Find the value of $p$ for which the quadratic equation $p(x - 4)(x - 2) + (x - 1)^2 = 0$ has real and equal roots.

Question

Accepted Answer

Step 1: $p(x - 4)(x - 2) + (x - 1)^2 = 0$
$p(x^2 - 6x + 8) + x^2 - 2x + 1 = 0$
$(p + 1)x^2 - (6p + 2)x + (8p + 1) = 0$
Step 2: $a = p + 1$, $b = 6p + 2$, $c = 8p + 1$
For real and equal roots:
$D = 0 \Rightarrow b^2 - 4ac = 0$
Step 3: $\Rightarrow (6p + 2)^2 - 4(p + 1)(8p + 1) = 0$
Step 4: $36p^2 + 24p + 4 - 4(8p^2 + 9p + 1) = 0$
$4p^2 - 12p = 0 \Rightarrow 4p(p - 3) = 0$
Step 5: $\Rightarrow p = 0, 3$

Question

Solution