Find the roots of the following quadratic equ | vedclass

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(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} =

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

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: $x^{2}-4 \sqrt{2} x+6=0$

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