√3+√2/√3-√2 is approximately between which two numbers?

Hey,

First of all, we need to rationalize the given fraction.

\(\frac{√3+√2}{√3-√2}\)

So we will multiply the numerator and denominator by the conjugate of √3-√2, which is √3+√2.

Thus, we will get -

\(\frac{(√3+√2)(√3+√2)}{(√3-√2)(√3+√2)}\)
=\(\frac{(√3+√2)^2}{(√3)^2-(√2)^2}\)
=\(\frac{(√3)^2+(√2)^2+2.√3.√2}{(√3)^2-(√2)^2}\)
= \(\frac{(5+2.√6)}{(3-2)}\)

Now √6 lies between 2 and 3
i.e. \(2 <√6 <3\)
Taking the extreme cases, we can conclude that the fraction will lie between -

\(5 + 2*2<\frac{√3+√2}{√3-√2}<5 + 2*3\)

\(9<\frac{√3+√2}{√3-√2}< 11\)


Thus the correct answer is Option D

Thanks,
Saquib
Quant Expert
e-GMAT

To practise ten 700+ Level Number Properties Questions attempt the The E-GMAT Number Properties Knockout

MathRevolution Shouldn't this be written as (√3+√2)/(√3-√2) or \(\frac{√3+√2}{√3-√2}\)? What you have written appears to be a completely different math problem or am I missing something?

You are right mate!

==>From \(√3+√2/√3-√2=(√3+√2)^2/(√3-√2)( √3+√2)=3+2+2√6/3-2=5+2√6=5+2(2.xxx)=9.xxx\),

the answer is D.
Answer: D

\(\frac{(√3+√2)}{(√3-√2)}\)

= \(\frac{(√3+√2)(√3+√2)}{(√3-√2)(√3+√2)}\)

= \(\frac{(√3^2 + √2^2 + 2√3√2)}{( 3 - 2 )}\)

= \(3 + 2 + 2√6\)

= \(5 + 2√6\)

= \(5 + 2(2.45)\)

= \(5 + 4.90\)

= \(9.90\)

Hence, the correct answer must be (D) 9 and 11

↧↧↧ Weekly Video Solution to the Problem Series ↧↧↧




We need to find the approximate value of \(\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\)

\(\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\)

Now to simplify this we will multiply the numerator and the denominator with the conjugate of the denominator, which is \(\sqrt3 + \sqrt2\)

=> \(\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\) = \(\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\) * \(\frac{\sqrt3 + \sqrt2}{\sqrt3 + \sqrt2}\) = \(\frac{(\sqrt3+\sqrt2) * (\sqrt3 + \sqrt2)}{(\sqrt3-\sqrt2) * (\sqrt3 + \sqrt2)}\)

Now, numerator becomes \((a+b)^2 = a^2 + 2ab + b^2\) and the denominator becomes (a-b) * (a+b) = \(a^2 - b^2\)

=> \(\frac{(\sqrt3+\sqrt2) * (\sqrt3 + \sqrt2)}{(\sqrt3-\sqrt2) * (\sqrt3 + \sqrt2)}\) = \(\frac{(\sqrt3)^2 + 2 *\sqrt3*\sqrt2 + (\sqrt2)^2 }{(\sqrt3)^2 - (\sqrt2)^2}\) = \(\frac{3 + 2\sqrt6 + 2 }{ 1}\) = 5 + \(2\sqrt6\) ~ 5 + 2*2.45 = 5 + 4.9 = 9.9
=> Between 9 and 11

So, Answer will be D
Hope it helps!

Watch the following video to learn How to Rationalize Roots