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√(x+7)`

right side is -10

so, we can set them equal

`5√(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√(x+7))^2=(-10)^2`

`25(x+7)=100`

Divide both sides by 25

`(25(x+7))/(25) =(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√(-3+7)=-10`

`5√(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