Question

The number obtained on ratoonalising the denominator of 1/under root 7 - 2 is

please help me it's very urgent
my STANDARD is 6​

Answer 1

Solution -

We have,

• $\sf{\dfrac{1}{\sqrt{7} - 2}}$

⠀

Rationalising the denominator

$\tt:\implies\: \: \: \: \: \: \: \: {\dfrac{1}{\sqrt{7} - 2} \times \dfrac{\sqrt{7} + 2}{\sqrt{7} + 2}}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {\dfrac{\sqrt{7} + 2}{(\sqrt{7})^2 - (2)^2}}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {\dfrac{\sqrt{7} + 2}{7 - 4}}$

⠀

$\bf:\implies\: \: \: \: \: \: \: \: {\purple{\dfrac{\sqrt{7} + 2}{3}}}$

⠀

★ $\underline{\sf{Hence,\: the\: required\: number\: is}}$

⠀⠀⠀⠀⠀⠀⠀•⠀ $\bf{\dfrac{\sqrt{7} + 2}{3}}$

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Answer 2

Given:

The expression -

$\bf \longmapsto \dfrac{1}{ \sqrt{7}-2}$

What To Find:

We have to

• Rationalise the denominator.

How To Find:

To find it, we have to

• First, find the conjugate or the multiple inverses of the denominator.
• Next, multiply the conjugate with the denominator and numerator.
• Then, simplify the expression using identities.
• Then, if the denominator is rational then, it is rationalised.

Solution:

$\sf \longmapsto \dfrac{1}{ \sqrt{7}-2}$

The conjugate is,

$\sf \longmapsto \sqrt{7} + 2$

Multiply the expression with the conjugate.

$\sf \longmapsto \dfrac{1}{ \sqrt{7} - 2} \times \dfrac{\sqrt{7}+ 2}{ \sqrt{7} +2}$

Multiply the numerator,

$\sf \longmapsto \dfrac{\sqrt{7} + 2}{ \sqrt{7} - 2 \times \sqrt{7} +2}$

Using the identities:- (a - b) (a + b) = a² - b²

$\sf \longmapsto \dfrac{\sqrt{7} + 2}{(\sqrt{7})^2-2^2}$

Find the squares,

$\sf \longmapsto \dfrac{\sqrt{7} + 2}{7 - 4}$

Subtract 4 from 7,

$\sf \longmapsto \dfrac{\sqrt{7} + 2}{3}$

Since 3 is a rational number,

∴ Hence rationalised.

Final Answer:

∴ Thus, the answer is $\bf \dfrac{\sqrt{7}+2}{3}$ after rationalising the denominator.