Quadratic Equations Question 9

Which of the following is a solution to the equation $ \sqrt{19+x}+1=x $?

$-3$
$5$
$6$

I only

II only

III only

I and III

Approach 1

We want to square both sides to get rid of the radical. Isolate the radical first to make the process easier.

$$\sqrt{19+x}+1=x $$ $$ \sqrt{19+x}=x-1$$ $$ (\sqrt{19+x})^2=(x-1)^2$$ $$ 19+x=x^2-2x+1$$ $$ x^2-3x-18=0$$ $$ (x-6)(x+3)=0 $$ $$ x=6 \text{ and }-3 $$

We need to check for extraneous solutions since we squared a variable.

$x=6$

$$\sqrt{19+x}+1=x $$ $$\sqrt{19+6}+1=6 $$ $$\sqrt{25}+1=6$$ $$ 5+1=\boxed{6} \checkmark$$

$x=-3$

$$\sqrt{19+x}+1=x $$ $$\sqrt{19+(-3)}+1=-3 $$ $$\sqrt{16}+1=-3 $$ $$ 4+1 \ne -3 $$

Approach 2

For multiple choice questions like these, it may seem more reasonable to just check all options since we need to check for extraneous solutions anyways. As shown previously, $6$ is a solution and $-3$ is not a solution. Therefore, the only possible option that is correct is III only. There is no need to test II since no other options have both II and III. We can infer that II does not work.