**If `sqrt (x+3) + sqrt (x-2) = 5` , then the value of x is ____ ?**

This is a question I got while preparing for NTSE, PLZ sove it ASAP...

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`sqrt(x+3) + sqrt(x-2) = 5`

To remove the radicals, perform the opposite operation of square root. So, take the square of both sides.

`(sqrt(x+3) + sqrt(x-2))^2 = 5^2`

`x+3 + 2sqrt(x+3)*sqrt(x-2) + x-2=25`

`2x+1+2sqrt(x+3)*sqrt(x-2)=25`

Then, isolate the radical term.

`2x+1-25=-2sqrt(x+3)*sqrt(x-2)`

`2x-24=-2sqrt(x+3)*sqrt(x-2)`

Take the square of both sides again to eliminate the radicals.

`(2x-24)^2=(-2sqrt(x+3)*sqrt(x-2))^2`

`4x^2-96x+576=4(x+3)(x-2)`

`4x^2-96x+576=4(x^2+x-6)`

`4x^2-96x+576=4x^2+4x-24`

Note that the terms `4x^2` appears on both sides of the equation. So if we move the `4x^2` from the right to left, the resulting expression is:

`-96x+576=4x-24`

Then, bring together the terms with x on one side of the equation. Also, bring together the terms without x on the opposite side of the equation.

`576+24=4x+96x`

`600=100x`

Then, isolate x.

`600/100=x`

`6=x`

**Hence, the value of x is 6.**

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