Question

How to solve question 10

Answer 1

Answer:

Option E $x=x^2-10x+21$

Step-by-step explanation:

we have

$x-5=\sqrt{x+4}$

we know that

For x> 5 both sides of the equation will be positive

Squared both sides

$(x-5)^2=(\sqrt{x+4})^2$

$(x-5)^2=(x+4)$

$x^2-10x+25=(x+4)$

Subtract 4 both sides

$x^2-10x+25-4=(x+4)-4$

$x^2-10x+21=x$

Rewrite

$x=x^2-10x+21$