If \(3x^2 + p x + 12 = 0\) has equal roots, find the values…

Explanation

The general form of a quadratic equation is:

\(x^2\)-(sum of roots) = 0

For equal roots it's:\(x^2 - 2(α)+(α)^2=0\)

\(3x^2+px+12=0\)

Divide through by 3

= \(x^2+\frac{p}{3}x+4=0\)

=\(x^2-(-\frac{p}{3})x+4=0\)

\(So,α^2=4\)

= α = √4 = ±2

Also,-\(\frac{p}{3}= 2α\)

When α = 2

-\(\frac{p}{3} = 2(2)=4\)

=p=-12

When α =-2

-\(\frac{p}{3}=2(-2)=-4\)

= p = 12

∴ values of p = ±12