Question

If x = 5 - √21, then the value of

$\sf \frac{\sqrt{x}}{\sqrt{32 - 2x} - \sqrt{21}}$ is $\sf\frac{\sqrt{a}\pm\sqrt{b}}{\sqrt{c}}$;

then a + b + c is? ​

Answer 1

Given :-

• x = (5 - √21)

To Find :-

• (a + b + c) = ?

Formula used :-

• (a² + b² - 2ab) = (a - b)²
• √(a²) = a
• (a² + b² + 2ab) = (a + b)²
• (√a * √b) = √(ab)

Solution :-

❁❁ Refer To Image Now .. ❁❁

if we take (√a + √b) /√c , we get, (a + b + c) = 12.

if we take (√a - √b) /√c , we get, (a + b + c) = 6.

______________________________

Answer 2

Given that, x = 5 - √21

Now, multiply it with 2

$\implies\:\sf{x\:=\:5-\sqrt{21}}$ (×2)

$\implies\:\sf{2x\:=\:10-2\sqrt{21}}$

$\implies\:\sf{2x\:=\:10-2\sqrt{7}\sqrt{3}}$

Now, 7 + 3 = 10. Write it in the square root form

$\implies\:\sf{2x\:=\:(\sqrt{7})^2+(\sqrt{3})^2-2\sqrt{7}\sqrt{3}}$

Used identity: (a - b)²= a² + b² - 2ab

$\implies\:\sf{x\:=\:\frac{1}{2}(\sqrt{7} - \sqrt{3})^2}$

Take square root on both sides

$\implies\:\sf{\sqrt{x}\:=\:\frac{1}{\sqrt{2}} \sqrt{(\sqrt{7}-\sqrt{3})^2}}$

$\implies\:\sf{\sqrt{x}\:=\:\frac{1}{\sqrt{2}} (\sqrt{7}-\sqrt{3})}$

Now, substitute value of $\sf{\sqrt{x}}$ in $\sf \frac{\sqrt{x}}{\sqrt{32 - 2x} - \sqrt{21}}$

$\implies\:\sf \frac{\sqrt{7} - \sqrt{3} }{ \sqrt{2} (\sqrt{32 - 2x} - \sqrt{21})}$

Also, substitute value of x in it.

$\implies\:\sf \frac{\sqrt{7} - \sqrt{3} }{\sqrt{2} [\sqrt{32 - 2(5 - \sqrt{21}] } - \sqrt{21})}$

$\implies\:\sf \frac{\sqrt{7} - \sqrt{3} }{\sqrt{2} (\sqrt{32 - 10 +2 \sqrt{21}) } - \sqrt{21})}$

$\implies\:\sf \frac{\sqrt{7} - \sqrt{3} }{\sqrt{2} (\sqrt{22 +2 \sqrt{21} } - \sqrt{21})}$

Used identity: (a + b)² = a² + b² + 2ab

$\implies\:\sf \frac{\sqrt{7} - \sqrt{3} }{\sqrt{2} ( \sqrt{\sqrt{21 + 1}})^{2} - \sqrt{21} }$

$\implies\:\sf \frac{\sqrt{7} - \sqrt{3} }{\sqrt{2} ({\sqrt{21}} - \sqrt{21} + 1) }$

$\implies\:\sf \frac{\sqrt{7} - \sqrt{3} }{ \sqrt{2} ( + 1) }$

$\implies\:\sf \frac{\sqrt{7} - \sqrt{3} }{ \sqrt{2} }$

As per given condition,

$\sf \frac{\sqrt{x}}{\sqrt{32 - 2x} - \sqrt{21}}$ is $\sf\frac{\sqrt{a}\pm\sqrt{b}}{\sqrt{c}}$

On solving $\sf \frac{\sqrt{x}}{\sqrt{32 - 2x} - \sqrt{21}}$ we get, $\implies\:\sf \frac{\sqrt{7} - \sqrt{3} }{ \sqrt{2} }$

On comparing we get,

$\implies\:\sf{a\:=\:7,\:b\:=\:\pm 3\:and\:c\:=\:2}$

Now, we have to find a + b + c.

So,

$\implies\:\sf{a+b+c\:=\:7\pm 3+2\:=\:12 \:and\: 6}$