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Writing Perfect Square Trinomials

Note

If you completed “Extension 1 – Completing the Square” in Algebra 1: Principles of Secondary Mathematics, parts of this lesson should be a review. However, new and more advanced content is presented here as well.

When quadratic trinomials are perfect square trinomials, they can be written as: $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$
With a perfect square trinomial in standard form , $(a x^{2} + b x + c),$ the terms a and c are perfect squares and: $b = 2 \sqrt{a c}$

Note

Because a and c are perfect square terms, finding twice the product of their square roots is likely something you have already done without even realizing it.

You can prove this formula is true by using a perfect square trinomial when the value of c is known. For example: $x^{2} + 8 x + 16 = {(x + 4)}^{2}$

Plan

Determine a, b, and c using standard form

Find the value of b and c using the formula: $b = 2 \sqrt{a c}$

Implement

$a = 1, b = 8, c = 16 b = 2 \sqrt{(1) (16)} = 2 \sqrt{16} = 2 \cdot 4 = 8 c = {(\frac{b}{2})}^{2} = {(\frac{8}{2})}^{2} = {(4)}^{2} = 16$

When you know the values of a and b, you can use this formula to find the value of c: $c = {(\frac{b}{2})}^{2} when a = 1$

Example 1

Solve for the value that will make the expression a perfect square trinomial.

$x^{2} - 11 x + c$

$a = 1, b = - 11 c = {(\frac{b}{2})}^{2} c = {(- \frac{11}{2})}^{2} \frac{121}{4}$

$Check x^{2} - 11 x + \frac{121}{4} {(x - \frac{11}{2})}^{2}$

Example 2

Solve for the value that will make the expression a perfect square trinomial.

$x^{2} + b x + \frac{16}{49}$

$a = 1, c = \frac{16}{49} b = 2 \sqrt{a c} b = 2 \sqrt{(1) (\frac{16}{49})} = 2 \sqrt{\frac{16}{49}} = 2 (\frac{4}{7}) b = \frac{8}{7}$

$Check x^{2} + \frac{8}{7} x + \frac{16}{49} {(x + \frac{4}{7})}^{2}$