Find the value of $c$ for which the quadratic equation $(c + 1)x^2 - 6(c + 1)x + 3(c + 9) = 0$, $c
eq -1$ has real and equal roots.

Question

Find the value of $c$ for which the quadratic equation $(c + 1)x^2 - 6(c + 1)x + 3(c + 9) = 0$, $c 
eq -1$ has real and equal roots.

Accepted Answer

Step 1: For real and equal roots, $D = 0$: $[-6(c + 1)]^2 - 4(c + 1) 	imes 3(c + 9) = 0$
Step 2: $\Rightarrow 36(c + 1)^2 - 12(c + 1)(c + 9) = 0 \Rightarrow 12(c + 1)[3(c + 1) - (c + 9)] = 0$
Step 3: $\Rightarrow 12(c + 1)(2c - 6) = 0$. Since $c 
eq -1$, $2c - 6 = 0 \Rightarrow c = 3$

Question

Solution