Solve the following quadratic equation using | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For the quadratic equation $am^{2} + bm + c = 0$,the quadratic formula is $m = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.Here,$a = 48$,$b = -13$,and $

Vedclass · Accepted Answer

$\frac{1}{3}, -\frac{1}{16}$

Vedclass · Answer

(D) For the quadratic equation $am^{2} + bm + c = 0$,the quadratic formula is $m = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.Here,$a = 48$,$b = -13$,and $c = -1$.First,calculate the discriminant $D = b^{2} - 4ac = (-13)^{2} - 4(48)(-1) = 169 + 192 = 361$.Since $D > 0$,real solutions exist.$m = \frac{-(-13) \pm \sqrt{361}}{2(48)} = \frac{13 \pm 19}{96}$.Case $1$: $m = \frac{13 + 19}{96} = \frac{32}{96} = \frac{1}{3}$.Case $2$: $m = \frac{13 - 19}{96} = \frac{-6}{96} = -\frac{1}{16}$.Thus,the solutions are $\frac{1}{3}$ and $-\frac{1}{16}$.

Solve the following quadratic equation using the quadratic formula,if the equation has a solution in $R$: $48m^{2} - 13m - 1 = 0$.

Similar Questions

If the value of the discriminant of the quadratic equation $x^{2}-10x+(2k-1)=0$ is $40$,then $k = \ldots$.

The value of the discriminant of a quadratic equation $kx^{2} - 4x - 4 = 0$ is $64$,then $k = \dots$

The roots of the quadratic equation $x^{2}-2x-15=0$ are ....... .

Solve the following equation by using the general formula,if the equation has a solution in $R$: $5x^{2} + 12x + 10 = 0$.

Find the discriminant of the following quadratic equation and hence determine the nature of the roots of the equation: $5x^{2} - 4\sqrt{5}x + 4 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module