For how many values of m the equation x^2- 2x (1 + 3m) + 7 (3 + 2m) =

1.	For how many values of m the equation x^2- 2x (1 + 3m) + 7 (3 + 2m) = 0 will have equal roots?
Answer» Answer: $\sf{Values \ of \ m \ are \ \dfrac{4+9\sqrt3}{9}}$ $\sf{or \ \dfrac{4-9\sqrt3}{9}.}$ Given: The given quadratic EQUATION is x²-2(1+3m)x+7(3+2m)=0 To find: The value of m if equation has EQUAL roots. Solution: $\sf{The \ given \ quadratic \ equation \ is}$ $\sf{\longmapsto{x^{2}-2(1+3m)x+7(3+2m)=0}}$ $\sf{Here, \ a=1, \ b=-2(1+3m) \ and \ c=7(3+2m)}$ $\sf{Roots \ are \ equal.}$ $\sf{Hence,}$ $\sf{b^{2}-4ac=0}$ $\sf{\therefore{[2(1+3m)]^{2}-4\times(1)[7(4+2m)]=0}}$ $\sf{\therefore{(2+6m)^{2}-4(28+14m)=0}}$ $\sf{\therefore{4+24m+36m^{2}-112-56m=0}}$ $\sf{\therefore{36m^{2}-32m-108=0}}$ $\sf{\therefore{4(9m^{2}-8m-27)=0}}$ $\sf{\therefore{9m^{2}-8m-27=\dfrac{0}{4}}}$ $\sf{\therefore{9m^{2}-8m-27=0}}$ $\sf{Here, \ a=9, \ b=-8 \ and \ c=-27}$ $\sf{By \ quadratic \ formula}$ $\sf{m=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}}$ $\sf{\therefore{m=\dfrac{8\pm18\sqrt3}{2(9)}}}$ $\sf{\therefore{m=\dfrac{4\pm9\sqrt3}{9}}}$ $\sf\purple{\tt{\therefore{Values \ of \ m \ are \ \dfrac{4+9\sqrt3}{9}}}}$ $\sf\purple{\tt{or \ \dfrac{4-9\sqrt3}{9}.}}$