Question

If 1/√a-√b=1/3 and 1/√a+√b=1/2, then find the difference of a and b.​

Answer 1

ANSWER:

If $\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{1}{3}$ and $\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{2}$ then the difference of a and b is 6

SOLUTION:

Given, $\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{1}{3}$ →$\sqrt{a}-\sqrt{b}=3$ ----- (1)

And $\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{2}$ → $\sqrt{a}+\sqrt{b}=2$ --- (2)

We have to find difference of a and b.

Now, add (1) and (2)

$\sqrt{a}-\sqrt{b}=3$

$\sqrt{a}+\sqrt{b}=2$

Adding above two equations, we get,

$2 \sqrt{a}+0=2+3$

$\begin{array}{l}{2 \sqrt{a}=5} \\\\ {\sqrt{a}=\frac{5}{2}} \\\\ {a=\frac{25}{4}}\end{array}$

substitute $\sqrt{a}$ value in (2)

$\begin{array}{l}{\frac{5}{2}+\sqrt{b}=2} \\\\ {\sqrt{b}=\frac{2}{\sin \frac{5}{2}}} \\\\ {\sqrt{b}=\frac{4-5}{2}} \\\\ {\sqrt{b}=\frac{-1}{2}} \\\\ {b=\frac{1}{4}}\end{array}$

Now, difference of a and b is a – b = $\frac{25}{4}-\frac{1}{4}=\frac{24}{4}=6$

Hence, the difference of a and b is 6.