Answer:
[tex]x=\frac{225}{16}[/tex] is an extraneous solution.
Step-by-step explanation:
Given : Equation [tex]-4\sqrt x-3=12[/tex]
To find : Solve for x and identify if it is an extraneous solution?
Solution :
Step 1 - Write the equation,
[tex]-4\sqrt x-3=12[/tex]
Step 2 - Add 3 both sides,
[tex]-4\sqrt x-3+3=12+3[/tex]
[tex]-4\sqrt x=15[/tex]
Step 3 - Squaring both sides,
[tex](-4\sqrt x)^2=(15)^2[/tex]
[tex]16x=225[/tex]
Step 4 - Divide both side by 16,
[tex]\frac{16}{16}x=\frac{225}{16}[/tex]
[tex]x=\frac{225}{16}[/tex]
For extraneous solution, Substitute the value of x back in the equation.
[tex] -4\sqrt (\frac{225}{16})-3=12[/tex]
[tex]-4\times\frac{15}{4}-3=12[/tex]
[tex]-15-3=12[/tex]
[tex]-18\neq12[/tex]
The value of x does not satisfy the equation which means the value of x is an extraneous solution.
So, [tex]x=\frac{225}{16}[/tex] is an extraneous solution.
