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

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Answer 1 · 2017-08-15T11:36:02

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

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]

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