# Find the square root of

$5 - \sqrt {10} - \sqrt {15} + \sqrt 6 $

Last updated date: 21st Mar 2023

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Answer

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Hint: - Let the square root of the given equation be\[\left( {{\text{1 + }}\sqrt a - \sqrt b } \right)\], and also use the property which is \[{\left( {a + b - c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab - 2bc - 2ca\]

Given equation

$5 - \sqrt {10} - \sqrt {15} + \sqrt 6 $

Square root of given equation is

\[\sqrt {5 - \sqrt {10} - \sqrt {15} + \sqrt 6 } \]

There are four terms in the given equation therefore in the square root of this it has three terms.

So, let \[{\text{1 + }}\sqrt a - \sqrt b = \sqrt {5 - \sqrt {10} - \sqrt {15} + \sqrt 6 } ................\left( 1 \right)\]

Squaring both sides

\[{\left( {{\text{1 + }}\sqrt a - \sqrt b } \right)^2} = {\left( {\sqrt {5 - \sqrt {10} - \sqrt {15} + \sqrt 6 } } \right)^2}\]

Now, as we know that \[{\left( {a + b - c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab - 2bc - 2ca\]

\[

\Rightarrow 1 + a + b + 2\sqrt a - 2\sqrt {ab} - 2\sqrt b = 5 - \sqrt {10} - \sqrt {15} + \sqrt 6 \\

\Rightarrow 1 + a + b + 2\sqrt a - 2\sqrt {ab} - 2\sqrt b = 5 + \sqrt 6 - \sqrt {15} - \sqrt {10} \\

\]

So, on comparing

\[

1 + a + b = 5{\text{, }}\sqrt a = \dfrac{{\sqrt 6 }}{2},{\text{ }}\sqrt {ab} {\text{ = }}\dfrac{{\sqrt {15} }}{2}{\text{, and }}\sqrt b = \dfrac{{\sqrt {10} }}{2} \\

\sqrt a = \dfrac{{\sqrt 6 }}{2} \Rightarrow a = \dfrac{6}{4} = \dfrac{3}{2} \\

\sqrt b = \dfrac{{\sqrt {10} }}{2} \Rightarrow b = \dfrac{{10}}{4} = \dfrac{5}{2} \\

\]

From equation (1)

\[

{\text{1 + }}\sqrt a - \sqrt b = \sqrt {5 - \sqrt {10} - \sqrt {15} + \sqrt 6 } \\

\Rightarrow \sqrt {5 - \sqrt {10} - \sqrt {15} + \sqrt 6 } = 1 + \sqrt {\dfrac{3}{2}} - \sqrt {\dfrac{5}{2}} \\

\]

As we know perfect square root always gives us \[ \pm \]condition

\[ \Rightarrow \sqrt {5 - \sqrt {10} - \sqrt {15} + \sqrt 6 } = \pm \left( {1 + \sqrt {\dfrac{3}{2}} - \sqrt {\dfrac{5}{2}} } \right)\]

So, this is the required square root.

Note: -In such types of questions always assume the required square root as above then take square on both sides and compare its terms and calculate the values of $a$ and $b$ then from the equation which we assumed substitute the values of $a$ and $b$ we will get the required square root.

Given equation

$5 - \sqrt {10} - \sqrt {15} + \sqrt 6 $

Square root of given equation is

\[\sqrt {5 - \sqrt {10} - \sqrt {15} + \sqrt 6 } \]

There are four terms in the given equation therefore in the square root of this it has three terms.

So, let \[{\text{1 + }}\sqrt a - \sqrt b = \sqrt {5 - \sqrt {10} - \sqrt {15} + \sqrt 6 } ................\left( 1 \right)\]

Squaring both sides

\[{\left( {{\text{1 + }}\sqrt a - \sqrt b } \right)^2} = {\left( {\sqrt {5 - \sqrt {10} - \sqrt {15} + \sqrt 6 } } \right)^2}\]

Now, as we know that \[{\left( {a + b - c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab - 2bc - 2ca\]

\[

\Rightarrow 1 + a + b + 2\sqrt a - 2\sqrt {ab} - 2\sqrt b = 5 - \sqrt {10} - \sqrt {15} + \sqrt 6 \\

\Rightarrow 1 + a + b + 2\sqrt a - 2\sqrt {ab} - 2\sqrt b = 5 + \sqrt 6 - \sqrt {15} - \sqrt {10} \\

\]

So, on comparing

\[

1 + a + b = 5{\text{, }}\sqrt a = \dfrac{{\sqrt 6 }}{2},{\text{ }}\sqrt {ab} {\text{ = }}\dfrac{{\sqrt {15} }}{2}{\text{, and }}\sqrt b = \dfrac{{\sqrt {10} }}{2} \\

\sqrt a = \dfrac{{\sqrt 6 }}{2} \Rightarrow a = \dfrac{6}{4} = \dfrac{3}{2} \\

\sqrt b = \dfrac{{\sqrt {10} }}{2} \Rightarrow b = \dfrac{{10}}{4} = \dfrac{5}{2} \\

\]

From equation (1)

\[

{\text{1 + }}\sqrt a - \sqrt b = \sqrt {5 - \sqrt {10} - \sqrt {15} + \sqrt 6 } \\

\Rightarrow \sqrt {5 - \sqrt {10} - \sqrt {15} + \sqrt 6 } = 1 + \sqrt {\dfrac{3}{2}} - \sqrt {\dfrac{5}{2}} \\

\]

As we know perfect square root always gives us \[ \pm \]condition

\[ \Rightarrow \sqrt {5 - \sqrt {10} - \sqrt {15} + \sqrt 6 } = \pm \left( {1 + \sqrt {\dfrac{3}{2}} - \sqrt {\dfrac{5}{2}} } \right)\]

So, this is the required square root.

Note: -In such types of questions always assume the required square root as above then take square on both sides and compare its terms and calculate the values of $a$ and $b$ then from the equation which we assumed substitute the values of $a$ and $b$ we will get the required square root.

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