# How do you solve sqrt(a-2)+4=a and check your solution?

$a = 6$

#### Explanation:

$\sqrt{a - 2} + 4 = a$

Let's bring the 4 over to the right side so that we can square the square root:

$\sqrt{a - 2} = a - 4$

${\left(\sqrt{a - 2}\right)}^{2} = {\left(a - 4\right)}^{2}$

$a - 2 = {a}^{2} - 8 a + 16$

Now let's bring all the terms to one side and set it to 0:

${a}^{2} - 9 a + 18 = 0$

$\left(a - 6\right) \left(a - 3\right) = 0$

therefore

$a = 6 , a = 3$

And now let's check our solutions:

$\sqrt{6 - 2} + 4 = 6$

$\sqrt{4} + 4 = 6$

$2 + 4 = 6$

$6 = 6$ color(green)(Check!)

~~~~~

$\sqrt{3 - 2} + 4 = 3$

$\sqrt{1} + 4 = 3$

$1 + 4 = 3$

$5 \ne 3$ color(red)(No!)

And so our final solution is $a = 6$

We can also check this solution by graphing the left side and the right side:

graph{(y-sqrt(x-2)-4)(y-x)=0 [-4.79, 15.21, -1, 9]}