Question

© houghton mifflin harcourt publis

1. math on the spot solve each equation for x.

a. $x^{2}=81$
b. $x^{2}=\frac{25}{144}$

for problems 5–10, find each root.

1. $\sqrt{289}$
2. $\sqrt3{512}$
3. $\sqrt3{\frac{1}{1000}}$
4. $z^{2}=\frac{81}{121}$
5. $x^{3}=343$
6. $y^{3}=\frac{8}{729}$

Explanation:

Problem 4A

Step1: Take square root of both sides

$x = \pm\sqrt{81}$

Step2: Evaluate square root

$\sqrt{81}=9$, so $x=\pm9$

Problem 4B

Step1: Take square root of both sides

$x = \pm\sqrt{\frac{25}{144}}$

Step2: Evaluate fraction square root

$\sqrt{\frac{25}{144}}=\frac{\sqrt{25}}{\sqrt{144}}=\frac{5}{12}$, so $x=\pm\frac{5}{12}$

Problem 5

Step1: Evaluate positive square root

$\sqrt{289}=17$ (since $17^2=289$)

Problem 6

Step1: Evaluate cube root

$\sqrt[3]{512}=8$ (since $8^3=512$)

Problem 7

Step1: Evaluate fraction cube root

$\sqrt[3]{\frac{1}{1000}}=\frac{\sqrt[3]{1}}{\sqrt[3]{1000}}=\frac{1}{10}$ (since $10^3=1000$)

Problem 8

Step1: Take square root of both sides

$z = \pm\sqrt{\frac{81}{121}}$

Step2: Evaluate fraction square root

$\sqrt{\frac{81}{121}}=\frac{\sqrt{81}}{\sqrt{121}}=\frac{9}{11}$, so $z=\pm\frac{9}{11}$

Problem 9

Step1: Take cube root of both sides

$x = \sqrt[3]{343}$

Step2: Evaluate cube root

$\sqrt[3]{343}=7$ (since $7^3=343$)

Problem 10

Step1: Take cube root of both sides

$y = \sqrt[3]{\frac{8}{729}}$

Step2: Evaluate fraction cube root

$\sqrt[3]{\frac{8}{729}}=\frac{\sqrt[3]{8}}{\sqrt[3]{729}}=\frac{2}{9}$ (since $2^3=8, 9^3=729$)

Answer:

4A. $x = 9$ or $x = -9$
4B. $x = \frac{5}{12}$ or $x = -\frac{5}{12}$

1. $17$
2. $8$
3. $\frac{1}{10}$
4. $z = \frac{9}{11}$ or $z = -\frac{9}{11}$
5. $x = 7$
6. $y = \frac{2}{9}$