Find the roots of 6x^{2} - sqrt{2}x - 2 = 0 b | vedclass

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

Question

(D) To find the roots of $6x^{2} - \sqrt{2}x - 2 = 0$,we factorize the quadratic polynomial.We need two numbers whose product is $6 	imes (-2) = -12$

Vedclass · Accepted Answer

$\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{3}$

Vedclass · Answer

(D) To find the roots of $6x^{2} - \sqrt{2}x - 2 = 0$,we factorize the quadratic polynomial.We need two numbers whose product is $6 	imes (-2) = -12$ and whose sum is $-\sqrt{2}$.These numbers are $-3\sqrt{2}$ and $2\sqrt{2}$.$6x^{2} - 3\sqrt{2}x + 2\sqrt{2}x - 2 = 0$$3x(2x - \sqrt{2}) + \sqrt{2}(2x - \sqrt{2}) = 0$$(3x + \sqrt{2})(2x - \sqrt{2}) = 0$Setting each factor to zero:$3x + \sqrt{2} = 0 \implies x = -\frac{\sqrt{2}}{3}$$2x - \sqrt{2} = 0 \implies x = \frac{\sqrt{2}}{2}$Thus,the roots are $-\frac{\sqrt{2}}{3}$ and $\frac{\sqrt{2}}{2}$.

Find the roots of $6x^{2} - \sqrt{2}x - 2 = 0$ by the factorisation of the corresponding quadratic polynomial.

Similar Questions

If the following quadratic equation has two equal and real roots,then find the value of $k$: $x(4 - kx) = 3 - 2x$.

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

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

The product of the digits of a two-digit number is $14$. The number obtained by interchanging the digits is $45$ more than the original number. Find the original number.

Find the discriminant of the following quadratic equation and hence determine the nature of the roots of the equation: $x^{2} = 9$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module