Question

Find the non-extraneous solutions of the square root of the quantity x plus 9 minus 5 equals quantity x plus 4

Answer 1

Answer:

x=8 and x=9

Step-by-step explanation:

The given equation is:

$\sqrt{x+9}-5=x+4$

On solving this equation, we get

⇒$\sqrt{x+9}=x+9$

Squaring on both sides, we get

⇒$x+9=x^2+81+18x$

⇒$x^2+18x-x-9+81=0$

⇒$x^2+17x+72=0$

⇒$x^2+8x+9x+72=0$

⇒$(x+8)(x+9)=0$

Thus, the solution of the above equations are x= 8 and x=9.

Answer 2

I"m guessing you mean
$\sqrt{x+9} -5=x+4$
add 5 to both sides
$\sqrt{x+9}=x+9$
square both sides
x+9=x²+18x+81
minus x+9 both sides
0=x²+17x+72
factor
what 2 numbres multiply to get 72 and add to get 17?
8 and 9
0=(x+8)(x+9)
set to zer

0=x+8
x=-8

x+9=0
x=-9

test each

$\sqrt{-8+9} -5=-8+4$
$\sqrt{1} -5=-4$
1-5=-4
-4=-4
true

$\sqrt{-9+9} -5=-9+4$
$\sqrt{0} -5=-5$
-5=-5
true


the solutions are x=-9 and -8