. This strategy is efficient under time constraints. To ensure accuracy, verify that no other answer choice yields 84.
