# Find the value of $\sqrt {\sqrt[3]{{125}} + \sqrt {24} } $.

(A) $\sqrt 5 + 1$ (B) $\sqrt 3 + \sqrt 2 $ (C) $\sqrt 3 + 1$ (D) $\sqrt 5 + \sqrt 2 $

Last updated date: 25th Mar 2023

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Answer

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Hint: The number $\sqrt {\sqrt[3]{{125}} + \sqrt {24} } $ is an irrational number. Assume it to be some variable irrational number and then solve by squaring both sides.

The number given in the question is an irrational number. Its square root is also an irrational number.

Thus, let $\sqrt {\sqrt[3]{{125}} + \sqrt {24} } = \sqrt a + \sqrt b $. We know that the cube root of 125 is 5. So, we’ll get:

$ \Rightarrow \sqrt a + \sqrt b = \sqrt {5 + \sqrt {24} } $

Squaring both sides, we’ll get:

$

\Rightarrow {\left( {\sqrt a + \sqrt b } \right)^2} = {\left( {\sqrt {5 + \sqrt {24} } } \right)^2}, \\

\Rightarrow a + b + 2\sqrt {ab} = 5 + \sqrt {24} .....(i) \\

$

Now, equating rational parts on both sides, we’ll get:

$ \Rightarrow a + b = 5$,

Similarly equating irrational parts on both sides, we have:

$

\Rightarrow 2\sqrt {ab} = \sqrt {24} , \\

\Rightarrow 4ab = 24, \\

\Rightarrow ab = 6 .....(ii) \\

$

Putting $b = 5 - a$ from equation $(i)$, we’ll get:

$

\Rightarrow a\left( {5 - a} \right) = 6, \\

\Rightarrow 5a - {a^2} = 6, \\

\Rightarrow {a^2} - 5a + 6 = 0 \\

$

This is a quadratic equation in a. We will use factorization method to solve it:

$

\Rightarrow {a^2} - 5a + 6 = 0, \\

\Rightarrow {a^2} - 3a - 2a + 6 = 0, \\

\Rightarrow a\left( {a - 3} \right) - 2\left( {a - 3} \right) - 0, \\

\Rightarrow \left( {a - 2} \right)\left( {a - 3} \right) = 0, \\

$

$ \Rightarrow a = 2$ or $a = 3$

From equation $(ii)$, we have $ab = 6$.

If we consider $a = 2$, we will get $b = 3$ and our number will be:

$ \Rightarrow \sqrt a + \sqrt b = \sqrt 2 + \sqrt 3 $

And if we consider $a = 3$, we will get $b = 2$ and our number will be:

$ \Rightarrow \sqrt a + \sqrt b = \sqrt 3 + \sqrt 2 $

Therefore the square root $\sqrt {\sqrt[3]{{125}} + \sqrt {24} } $ is $\sqrt 3 + \sqrt 2 $. (B) is the correct option.

Note: If we are facing some difficulty over solving the quadratic equation $a{x^2} + bx + c = 0$by factorization method, we can also use formula for finding its roots:

$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

The number given in the question is an irrational number. Its square root is also an irrational number.

Thus, let $\sqrt {\sqrt[3]{{125}} + \sqrt {24} } = \sqrt a + \sqrt b $. We know that the cube root of 125 is 5. So, we’ll get:

$ \Rightarrow \sqrt a + \sqrt b = \sqrt {5 + \sqrt {24} } $

Squaring both sides, we’ll get:

$

\Rightarrow {\left( {\sqrt a + \sqrt b } \right)^2} = {\left( {\sqrt {5 + \sqrt {24} } } \right)^2}, \\

\Rightarrow a + b + 2\sqrt {ab} = 5 + \sqrt {24} .....(i) \\

$

Now, equating rational parts on both sides, we’ll get:

$ \Rightarrow a + b = 5$,

Similarly equating irrational parts on both sides, we have:

$

\Rightarrow 2\sqrt {ab} = \sqrt {24} , \\

\Rightarrow 4ab = 24, \\

\Rightarrow ab = 6 .....(ii) \\

$

Putting $b = 5 - a$ from equation $(i)$, we’ll get:

$

\Rightarrow a\left( {5 - a} \right) = 6, \\

\Rightarrow 5a - {a^2} = 6, \\

\Rightarrow {a^2} - 5a + 6 = 0 \\

$

This is a quadratic equation in a. We will use factorization method to solve it:

$

\Rightarrow {a^2} - 5a + 6 = 0, \\

\Rightarrow {a^2} - 3a - 2a + 6 = 0, \\

\Rightarrow a\left( {a - 3} \right) - 2\left( {a - 3} \right) - 0, \\

\Rightarrow \left( {a - 2} \right)\left( {a - 3} \right) = 0, \\

$

$ \Rightarrow a = 2$ or $a = 3$

From equation $(ii)$, we have $ab = 6$.

If we consider $a = 2$, we will get $b = 3$ and our number will be:

$ \Rightarrow \sqrt a + \sqrt b = \sqrt 2 + \sqrt 3 $

And if we consider $a = 3$, we will get $b = 2$ and our number will be:

$ \Rightarrow \sqrt a + \sqrt b = \sqrt 3 + \sqrt 2 $

Therefore the square root $\sqrt {\sqrt[3]{{125}} + \sqrt {24} } $ is $\sqrt 3 + \sqrt 2 $. (B) is the correct option.

Note: If we are facing some difficulty over solving the quadratic equation $a{x^2} + bx + c = 0$by factorization method, we can also use formula for finding its roots:

$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

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