Question

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

• A. $\frac{49}{4}$
• B. $\frac{4}{49}$
• C. 4
• D. $\frac{1}{4}$

Answer 1

The correct option is A $\frac{49}{4}$

Given $x^2+px+12=0$
Since, x = 4 is one root of the equation, therefore x = 4 will satisfy this equation
$\therefore 16+4p+12=0\Rightarrow p=-7$
Other quadratic equation becomes $x^2-7x+q=0$
(By putting value of p)
Its roots are equal, so, $b^2=4ac$
$\Rightarrow 49=4qorq=\frac{49}{4}$