Question

How do you solve `\sqrt { 3- x } + \sqrt { 2x + 3} = 3`?

Answer 1

`x=2sqrt15-5`

Explanation:

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

Observe that domain is limited to `x<=3` or `x>= -3/2` i.e. `[-3/2,3]`

`hArrsqrt(2x+3)=3-sqrt(3-x)`, now squaring both sides

`2x+3=3+3-x-6sqrt(9-3x)`

or `2x+3=6-x-6sqrt(9-3x)`

or `6sqrt(9-3x)=3-3x` or `2sqrt(9-3x)=1-x`

and squaring again, we get

`4(9-3x)=1+x^2-2x`

or `36-12x=1+x^2-2x`

or `x^2+10x-35=0` and using quadratic formula

i.e. `x=(-10+-sqrt(100-4xx1xx(-35)))/2`

= `-5+-sqrt240/2=-5+-2sqrt15`

But `-5-2sqrt15` is not in domain and hence `x=2sqrt15-5`