Find whether the following equation has real | vedclass

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

Question

(C) Given equation is $8 x^{2}+2 x-3=0$.On comparing with $a x^{2}+b x+c=0$,we get $a=8, b=2$ and $c=-3$.Discriminant,$D = b^{2}-4 a c = (2)^{2}-4(8)(

Vedclass · Accepted Answer

$\frac{1}{2}, -\frac{3}{4}$

Vedclass · Answer

(C) Given equation is $8 x^{2}+2 x-3=0$.On comparing with $a x^{2}+b x+c=0$,we get $a=8, b=2$ and $c=-3$.Discriminant,$D = b^{2}-4 a c = (2)^{2}-4(8)(-3) = 4+96 = 100$.Since $D > 0$,the equation has two distinct real roots.Using the quadratic formula,$x = \frac{-b \pm \sqrt{D}}{2 a}$.$x = \frac{-2 \pm \sqrt{100}}{2(8)} = \frac{-2 \pm 10}{16}$.Case $1$: $x = \frac{-2+10}{16} = \frac{8}{16} = \frac{1}{2}$.Case $2$: $x = \frac{-2-10}{16} = \frac{-12}{16} = -\frac{3}{4}$.Thus,the roots are $\frac{1}{2}$ and $-\frac{3}{4}$.

Find whether the following equation has real roots. If real roots exist,find them.
$8 x^{2}+2 x-3=0$

Similar Questions

Obtain the roots of the following quadratic equation by using the quadratic formula: $6x^{2} + x - 2 = 0$.

The sum of two natural numbers is $5$ and the sum of their reciprocals is $\frac{5}{6}$. Find those numbers.

Find the roots of the following quadratic equation by the method of completing the square: $m^{2}-18m+81=0$

Solve the following equation using the method of factorization and write its solution set: $\frac{x+1}{x-3}+\frac{x-3}{x+1}=\frac{5}{2}$

In a right triangle,the length of the hypotenuse is $2\,cm$ more than the length of the base and $1\,cm$ more than twice the length of the altitude. Find the length of the hypotenuse. (in $,cm$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Find whether the following equation has real roots. If real roots exist,find them.$8 x^{2}+2 x-3=0$