Answer:
False.
because in the process of completing the square, \(3x^2+7x-12 = 0\) become \(x^2+\frac{7}{3}x = 4\) but not \(x^2+\frac{7}{4}x = 4\)
Step-by-step explanation:
Given the equation: \(3x^2+7x-12 = 0\)
Divide both sides by 3 we get;
\(x^2+\frac{7}{3}x -4=0\)
Add 4 to both sides we have;
\(x^2+\frac{7}{3}x -4+4=0+4\)
Simplify:
\(x^2+\frac{7}{3}x = 4\)
In the process of completing the square, \(3x^2+7x-12 = 0\) becomes \(x^2+\frac{7}{3}x = 4\) not \(x^2+\frac{7}{4}x = 4\)
Therefore, the statement is FALSE.
