Question

Find all the real square roots of -9/16.

Answer 1

Find all the real square roots of -9/16. 
= 3/4

Find all the real cube roots of -0.000125.
= -0.05

Simplify √(36g^6).

= 6 g^3

Simplify ∛(125x^21y^24).
= 5x^7y^8

Simplify √((27x^4)/(75y^2)).

    3x^2
-----------
     5y

Answer 2

Answer:

1)  $\sqrt{\frac{-9}{16}}=\pm\frac{3}{4}i$

2)  $\sqrt[3]{-0.000125}=-0.05$

3) $\sqrt{(36g^6)}=6(g^3)$

4) $\sqrt[3]{125x^{21}y^{24}}=5(x^{7})(y^{8})$

5) $\sqrt{\frac{27x^4}{75y^2}}=\frac{3(x^2)}{5y}$

Step-by-step explanation:

To find : The following expression

1) Find all the real square roots of -9/16.

$\sqrt{\frac{-9}{16}}$

There is no real roots because $\sqrt{-1}=i$ is an imaginary number

Therefore,  $\sqrt{\frac{-9}{16}}=\pm\frac{3}{4}i$

2) Find all the real cube roots of -0.000125.

$\sqrt[3]{-0.000125}$

There is a real roots.

Therefore,  $\sqrt[3]{-0.000125}=-0.05$

3) Simplify $\sqrt{(36g^6)}$

$=\sqrt{(36g^6)}$

$=\sqrt{6^2(g^3)^2}$

$=6(g^3)$

Therefore, simplified form is $\sqrt{(36g^6)}=6(g^3)$

4) Simplify $\sqrt[3]{125x^{21}y^{24}}$

$=\sqrt[3]{125x^{21}y^{24}}$

$=\sqrt[3]{5^3(x^{7})^3(y^{8})^3}$

$=5(x^{7})(y^{8})$

Therefore, simplified form is $\sqrt[3]{125x^{21}y^{24}}=5(x^{7})(y^{8})$

5) Simplify $\sqrt{\frac{27x^4}{75y^2}$

$=\sqrt{\frac{27x^4}{75y^2}$

$=\sqrt{\frac{9(x^2)^2}{25y^2}$ (Divide nr. and dr. by 3)

$=\sqrt{\frac{3^2(x^2)^2}{5^2y^2}$

$=\frac{3(x^2)}{5y}$

Therefore, simplified form is $\sqrt{\frac{27x^4}{75y^2}}=\frac{3(x^2)}{5y}$