If one root of the equation x^2 + px + 12 = 0 is 4, while the equation x^2 + px + q = 0 ha…

Question

If one root of the equation ${x^2} + px + 12 = 0$  is 4, while the equation ${x^2} + px + q = 0$ has equal roots,

then the value of $'q'$ is

Accepted Answer

$4$ is a root of ${x^2} + px + 12 = 0$

$$ \Rightarrow 16 + 4p + 12 = 0$$

$$ \Rightarrow p =  - 7$$

Now, the equation ${x^2} + px + q = 0$

has equal roots.

$$\therefore$$ ${p^2} - 4q = 0$ $$ \Rightarrow q = {{{p^2}} \over 4} = {{49} \over 4}$$

Solutions · 1