Find the roots of the following quadratic equ | vedclass

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

Question

(B) Given equation: $x^{2}-4x-8=0$Step $1$: Shift the constant term to the right side: $x^{2}-4x=8$Step $2$: Add the square of half the coefficient of

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given equation: $x^{2}-4x-8=0$Step $1$: Shift the constant term to the right side: $x^{2}-4x=8$Step $2$: Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $-4$,so half of it is $-2$,and its square is $(-2)^{2}=4$.$x^{2}-4x+4=8+4$Step $3$: Write the left side as a perfect square: $(x-2)^{2}=12$Step $4$: Take the square root on both sides: $x-2 = \pm\sqrt{12}$$x-2 = \pm 2\sqrt{3}$Step $5$: Solve for $x$: $x = 2 \pm 2\sqrt{3}$Thus,the roots are $2+2\sqrt{3}$ and $2-2\sqrt{3}$.

Find the roots of the following quadratic equation by the method of completing the square: $x^{2}-4x-8=0$

Similar Questions

Solve the following equation using the quadratic formula,if the equation has a solution in $R$: $\sqrt{2} x^{2} + 7x + 5\sqrt{2} = 0$

Examine whether the following equation is quadratic or not: $16x^2 - 3 = (2x + 5)(5x - 3)$

The quadratic equation of variable $x$ having roots $-8$ and $8$ is $\ldots \ldots \ldots \ldots .$

Examine whether the following equation is quadratic or not: $\sqrt{2} x^{2}-5 \sqrt{3} x+6=0$

Find the discriminant of the following quadratic equation and hence determine the nature of its roots: $x^{2}-2 \sqrt{2} x+1=0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module