Question

Find the solution of 5 times the square root of the quantity of x plus 7 equals negative 10, and determine if it is an extraneous solution

Answer 1

we are given

5 times the square root of the quantity of x plus 7

so, we get

left side is

$5\sqrt{x+7}$

right side is -10

so, we can set them equal

$5\sqrt{x+7}=-10$

Since, we have to solve for x

so, we will isolate x one anyone side

and then we can solve for x

we take square both sides

$(5\sqrt{x+7})^2=(-10)^2$

$25(x+7)=100$

Divide both sides by 25

$\frac{25(x+7)}{25} =\frac{100}{25}$

$x+7 =4$

Subtract both sides by 7

$x+7-7 =4-7$

$x =-3$

Extraneous solution:

we can plug x=-3 into original

$5\sqrt{-3+7}=-10$

$5\sqrt{4}=-10$

$5*2=-10$

$10=-10$

we know that 10 is not equal to -10

so, x=-3 is not valid solution

Hence, this is extraneous solution...........Answer

Answer 2

The solution of the information given will be as follows:
5√(x+7)=10
dividing both sides by 5 we get:
√(x+7)=2
squaring both sides we get:
(x+7)=2²
x+7=4
solving for x we get
x=4-7
x=-3
thus the answer is x=-3
this is an extraneous solution.<span />