Question

The fractions below are possible values of X in the given equations.write the correct fraction inside the box for each equation.

Answer 1

Given:

$\text { a. } x^{2}=\frac{4}{9}$

$\text { b. } x^{3}=\frac{27}{64}$

$\text { c. } x^{2}=\frac{81}{64}$

$\text { d. } x^{3}=\frac{1}{8}$

To find:

The value of x for each equation.

Solution:

a) $x^{2}=\frac{4}{9}$

$x^{2}=\frac{2^2}{3^2}$

Taking square root on both sides.

$\sqrt{x^{2}}=\sqrt{ \frac{2^2}{3^2}}$

Square and square roots get canceled.

$x =\frac{2}{3}$

b) $x^{3}=\frac{27}{64}$

$x^{3}=\frac{3^3}{4^3}$

Taking cube root on both sides.

$\sqrt[3]{x^{3}} =\sqrt[3]{\frac{3^3}{4^3} }$

cube and cube roots get canceled.

$x=\frac{3}{4}$

c) $x^{2}=\frac{81}{64}$

$x^{2}=\frac{9^2}{8^2}$

Taking square root on both sides.

$\sqrt{x^{2}}=\sqrt{ \frac{9^2}{8^2}}$

Square and square roots get canceled.

$x =\frac{9}{8}$

d) $x^{3}=\frac{1}{8}$

$x^{3}=\frac{1^3}{2^3}$

Taking cube root on both sides.

$\sqrt[3]{x^{3}} =\sqrt[3]{\frac{1^3}{2^3} }$

cube and cube roots get canceled.

$x=\frac{1}{2}$