If m and ( m + 4) are the roots of \(4x^2 - 4x - 15 = 0\), find the equation whose roots are 2 m and (2 m + 8)

A.

\(x^2+8x-15=0\)

B.

\(x^2-2x-15=0\)

C.

\(x^2-8x-15=0\)

D.

\(x^2+2x+15=0\)

Correct answer is B

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

\(4x^2-4x-15=0\)

Divide through by 4

\(=x^2-x-\frac{15}{4}=0\)

\(=x^2-x+(-\frac{15}{4})=0\)

\(=x^2-(1)x+(-\frac{15}{4})=0\)

sum of roots =1

= m + (m + 4) = 1
=2m+4=1

=2m=-3

=m=-\(\frac{3}{2}\)

The equation whose roots are 2m and 2m+8

2m=2×-\(\frac{3}{2}=-3\)and \(2m+8=2×-\frac{3}{2}+8=5\)

\(=x^2-(-3+5)x+(-3)(5)=0\)

\(=x^2-2x+(-15)=0\)

\(∴x^2-2x-15=0\)