Question

b) In the equation (3m-4)x²- (2m + 1) x-3m-1 = 0.1. Determine for which values of m we have two distinct opposing solutions.2. Determine m such that x'²+ x"² = 13

Answer 1

The given equation is:

$(3m-4)x^2-(2m+1)x-3m-1=0$

The equation will have distinct roots if:

$\begin{gathered} (2m+1)^2-4(3m-4)(-3m-1)\ge0_{} \\ 4m^2+4m+1-4(-9m^2+12m-3m+4)\ge0 \\ 4m^2+4m+1+36m^2-36m-16\ge0 \\ 40m^2-32m-15\ge0 \end{gathered}$

Solve for zero to get:

$m=\frac{8+\sqrt[]{214}}{20},\frac{8-\sqrt[]{214}}{20}$