Lynn Marecek; MaryAnne Anthony-Smith; Andrea Honeycutt Mathis; Jay Abramson; Sharon North; Amy Baldwin; Alyssa Howell

9.4.3 Standard Form and Squared Factors

9.4.3 • Standard Form and Squared Factors

Activity

1. For each expression, find the value of $c$ to make a perfect square in standard form.

a. $100 x^{2} + 80 x + c$

16. Standard form: $100 x^{2} + 80 x + 16$ .

b. $36 x^{2} - 60 x + c$

25. Standard form: $36 x^{2} - 60 x + 25$ .

c. $25 x^{2} + 40 x + c$

16. Standard form: $25 x^{2} + 40 x + 16$ .

d. $0.25 x^{2} - 14 x + c$

196. Standard form: $0.25 x^{2} - 14 x + 196$ .

2. For each expression, refer to your responses in question 1 to help write an equivalent expression in the form of squared factors.

a. $100 x^{2} + 80 x + c$

Compare your answer:

$(10 x + 4)^{2}$

b. $36 x^{2} - 60 x + c$

Compare your answer:

$(6 x - 5)^{2}$

c. $25 x^{2} + 40 x + c$

Compare your answer:

$(5 x + 4)^{2}$

d. $0.25 x^{2} - 14 x + c$

Compare your answer:

$(0.5 x - 14)^{2}$

3. Write your own pair of equivalent expressions.

a. Create an expression that is a perfect square trinomial in standard form $(a x^{2} + b x + c)$ .

Compare your answer:

Answers will vary. One expression in standard form is: $49 x^{2} + 14 x + 1$ .

b. Now write the same equivalent expression in the form of squared factors $(k x + m)^{2}$ .

Compare your answer:

Answers will vary. Using the expression above, for example, the squared factor of $49 x^{2} + 14 x + 1$ would be $(7 x + 1)^{2}$ .

4. Solve each equation by completing the square.

a. $25 x^{2} + 40 x = - 12$

Compare your answer:

$25 x^{2} + 40 x = - 12 25 x^{2} + 40 x + 12 = 0 25 x^{2} + 40 x + 12 + 4 = 4 25 x^{2} + 40 x + 16 = 4 (5 x + 4)^{2} = 4 \sqrt{(5 x + 4)^{2}} = \sqrt{4} 5 x + 4 = 2 5 x + 4 = - 2 5 x = - 2 5 x = - 6 x = - \frac{2}{5} and x = - \frac{6}{5}$

b. $36 x^{2} - 60 x + 10 = - 6$

Compare your answer:

$36 x^{2} - 60 x + 10 = - 6 36 x^{2} - 60 x + 10 + 15 = - 6 + 15 36 x^{2} - 60 x + 25 = 9 (6 x - 5)^{2} = 9 \sqrt{(6 x - 5)^{2}} = \sqrt{9} 6 x - 5 = 3 6 x - 5 = - 3 6 x = 8 6 x = 2 x = \frac{8}{6} = \frac{4}{3} and x = \frac{2}{6} = \frac{1}{3}$

Self Check

Solve by completing the square:

25x^2+30x=7

.

$x=-\frac{3}{5}$
$x=\frac{13}{5}$ , $x=-\frac{19}{5}$
$x=\frac15$ , $x= -\frac75$
$x=9$

Find the Value to Make a Perfect Square and Solve

In earlier lessons, we worked with perfect squares such as $(x + 1)^{2}$ and $(x - 5) (x - 5)$ . We learned that their equivalent expressions in standard form follow a predictable pattern:

In general, $(x + m)^{2}$ can be written as $x^{2} + 2 m x + m^{2}$ .
If a quadratic expression of the form $a x^{2} + b x + c$ is a perfect square, and the value of $a$ is 1, then the value of $b$ is $2 m$ , and the value of $c$ is $m^{2}$ for some value of $m$ .

In this lesson, the variable in the factors being squared had a coefficient other than 1, for example $(3 x + 1)^{2}$ and $(2 x - 5) (2 x - 5)$ . Their equivalent expression in standard form also followed the same pattern we saw earlier.

