Solution: The value of (?) is 16 and the other value of x is 0.
Explanation:
Let the value of (?) is C.
It the given equation is [tex](6x+4)^2=C[/tex]. It can be written as [tex](6x+4)^2-C=0[/tex]
Since [tex]x=\frac{-4}{3}[/tex] is the solution of the given equation, therefore [tex]x=\frac{-4}{3}[/tex] will satisfies the given equation.
Put [tex]x=\frac{-4}{3}[/tex]
[tex](6(\frac{-4}{3}) +4)^2-C=0\\(-8+4)^2-C=0\\{-4}^2-C=0\\C=16[/tex].
The given equation become [tex](6x+4)^2+16=0[/tex].
[tex](6x+4)^2=16[/tex]
[tex]6x+4=4\\6x=0\\x=0[/tex]
[tex]6x+4=-4\\6x=-8\\x=\frac{-4}{3}[/tex]
Therefore the value of (?) is 16 and the other value of x is 0.
