Question

a. how are the contents of the three diagrams alike? this diagram represents (term_1 + term_2)^2. describe your observations about cells 1, 2, 3, and 4.

b. rewrite the perfect - square expressions (n + 7)^2, (5 - m)^2, and (h + 1/3)^2 in standard form: ax^2 + bx + c.

c. how are the ax^2, bx, and c of a perfect square in standard form related to the two terms in (term_1 + term_2)^2?

1. solve each equation.

a. (x - 1)^2 = 4

b. (x + 5)^2 = 81

c. (x - 2)^2 = 0

d. (x + 11)^2 = 121

e. (x - 7)^2 = 64/49

1. explain or show why the product of a sum and a difference, such as (2x + 1)(2x - 1), has no linear term when written in standard form.

Explanation:

Step1: Take square root on both sides

For \((x - 1)^2=4\), we have \(x - 1=\pm\sqrt{4}=\pm2\)

Step2: Solve for \(x\)

When \(x - 1 = 2\), then \(x=2 + 1=3\); when \(x - 1=-2\), then \(x=-2 + 1=-1\)

Step3: Take square root on both sides for \((x + 5)^2=81\)

\(x + 5=\pm\sqrt{81}=\pm9\)

Step4: Solve for \(x\)

When \(x + 5 = 9\), \(x=9 - 5 = 4\); when \(x+5=-9\), \(x=-9 - 5=-14\)

Step5: Take square root on both sides for \((x - 2)^2=0\)

\(x - 2 = 0\), so \(x=2\)

Step6: Take square root on both sides for \((x + 11)^2=121\)

\(x + 11=\pm\sqrt{121}=\pm11\)

Step7: Solve for \(x\)

When \(x + 11 = 11\), \(x=11-11 = 0\); when \(x + 11=-11\), \(x=-11-11=-22\)

Step8: Take square root on both sides for \((x - 7)^2=\frac{64}{49}\)

\(x - 7=\pm\sqrt{\frac{64}{49}}=\pm\frac{8}{7}\)

Step9: Solve for \(x\)

When \(x - 7=\frac{8}{7}\), \(x=\frac{8}{7}+7=\frac{8 + 49}{7}=\frac{57}{7}\); when \(x - 7=-\frac{8}{7}\), \(x=-\frac{8}{7}+7=\frac{- 8+49}{7}=\frac{41}{7}\)

Answer:

a. \(x = 3\) or \(x=-1\)
b. \(x = 4\) or \(x=-14\)
c. \(x = 2\)
d. \(x = 0\) or \(x=-22\)
e. \(x=\frac{57}{7}\) or \(x=\frac{41}{7}\)