Question

Taking √2=1.414 and √3=1.732 find without using tables or long division the value of 1/3-√2

Answer 1

Given:

$\dfrac{1}{\sqrt{3}-\sqrt{2}}$

Here, $\sqrt{2}$ = 1.414 and $\sqrt{3}$ = 1.732

We have to find, the value of $\dfrac{1}{\sqrt{3}-\sqrt{2}}$ is:

Solution:

∴ $\dfrac{1}{\sqrt{3}-\sqrt{2}}$

Rationalising numerator and denominator, we get

$=\dfrac{1}{\sqrt{3}-\sqrt{2}}\times \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

Using the algebraic identity:

(a + b)(a - b) = $a^{2} -b^{2}$

$=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}^2-\sqrt{2}^2}$

$=\dfrac{\sqrt{3}+\sqrt{2}}{3-2}$

$=\dfrac{\sqrt{3}+\sqrt{2}}{1}$

= $\sqrt{3}$ + $\sqrt{2}$

Put $\sqrt{2}$ = 1.414 and $\sqrt{3}$ = 1.732, we get

= 1.732 + 1.414

= 3.146

∴ $\dfrac{1}{\sqrt{3}-\sqrt{2}}$ = 3.146

Thus, the value of $\dfrac{1}{\sqrt{3}-\sqrt{2}}$ is "equal to 3.146".

Answer 2

Answer:

The answer is 6.292