MATH SOLVE

4 months ago

Q:
# 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

Accepted Solution

A:

we are given 5 times the square root of the quantity of x plus 7so, we get left side is [tex]5\sqrt{x+7}[/tex]right side is -10so, we can set them equal [tex]5\sqrt{x+7}=-10[/tex]Since, we have to solve for xso, we will isolate x one anyone side and then we can solve for xwe take square both sides [tex](5\sqrt{x+7})^2=(-10)^2[/tex][tex]25(x+7)=100[/tex]Divide both sides by 25 [tex]\frac{25(x+7)}{25} =\frac{100}{25}[/tex][tex]x+7 =4[/tex]Subtract both sides by 7[tex]x+7-7 =4-7[/tex][tex]x =-3[/tex]Extraneous solution:we can plug x=-3 into original [tex]5\sqrt{-3+7}=-10[/tex][tex]5\sqrt{4}=-10[/tex][tex]5*2=-10[/tex][tex]10=-10[/tex]we know that 10 is not equal to -10so, x=-3 is not valid solution Hence, this is extraneous solution...........Answer