Question

If root3-root2/root3-root2 =a+b root6
Then find the value of a and b where a and b are rational numbers

Answer 1

Answer:

a= 5 b=2

Step-by-step explanation:

Given:

(√3+√2)/(√3-√2) =a+b√6

Multiplying numerator and denominator of LHS with (√3+√2) we get―

=(√3+√2)(√3+√2)/[(√3-√2)(√3+√2)]

=[(√3+√2)^2]/(3-2) [using (a+b)(a-b)=a^2-b^2]

=(3+2+2√6)/1

=5+√6

On comparision with a+b√6, we see–

a=5 and b=2

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Answer 2

Correct Question:

$[tex]\dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2}} = a + b \sqrt{6}$

To find:

The value of a and b.

Solution:

Just rationalize the denominator of L.H.S term and we will get the answer...

$\dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2}} \times\dfrac{\sqrt{3}+\sqrt{2} }{ \sqrt{3}+\sqrt{2}}\\ \\ = \dfrac{{( \sqrt{3} + \sqrt{2} )}^{2} }{ {(\sqrt{3})}^{2} - {(\sqrt{2} )}^{2} } \\ \\ = \frac{ {(\sqrt{3}) }^{2} + 2.\sqrt{3} . \sqrt{2} + {(\sqrt{2})}^{2}}{3-2}\\ \\ = \frac{3 + 2 + 2 \sqrt{6} }{1} \\ \\= 5 + 2\sqrt{6}$

Comparing the R.H.S from L.H.S I get,

a = 5 and b√6 = 2√6 i.e b = 2

$\text{Hence, the required answer is}$$\boxed{a = 5, b = 2}$

${{\boxed{ \text{\blue{Important Algebra Rules}}}}}$

${(x+y)}^{2}={x}^{2}+{y}^{2}+2xy\\ \\{(x-y)}^{2}={x}^{2}+{y}^{2}-2xy\\ \\{(x+y)}^{2}= (x - y) {}^{2}+4xy\\ \\{(x-y)}^{2}=(x+y){}^{2}-4xy\\ \\ (x + y)^{2}+(x-y)^{2}=2( {x}^{2}+{y}^{2} )\\ \\(x+y) (x-y)={x}^2-{y}^2$