Question

Identify all of the following solutions of square root of x minus 8 end root plus 8 equals x.

Answer 1

$\sqrt{x-8} +8=x \\ \sqrt{x-8} =x-8 \\ {( \sqrt{x-8} )}^{2} = {(x-8)}^{2} \\ x-8= x^{2} -16x+64 \\ x^{2} -16x-x+64+8=0 \\ x^{2} -17x+72=0$

$(x-8)(x-9)=0 \\ x-8=0 \ and \ x-9=0 \\ x=8 \ and \ x=9$

Answer 2

Answer:

Step-by-step explanation:

According to the question, the equation becomes,

$\sqrt{x-8}+8=x$

⇒$\sqrt{x-8}=x-8$

Squaring on both the sides, we get

⇒$x-8=(x-8)^2$

⇒$x-8=x^2-16x+64$

⇒$x^2-16x-x+64+8=0$

⇒$x^2-17x+72=0$

⇒$x^2-8x-9x+72=0$

⇒$x(x-8)-9(x-8)=0$

⇒$(x-8)(x-9)=0$

Then, (x-8)=0

⇒x=8

and (x-9)=0

⇒x=9

Thus, the solution of the given equation is x=8 and x=9.