Question

2+root7/ 3-2root7 answer pls

Answer 1

Appropriate Question:

Rationalize the denominator of :

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

Step-by-step explanation:

As per the provided information in the given question, we have to rationalise the denominator of $\rm{\dfrac{2 +\sqrt{7}}{3-2\sqrt{7}} }$.

In order to rationalise the denominator of any fraction, we multiply the rationalising factor of the denominator with both the numerator and the denominator of the fraction. Then, by using identities, we further perform arithmetic operations to rationalise it.

Here, the denominator is in the form of (a – b) and the rationalising factor of (a – b) is (a + b). So, the rationalising factor of (3 – 2√7) is (3 + 2√7).

Multiplying (3 + 2√7) with both the numerator and the denominator of the fraction.

$\dashrightarrow \quad \rm {\dfrac{2 +\sqrt{7}}{3-2\sqrt{7}} \times \dfrac{3+2\sqrt{7}}{3+2\sqrt{7}} }$

As we know that fraction rules state that a/b × c/d is equivalent to ac/bd. So, we can also rearrange the terms here in the same way.

$\dashrightarrow \quad \rm {\dfrac{(2 +\sqrt{7})(3+2\sqrt{7})}{(3-2\sqrt{7})(3+2\sqrt{7})} }$

In the numerator, we'll use distributive property of multiplication over addition to perform addition. Whereas, in the denominator we'll use the following identity,

⠀⠀⠀⠀⠀⠀★ (a – b)(a + b) = a² – b²

$\dashrightarrow \quad \rm {\dfrac{2(3+2\sqrt{7}) +\sqrt{7}(3+2\sqrt{7})}{(3)^2-(2\sqrt{7})^2 } }$

Again using distributive property of multiplication over addition to perform addition in the numerator and writing the squares of the terms in the denominator.

$\dashrightarrow \quad \rm {\dfrac{6 +4\sqrt{7} +3\sqrt{7} + 14 }{9- 28 } }$

Now, performing addition of the like terms in the numerator of the fraction and performing subtraction in the denominator.

$\dashrightarrow \quad \rm {\dfrac{20 +7\sqrt{7} }{-19} }$

Multiplying –1 with both numerator and the denominator.

$\dashrightarrow \quad \rm {\dfrac{-1(20 +7\sqrt{7}) }{-1(-19)} }$

Performing multiplication in the both the numerator and the denominator.

$\dashrightarrow \quad\underline{\boxed{ \bf {\dfrac{-20 -7 \sqrt{7} }{19} } }}$

Hence, rationalised!

Answer 2

$\purple{ \huge \underline{ \boxed{ \mathbb \red{QUESTION: }}}}$

$\huge \implies\bf \frac{2 + \sqrt{7} }{3 - 2 \sqrt{7} }$

$\purple{ \huge\underline{ \boxed{ \mathbb \red{ANSWER: }}}}$

$\purple{ \bf\underline{ \underline{ \mathbb \purple{RATIONALISING : }}}}$

$\implies\bf \frac{2 + \sqrt{7} }{3 - 2 \sqrt{7} } \times \frac{3 + 2 \sqrt{7} }{3 + 2 \sqrt{7} } \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies\bf \frac{(2 + \sqrt{7})(3 + 2 \sqrt{7} ) }{ ({3})^{2} - ( {2 \sqrt{7} })^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies\bf \frac{6 + 4 \sqrt{7} + 3 \sqrt{7} + 2 ({\sqrt{7} })^{2} }{9 - 4\times 7} \\ \\ \implies\bf \frac{6 + 7 \sqrt{7} + 2 \times 7 }{9 - 28} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies\bf\frac{6 + 7 \sqrt{7} + 14}{ - 19} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies\bf \frac{20 + 7 \sqrt{7} }{ - 19} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies\bf \frac{ \cancel- ( - 20 - 7 \sqrt{7} )}{ \cancel- 19} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies\bf \frac{ - 20 - 7 \sqrt{7} }{19} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\purple{ \bf\underline{ \underline{ \mathbb \purple{HENCE \: \: \: RATIONALISED \: . }}}}$