Question

If x = 7 + 7√40 , find the value of √x+
1
√x​

Answer 1

Answer:

Given x = 7+√40

=> x = 7+√2×2×10

=> x = 7+2√10

=> x = 7+2×√5×√2

=> x = 5+2+2×√5×√2

=> x = (√5)²+(√2)²+2×√5×√2

=> x = (√5+√2)²

=> √x = √5+√2 ----(1)

Now ,

ii)\frac{1}{\sqrt{x}}

= \frac{1}{\sqrt{5}+\sqrt{2}}

/* Rationalising the denominator, we get

=\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}

=\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}

=\frac{\sqrt{5}-\sqrt{2}}{5-2}

=\frac{\sqrt{5}-\sqrt{2}}{3} ----(2)

iii) \sqrt{x}+\frac{1}{\sqrt{x}}

=\sqrt{5}+\sqrt{2}+\frac{\sqrt{5}-\sqrt{2}}{3}

=\frac{3(\sqrt{5}+\sqrt{2})+\sqrt{5}-\sqrt{2}}{3}

=\frac{3\sqrt{5}+3\sqrt{2}+\sqrt{5}-\sqrt{2}}{3}

=\frac{4\sqrt{5}+2\sqrt{2}}{3}

Therefore,

\sqrt{x}+\frac{1}{\sqrt{x}}=\frac{4\sqrt{5}+2\sqrt{2}}{3}

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