Answered

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


Find all the real cube roots of -0.000125.


Simplify √(36g^6).


Simplify ∛(125x^21y^24).


Simplify √((27x^4)/(75y^2)).

Verified Answer :

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

A real square root of number is the value which is when multiplicand by itself gives the same value as the number posses.

  • a) All the real square roots of -9/16 are \(\dfrac{3i}{4}\) and \(-\dfrac{3i}{4}\).
  • b) All the real square roots of -9/16 are 0.05\(i\) and -0.05 \(i\).
  • c) The simplified value of  \(\sqrt{36g^6}\) is  \(6g^3\).
  • e) The  Simplified value of  \(\sqrt[]{125x^{21}y^{24}}\) is  \(5x^7y^8\)..
  • d) The simplified value of   \(\sqrt{\dfrac{27x^4}{75y^2}}\) is  \({\dfrac{9x^2}{25y} }\).

What is real square root?

A real square root of number is the value which is when multiplicand by itself gives the same value as the number posses.

Given information-

  • a) All the real square roots of -9/16-

Put the square root on the number to find the real square roots,

\(\sqrt{\dfrac{-9}{16} } =\sqrt{\dfrac{(-1)3^2}{4^2} } \\\sqrt{\dfrac{-9}{16} } =\pm \dfrac{3}{4} \times\sqrt{-1} \\\sqrt{\dfrac{-9}{16} } =\pm \dfrac{3}{4} (i)\\\sqrt{\dfrac{-9}{16} } =\pm\dfrac{3i}{4}\)

Thus all the real square roots of -9/16 are \(\dfrac{3i}{4}\) and \(-\dfrac{3i}{4}\).

  • b) All the real cube roots of -0.000125.

Put the square root on the number to find the real square roots,

\(\sqrt{-0.000125} =\sqrt{(-1)\times(0.05)^2} \\\sqrt{-0.000125} =\pm (0.05)\sqrt{(-1)} \\\sqrt{-0.000125} =\pm (0.05)(i)\\\sqrt{-0.000125} =\pm (0.05i)\)

Thus all the real square roots of -9/16 are 0.05\(i\) and-0.05 \(i\).

  • c) Simplify \(\sqrt{36g^6}\)

Simplify above square root as,

\(\sqrt{36g^6}=\sqrt{6^2\times g^6} \\\sqrt{36g^6}=(6^2\times g^6} )^\dfrac{1}{2}\\\sqrt{36g^6}=(6^2\times g^{(3)^2}} )^\dfrac{1}{2} \\\)

\(\sqrt{36g^6}=(6\times g^3} )^\dfrac{2}{2}\\\)

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

Hence the Simplified value of  \(\sqrt{36g^6}\) is  \(6g^3\).

  • d) Simplify \(\sqrt[]{125x^{21}y^{24}}\)

Simplify above square root as,

\(\sqrt[]{125x^{21}y^{24}} ={(5^3)(x^7)^3(y^8)^3} )^\dfrac{1}{3}}\\\sqrt[]{125x^{21}y^{24}} ={(5^3)(x^7)^3(y^8)^3} )^\dfrac{3}{3}}\\\sqrt[]{125x^{21}y^{24}} =5^3x^7y^8}\)

Hence the  Simplified value of  \(\sqrt[]{125x^{21}y^{24}}\) is  \(5x^7y^8\)..

  • e) Simplify \(\sqrt{\dfrac{27x^4}{75y^2}}\)

Simplify above square root as,

\(\sqrt{\dfrac{27x^4}{75y^2} } =\sqrt{\dfrac{9\times3\times x^4}{15\times3\times y^2} }\\\sqrt{\dfrac{27x^4}{75y^2} } =({\dfrac{9\times (x^2)^2}{25\times y^2} } )^\dfrac{1}{2}} \\\sqrt{\dfrac{27x^4}{75y^2} } ={\dfrac{9x^2}{25y} }\)

Hence the Simplified value of   \(\sqrt{\dfrac{27x^4}{75y^2}}\) is  \({\dfrac{9x^2}{25y} }\).

Thus,

  • a) All the real square roots of -9/16 are \(\dfrac{3i}{4}\) and \(-\dfrac{3i}{4}\).
  • b) All the real square roots of -9/16 are 0.05\(i\) and -0.05 \(i\).
  • c) The simplified value of  \(\sqrt{36g^6}\) is  \(6g^3\).
  • d) The  Simplified value of  \(\sqrt[]{125x^{21}y^{24}}\) is  \(5x^7y^8\)..
  • e) The simplified value of   \(\sqrt{\dfrac{27x^4}{75y^2}}\) is  \({\dfrac{9x^2}{25y} }\).

Learn more about the real square root here;

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