Question

How do you solve `sqrt(13-4x^2)=2-x` and identify any restrictions?

Answer 1

Solution: `x = -1 or x =9/5`

Explanation:

Restriction : `13 - 4x^2 >=0 or 13 >= 4x^2`

`13/4 >= x^2 or x^2 <= 13/4 or x <= +- sqrt13/2`

`x <= 1.8027756 or x >= -1.8027756`

`-1.8028 <= x <= 1.8028 or [-1.8028 ,1.8028]`

`sqrt(13-4x^2) =2-x` Squaring both sides we get,

`(13-4x^2) =(2-x)^2 or 13-4x^2 =4-4x+x^2` or

`4x^2+x^2 -4x +4 -13=0 or 5x^2 - 4x -9 =0` or

`5x^2 +5x - 9x -9 =0 or 5x(x+1) -9(x+1) =0` or

`(x+1)(5x-9)=0` . Either `x+1 =0 :. x =-1` or

`5x-9=0 :. 5x=9 or x =9/5` . Solution: `x = -1 or x =9/5`

Check:

When `x= -1 ; sqrt(13-4x^2) =sqrt(13-4(-1)^2)=sqrt 9 =3`

`2-x = 2 - (-1) =3 :. LHS =RHS` (verified).

When `x= 9/5 ; sqrt(13-4x^2) =sqrt(13-4*81/25)=sqrt 0.04 =0.2`

`2-x = 2 - 9/5 =2-1.8 = 0.2 :. LHS =RHS` (verified).

So both `x = -1 and x =9/5` hold good. [Ans]