Name: Rationalisation of Surds
Uploaded: 2019-07-19
Duration: 5 min 28 s
Description: When the denominator of an expression is a surd that can be reduced to an expression with a rational denominator, this process is called the rationalisation of a surd.

Question

Find the cube root of

54 \sqrt{3} + 41 \sqrt{5}

Accepted Answer

Rewriting the given expression as

$54 \sqrt{3} + 41 \sqrt{5} = 3 \sqrt{3} (18 + \frac{41}{3} \times \frac{\sqrt{5}}{\sqrt{3}})$

$\Rightarrow {(54 \sqrt{3} + 41 \sqrt{5})}^{\frac{1}{3}} = {[3 \sqrt{3} (18 + \frac{41}{3} \sqrt{\frac{5}{3}})]}^{\frac{1}{3}}$

$= \sqrt{3} {(18 + \frac{41}{3} \sqrt{\frac{5}{3}})}^{\frac{1}{3}}$

Now, let ${(18 + \frac{41}{3} \sqrt{\frac{5}{3}})}^{\frac{1}{3}} = x + \sqrt{y} . . . (i)$

Then, ${(18 - \frac{41}{3} \sqrt{\frac{5}{3}})}^{\frac{1}{3}} = x - \sqrt{y} . . . (ii)$

Multiplying $(i)$ and $(ii)$ , we have

${[{(18)}^{2} - {(\frac{41}{3} \times \sqrt{\frac{5}{3}})}^{2}]}^{\frac{1}{3}} = x^{2} - y$

$\Rightarrow {(324 - \frac{1681 \times 5}{27})}^{\frac{1}{3}} = x^{2} - y$

$\Rightarrow {(\frac{343}{27})}^{\frac{1}{3}} = x^{2} - y$

$\Rightarrow \frac{7}{3} = x^{2} - y$

$\Rightarrow y = x^{2} - \frac{7}{3}$

Cubing $(i)$ both sides, we get

$18 + \frac{41}{3} \sqrt{\frac{5}{3}} = x^{3} + 3 x y + 3 x^{2} \sqrt{y} + y \sqrt{y}$

Equating rational part, we get

$x^{3} + 3 x y = 18$

or $x^{3} + 3 x (x^{2} - \frac{7}{3}) = 18; [∵ y = x^{2} - \frac{7}{3}]$

Solving it for $x$ , we get, $x = 2 \Rightarrow$ $y = 4 - \frac{7}{3} = \frac{5}{3}$

Hence, the required cube root is

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

Find the cube root of $54 \sqrt{3} + 41 \sqrt{5}$

Important Questions on Surds and Imaginary Quantities

Learn and Prepare for any exam you want !

Find the cube root of 543+415

Important Questions on Surds and Imaginary Quantities

Learn and Prepare for any exam you want !

Find the cube root of $54 \sqrt{3} + 41 \sqrt{5}$