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8. Solving Quadratic Equations
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United States of AmericaFL
Grade 912

8.04 Solving quadratic equations by completing the square

Lesson
Practice
Lesson

Concept summary

Completing the square is a method we use to rewrite a quadratic expression so that it contains a perfect square trinomial. A perfect square trinomial takes on the form A^2+2AB+B^2=\left(A+B\right)^2.

If we can rewrite an equation by completing the square, then we can solve it using square roots.

For quadratic equations where a=1, we can write them in perfect square form by following these steps:

1\displaystyle x^2+bx+c\displaystyle =\displaystyle 0
2\displaystyle x^2+bx\displaystyle =\displaystyle -cSubtract c from both sides
3\displaystyle x^2+2\left(\frac{b}{2}\right)x\displaystyle =\displaystyle -cRewrite the x term
4\displaystyle x^2+2\left(\frac{b}{2}\right)x+\left(\frac{b}{2}\right)^2\displaystyle =\displaystyle -c+\left(\frac{b}{2}\right)^2Add \left(\dfrac{b}{2}\right)^2 to both sides
5\displaystyle \left(x+\frac{b}{2}\right)^2\displaystyle =\displaystyle -c+\left(\frac{b}{2}\right)^2Factor the trinomial

If a \neq 1, we can first divide through by a to factor it out.

Worked examples

Example 1

Solve the following quadratic equation by completing the square: x^{2} + 18 x + 32 = 0

Approach

To solve an equation by completing the square, start by moving the constant term to the other side of the equation. We will complete the square by finding half the coefficient of the x term, squaring it, and adding it to both sides of the equation. Since the coefficient of the x term is 18, we will need to add \left(\dfrac{18}{2}\right)^2=81 to both sides of our equation. Once we've completed the square, we can solve.

Solution

\displaystyle x^2+18x+32\displaystyle =\displaystyle 0
\displaystyle x^2+18x\displaystyle =\displaystyle -32Subtract 32 from both sides
\displaystyle x^2+18x+81\displaystyle =\displaystyle -32+81Complete the square
\displaystyle (x+9)^2\displaystyle =\displaystyle 49Factor the perfect square trinomial and simplify
\displaystyle x+9\displaystyle =\displaystyle \pm 7Square root both sides

This leaves us with two equations x+9=7 and x+9=-7. We can solve both equations by subtracting 9 and so we get the solutions x=-2,\, x=-16.

Example 2

Solve the following quadratic equation by completing the square: 3x^2 -12x + 10 = 0

Approach

In this example, the coefficient of x^2 is 3 so we will need to divide this coefficient out before completing the square. We can then perform steps similar to the previous example.

Once 3 has been divided out, the coefficient of x will be 4. Taking \left(\dfrac{4}{2}\right)^2 gives us 4 so this is the value that completes the square.

Solution

\displaystyle 3x^2-12x+10\displaystyle =\displaystyle 0
\displaystyle x^2-4x+\frac{10}{3}\displaystyle =\displaystyle 0Divide by 3
\displaystyle x^2-4x\displaystyle =\displaystyle -\frac{10}{3}Subtract \frac{10}{3} from both sides
\displaystyle x^2-4x+4\displaystyle =\displaystyle -\frac{10}{3}+4Complete the square
\displaystyle (x-2)^2\displaystyle =\displaystyle \frac{2}{3}Factor and simplify
\displaystyle x-2\displaystyle =\displaystyle \pm \sqrt{\frac{2}{3}}Square root both sides

This leaves us with two equations x-2=\sqrt{\dfrac{2}{3}} and x-2=-\sqrt{\dfrac{2}{3}}.

We need to add 2 to solve both equations and we find that the solutions are x=2+\sqrt{\dfrac{2}{3}},\, x=2-\sqrt{\dfrac{2}{3}}.

Outcomes

MA.912.AR.1.1

Identify and interpret parts of an equation or expression that represent a quantity in terms of a mathematical or real-world context, including viewing one or more of its parts as a single entity.

MA.912.AR.1.2

Rearrange equations or formulas to isolate a quantity of interest.

MA.912.AR.3.1

Given a mathematical or real-world context, write and solve one-variable quadratic equations over the real number system.

MA.912.AR.3.8

Solve and graph mathematical and real-world problems that are modeled with quadratic functions. Interpret key features and determine constraints in terms of the context.

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