Standard Form and Squared Factors

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 cc to make a perfect square in standard form.

a. 100x2+80x+c100x2+80x+c

16. Standard form: 100x2+80x+16100x2+80x+16.

b. 36x2−60x+c36x2−60x+c

25. Standard form: 36x2−60x+2536x2−60x+25.

c. 25x2+40x+c25x2+40x+c

16. Standard form: 25x2+40x+1625x2+40x+16.

d. 0.25x2−14x+c0.25x2−14x+c

196. Standard form: 0.25x2−14x+1960.25x2−14x+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. 100x2+80x+c100x2+80x+c

Compare your answer:

(10x+4)2(10x+4)2

b. 36x2−60x+c36x2−60x+c

Compare your answer:

(6x−5)2(6x−5)2

c. 25x2+40x+c25x2+40x+c

Compare your answer:

(5x+4)2(5x+4)2

d. 0.25x2−14x+c0.25x2−14x+c

Compare your answer:

(0.5x−14)2(0.5x−14)2

3. Write your own pair of equivalent expressions.

a. Create an expression that is a perfect square trinomial in standard form (ax2+bx+c)(ax2+bx+c).

Compare your answer:

Answers will vary. One expression in standard form is: 49x2+14x+149x2+14x+1.

b. Now write the same equivalent expression in the form of squared factors (kx+m)2(kx+m)2.

Compare your answer:

Answers will vary. Using the expression above, for example, the squared factor of 49x2+14x+149x2+14x+1 would be (7x+1)2(7x+1)2.

4. Solve each equation by completing the square.

a. 25x2+40x=−1225x2+40x=−12

Compare your answer:

25x2+40x=−1225x2+40x+12=025x2+40x+12+4=425x2+40x+16=4(5x+4)2=4(5x+4)2=45x+4=25x+4=−25x=−25x=−6x=−25 and x=−6525x2+40x=−1225x2+40x+12=025x2+40x+12+4=425x2+40x+16=4(5x+4)2=4(5x+4)2=45x+4=25x+4=−25x=−25x=−6x=−25 and x=−65

b. 36x2−60x+10=−636x2−60x+10=−6

Compare your answer:

36x2−60x+10=−636x2−60x+10+15=−6+1536x2−60x+25=9(6x−5)2=9(6x−5)2=96x−5=36x−5=−36x=86x=2x=86=43 and x=26=1336x2−60x+10=−636x2−60x+10+15=−6+1536x2−60x+25=9(6x−5)2=9(6x−5)2=96x−5=36x−5=−36x=86x=2x=86=43 and x=26=13

Self Check

Solve by completing the square: 25 x 2 + 30 x = 7 .
  1. x = − 3 5
  2. x = 13 5 , x = − 19 5
  3. x = 1 5 , x = − 7 5
  4. 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(x+1)2 and (x−5)(x−5)(x−5)(x−5). We learned that their equivalent expressions in standard form follow a predictable pattern:

  • In general, (x+m)2(x+m)2 can be written as x2+2mx+m2x2+2mx+m2.
  • If a quadratic expression of the form ax2+bx+cax2+bx+c is a perfect square, and the value of aa is 1, then the value of bb is 2m2m, and the value of cc is m2m2 for some value of mm.

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

Squared factors Standard form
(3x+1)2(3x+1)2 (3x)2+2(3x)(1)+12(3x)2+2(3x)(1)+12 or 9x2+6x+19x2+6x+1
(2x−5)2(2x−5)2 (2x)2+2(2x)(−5)+(−5)2(2x)2+2(2x)(−5)+(−5)2  or
4x2−20x+254x2−20x+25

In general, (kx+m)2(kx+m)2 can be written as:

(kx)2+2(kx)(m)+m2(kx)2+2(kx)(m)+m2 or k2x2+2kmx+m2k2x2+2kmx+m2

If a quadratic expression is of the form ax2+bx+cax2+bx+c, then:

  • the value of aa is k2k2
  • the value of bb is 2km2km
  • the value of cc is m2m2

We can use this pattern to help us complete the square and solve equations when the squared term x2x2 has a coefficient other than 1—for example: 16x2+40x=1116x2+40x=11.

What constant term cc 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 4242, so the squared factors could be (4x+m)2(4x+m)2.
  • 40 is equal to 2(4m)2(4m), so 2(4m)=402(4m)=40 or 8m=408m=40. This means that m=5m=5.
  • If cc is m2m2, then c=52c=52 or c=25c=25.
  • So the expression 16x2+40x+2516x2+40x+25 is a perfect square and is equivalent to (4x+5)2(4x+5)2.

Let's solve the equation 16x2+40x=1116x2+40x=11 by completing the square!

Step 1 -

16x2+40x=1116x2+40x=11

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

We found the 4x4x by determining the square root of the first term, 16x216x2.

Step 2 -

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

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

40=2(4m)40=2(4m)

40=8m,40=8m,

so m=5.m=5.

Step 3 -

Find c=m2.c=m2.

c=m2=52=25c=m2=52=25

Step 4 -

Add cc to both sides of the equation.

16x2+40x+25=11+2516x2+40x+25=11+25

16x2+40x+25=3616x2+40x+25=36

Step 5 -

Factor and simplify.

(4x+5)2=62(4x+5)2=62

Step 6 -

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

(4x+5)2=62(4x+5)2=62

(4x+5)2=62(4x+5)2=62

4x+5=±64x+5=±6

Step 7 -

Write two equations and solve.

4x+5=64x+5=6      or    4x+5=−64x+5=−6

          4x=14x=1    or    4x=−114x=−11

          x=14x=14      or    x=−114x=−114

Try it

Find the Value to Make a Perfect Square and Solve

Solve 4x2+28x=−134x2+28x=−13.

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

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

4x2+28x=−134x2+28x=−13

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

2(2m)=282(2m)=28, so m=7m=7.

Step 3 - Find c=m2c=m2.

c=m2=72=49c=m2=72=49

Step 4 - Add cc to both sides.

4x2+28x+49=−13+494x2+28x+49=−13+49

Step 5 - Factor and simplify.

(2x+7)2=36(2x+7)2=36

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

(2x+7)2=36(2x+7)2=36

2x+7=±62x+7=±6

Step 7 - Write two equations and solve.

2x+7=62x+7=−6x=−12x=−1322x+7=62x+7=−6x=−12x=−132