Question

If roots of equation ${6x}^2-7x+k=0$ are equal. find $k$.

• A. $\frac{49}{24}$
• B. $-2$
• C. $-1,-2$
• D. $1,2$

Answer 1

The correct option is A $\frac{49}{24}$

Given that

$6x^2-7x+k=0$

and, $b^2-4ac=0$

now

$\Rightarrow {\left(-7\right)}^2-4\times 6\times k=0$

$\Rightarrow 49-24k=0$

$\Rightarrow -24k=-49$ [cancle the -ve sign both sides]

$\therefore k=\frac{49}{24}$