Squared factors	Standard form
$(3 x + 1)^{2}$	$(3 x)^{2} + 2 (3 x) (1) + 1^{2}$ or $9 x^{2} + 6 x + 1$
$(2 x - 5)^{2}$	$(2 x)^{2} + 2 (2 x) (- 5) + (- 5)^{2}$ or $4 x^{2} - 20 x + 25$

In general, $(k x + m)^{2}$ can be written as:

$(k x)^{2} + 2 (k x) (m) + m^{2}$ or $k^{2} x^{2} + 2 k m x + m^{2}$

If a quadratic expression is of the form $a x^{2} + b x + c$ , then:

the value of $a$ is $k^{2}$
the value of $b$ is $2 k m$
the value of $c$ is $m^{2}$

We can use this pattern to help us complete the square and solve equations when the squared term $x^{2}$ has a coefficient other than 1—for example: $16 x^{2} + 40 x = 11$ .

What constant term $c$ can we add to make the expression on the left of the equal sign a perfect square? And how do we write this expression as squared factors?

16 is $4^{2}$ , so the squared factors could be $(4 x + m)^{2}$ .
40 is equal to $2 (4 m)$ , so $2 (4 m) = 40$ or $8 m = 40$ . This means that $m = 5$ .
If $c$ is $m^{2}$ , then $c = 5^{2}$ or $c = 25$ .
So the expression $16 x^{2} + 40 x + 25$ is a perfect square and is equivalent to $(4 x + 5)^{2}$ .

Let's solve the equation $16 x^{2} + 40 x = 11$ by completing the square!

Step 1 -

$16 x^{2} + 40 x = 11$

To be a perfect square, $(4 x + m)^{2}$ is needed.

We found the $4 x$ by determining the square root of the first term, $16 x^{2}$ .

Step 2 -

Now, find the value of $m$ to make a perfect square.

Since $b = 2 k m$ , and we know that $k = 4$ (from step 1), and that $b = 40$ (from the coefficient of the linear term), we substitute the values and solve.

$40 = 2 (4 m)$

$40 = 8 m,$

so $m = 5 .$

Step 3 -

Find $c = m^{2} .$

$c = m^{2} = 5^{2} = 25$

Step 4 -

Add $c$ to both sides of the equation.

$16 x^{2} + 40 x + 25 = 11 + 25$

$16 x^{2} + 40 x + 25 = 36$

Step 5 -

Factor and simplify.

$(4 x + 5)^{2} = 6^{2}$

Step 6 -

Take the square root of both sides. Remember the plus or minus.

$(4 x + 5)^{2} = 6^{2}$

$\sqrt{(4 x + 5)^{2}} = \sqrt{6^{2}}$

$4 x + 5 = \pm 6$

Step 7 -

Write two equations and solve.

$4 x + 5 = 6$ or $4 x + 5 = - 6$

$4 x = 1$ or $4 x = - 11$

$x = \frac{1}{4}$ or $x = - \frac{11}{4}$

Try it

Find the Value to Make a Perfect Square and Solve

Solve $4 x^{2} + 28 x = - 13$ .

Here is how to solve this equation by completing the square:

Step 1 - To be a perfect square, $(2 x + m)^{2}$ is needed.

$4 x^{2} + 28 x = - 13$

Step 2 - Find the value of $m$ to make a perfect square.

$2 (2 m) = 28$ , so $m = 7$ .

Step 3 - Find $c = m^{2}$ .

$c = m^{2} = 7^{2} = 49$

Step 4 - Add $c$ to both sides.

$4 x^{2} + 28 x + 49 = - 13 + 49$

Step 5 - Factor and simplify.

$(2 x + 7)^{2} = 36$

Step 6 - Take the square root of both sides. Remember the plus or minus.

$\sqrt{(2 x + 7)^{2}} = \sqrt{36}$

$2 x + 7 = \pm 6$

Step 7 - Write two equations and solve.

$\begin{matrix} 2 x + 7 = 6 & 2 x + 7 = - 6 \\ x = - \frac{1}{2} & x = - \frac{13}{2} \end{matrix}$