JEE MAIN - Mathematics (2004 - No. 36)
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
then the value of $$'q'$$ is
4
12
3
$${{49} \over 4}$$
Explanation
$$4$$ 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.
$$\therefore$$ $${p^2} - 4q = 0$$ $$ \Rightarrow q = {{{p^2}} \over 4} = {{49} \over 4}$$
$$ \Rightarrow 16 + 4p + 12 = 0$$
$$ \Rightarrow p = - 7$$
Now, the equation $${x^2} + px + q = 0$$
has equal roots.
$$\therefore$$ $${p^2} - 4q = 0$$ $$ \Rightarrow q = {{{p^2}} \over 4} = {{49} \over 4}$$
Comments (0)
