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