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

Aug 15, 2017

Solution: $x = - 1 \mathmr{and} x = \frac{9}{5}$

#### Explanation:

Restriction : $13 - 4 {x}^{2} \ge 0 \mathmr{and} 13 \ge 4 {x}^{2}$

$\frac{13}{4} \ge {x}^{2} \mathmr{and} {x}^{2} \le \frac{13}{4} \mathmr{and} x \le \pm \frac{\sqrt{13}}{2}$

$x \le 1.8027756 \mathmr{and} x \ge - 1.8027756$

$- 1.8028 \le x \le 1.8028 \mathmr{and} \left[- 1.8028 , 1.8028\right]$

$\sqrt{13 - 4 {x}^{2}} = 2 - x$ Squaring both sides we get,

$\left(13 - 4 {x}^{2}\right) = {\left(2 - x\right)}^{2} \mathmr{and} 13 - 4 {x}^{2} = 4 - 4 x + {x}^{2}$ or

$4 {x}^{2} + {x}^{2} - 4 x + 4 - 13 = 0 \mathmr{and} 5 {x}^{2} - 4 x - 9 = 0$ or

$5 {x}^{2} + 5 x - 9 x - 9 = 0 \mathmr{and} 5 x \left(x + 1\right) - 9 \left(x + 1\right) = 0$ or

$\left(x + 1\right) \left(5 x - 9\right) = 0$ . Either $x + 1 = 0 \therefore x = - 1$ or

$5 x - 9 = 0 \therefore 5 x = 9 \mathmr{and} x = \frac{9}{5}$ . Solution: $x = - 1 \mathmr{and} x = \frac{9}{5}$

Check:

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

$2 - x = 2 - \left(- 1\right) = 3 \therefore L H S = R H S$ (verified).

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

$2 - x = 2 - \frac{9}{5} = 2 - 1.8 = 0.2 \therefore L H S = R H S$ (verified).

So both $x = - 1 \mathmr{and} x = \frac{9}{5}$ hold good. [Ans]