Sample questions
A few samples from this topic — try one, then show the answer.
Solve with the Quadratic Formula
Solve x 2 − 15 x + 36 = 0 x^{2} \; - \; 15x \; + \; 36 \; = \; 0 x 2 − 15 x + 36 = 0 using the quadratic formula. enter both solutions (e.g., type -2, 3 for x = -2 and x = 3) \quad {\scriptsize\color{gray}{\textrm{enter both solutions (e.g., type -2, 3 for x = -2 and x = 3)}}} enter both solutions (e.g., type -2, 3 for x = -2 and x = 3) Show answer
Build a Perfect Square
Complete the square to solve x 2 − 10 x − 11 = 0 x^{2} \; - \; 10x \; - \; 11 \; = \; 0 x 2 − 10 x − 11 = 0 . The solutions are:
x = − 1 x \; = \; -1 x = − 1 , x = 11 x \; = \; 11 x = 11 x = − 11 x \; = \; -11 x = − 11 , x = 1 x \; = \; 1 x = 1 x = 1 x \; = \; 1 x = 1 , x = 12 x \; = \; 12 x = 12 x = 0 x \; = \; 0 x = 0 , x = 10 x \; = \; 10 x = 10
Show answer Answer: x = − 1 x \; = \; -1 x = − 1 , x = 11 x \; = \; 11 x = 11
Convert Between Standard and Vertex Form
The vertex form of f ( x ) = x 2 + 2 x − 7 f(x) \; = \; x^{2} \; + \; 2x\, \; - \; 7 f ( x ) = x 2 + 2 x − 7 is:
( x + 1 ) 2 − 8 (x + 1)^2 - 8 ( x + 1 ) 2 − 8 ( x − 1 ) 2 − 8 (x - 1)^2 - 8 ( x − 1 ) 2 − 8 ( x + 2 ) 2 − 7 (x + 2)^2 - 7 ( x + 2 ) 2 − 7 ( x + 1 ) 2 − 7 (x + 1)^2 - 7 ( x + 1 ) 2 − 7
Show answer Answer: ( x + 1 ) 2 − 8 (x + 1)^2 - 8 ( x + 1 ) 2 − 8
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A worked example
Solve x 2 − 15 x + 36 = 0 x^{2} \; - \; 15x \; + \; 36 \; = \; 0 x 2 − 15 x + 36 = 0 using the quadratic formula. enter both solutions (e.g., type -2, 3 for x = -2 and x = 3) \quad {\scriptsize\color{gray}{\textrm{enter both solutions (e.g., type -2, 3 for x = -2 and x = 3)}}} enter both solutions (e.g., type -2, 3 for x = -2 and x = 3)
The equation x 2 − 15 x + 36 = 0 x^{2} \; - \; 15x \; + \; 36 \; = \; 0 x 2 − 15 x + 36 = 0 is already in standard form a x 2 + b x + c = 0 ax^{2} + bx + c = 0 \\ a x 2 + b x + c = 0 Read off the coefficients: a = 1 , b = − 15 , c = 36 a = 1, \; b = -15, \; c = 36 a = 1 , b = − 15 , c = 36 Write out the quadratic formula and substitute a , b , c x = − b ± b 2 − 4 a c 2 a = − ( − 15 ) ± ( − 15 ) 2 − 4 ( 1 ) ( 36 ) 2 ( 1 ) a, b, c \\ x = \frac{-b \; \pm \; \sqrt{b^{2} \; - \; 4ac}}{2a} = \frac{-(-15) \; \pm \; \sqrt{(-15)^{2} \; - \; 4(1)(36)}}{2(1)} a , b , c x = 2 a − b ± b 2 − 4 a c = 2 ( 1 ) − ( − 15 ) ± ( − 15 ) 2 − 4 ( 1 ) ( 36 ) Compute the discriminant (the value under the square root): b 2 − 4 a c = 225 − 144 = 81 \\ b^{2} \; - \; 4ac = 225 \; - \; 144 = 81 \\ b 2 − 4 a c = 225 − 144 = 81 Take the square root: 81 = 9 \sqrt{{81}} = 9 81 = 9 Now evaluate the two branches using + and − x = 15 + 9 2 = 24 2 = 12 x = 15 − 9 2 = 6 2 = 3 - \\ x = \frac{{15 + 9}}{{2}} = \frac{{24}}{{2}} = 12 \qquad x = \frac{{15 - 9}}{{2}} = \frac{{6}}{{2}} = 3 − x = 2 15 + 9 = 2 24 = 12 x = 2 15 − 9 = 2 6 = 3 The solutions are x = 3 and x = 12
Skills covered
Compute the Discriminant & Test Factorability
Decide When to Move Beyond Factoring
Build a Perfect Square
Solve Quadratics by Completing the Square
Extend Completing the Square
Rewrite with Completed-Square Form
Convert Between Standard and Vertex Form
Diagnose Completing-the-Square Errors
Solve with the Quadratic Formula
Use the Discriminant to Predict the Solutions
Connect Roots, Solutions & x-Intercepts
Compare and Choose Quadratic Solving Methods
Understand and Check the Quadratic Formula
Quadratic Equations in other grades
For parents & teachers
Your child computes the discriminant to judge whether a quadratic factors, learns to complete the square and convert between standard and vertex form, applies the quadratic formula, and connects the discriminant to the number of solutions and to the x-intercepts of the graph.
The discriminant as a factorability and root-nature test; completing the square from monic through non-monic cases; completed-square and vertex form in both directions; the quadratic formula, its derivation from completing the square, and its procedural checks; and the discriminant read graphically as the x-intercept count.