Question

If $y^2+6y-3m=0$ and $y^2-3y+m=0$ have a common root, then find the possible values of m.

• A. $0,\frac{-27}{16}$
• B. $0,\frac{-81}{16}$
  
• C. $0,\frac{81}{16}$
• D. $0,\frac{27}{16}$

Answer 1

Answer: (D)

$0,\frac{27}{16}$

Let the common root be c.

$c^2+6c-3m=0$ and $c^2-3c+m=0$

That is, $3m=c^2+6c$ and $m=3c-c^2$

$3m=c^2+6c=3(3c-c^2)$

$c^2+6c=9c-3c^2$

$4c^2-3c=0$

$c=0$ or $\frac{3}{4}$

If $c=0,m=0$

If $c=\frac{3}{4},m=\frac{27}{16}$.

Hence, D is the correct option.