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: $x^{2}-4 \sqrt{2} x+6=0$. To complete the square,we rewrite the equation as $x^{2}-2(x)(2 \sqrt{2}) = -6$. Add $(2 \sqrt{2})^{2} =

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given equation: $x^{2}-4 \sqrt{2} x+6=0$. To complete the square,we rewrite the equation as $x^{2}-2(x)(2 \sqrt{2}) = -6$. Add $(2 \sqrt{2})^{2} = 8$ to both sides: $x^{2}-2(x)(2 \sqrt{2}) + (2 \sqrt{2})^{2} = -6 + 8$. This simplifies to $(x-2 \sqrt{2})^{2} = 2$. Taking the square root on both sides: $x-2 \sqrt{2} = \pm \sqrt{2}$. Case $1$: $x = 2 \sqrt{2} + \sqrt{2} = 3 \sqrt{2}$. Case $2$: $x = 2 \sqrt{2} - \sqrt{2} = \sqrt{2}$. Thus,the roots are $\sqrt{2}$ and $3 \sqrt{2}$.

Find the roots of the following quadratic equation by the method of completing the square: $x^{2}-4 \sqrt{2} x+6=0$

Similar Questions

State whether the quadratic equation $2 x^{2}+x-1=0$ has two distinct real roots. Justify your answer.

Find the discriminant of the following quadratic equation and hence determine the nature of the roots of the equation: $3x^{2} + x - 2 = 0$.

State whether the quadratic equation $(x+1)(x-2)+x=0$ has two distinct real roots. Justify your answer.

Find the roots of the quadratic equation using the quadratic formula:
$2x^{2} - 3x - 5 = 0$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module