Question

X square minus 4 root 2 X + 6 is equal to zero

Answer 1

${x}^{2} - 4 \sqrt{2} x + 6 = 0 \\ = > {x}^{2} - 2 \times x \times 2 \sqrt{2} + (2 \sqrt{2} ) {}^{2} + 6 = (2 \sqrt{2} ) {}^{2} \\ = > (x - 2 \sqrt{2} ) {}^{2} + 6 = 8 \\ = > (x - 2 \sqrt{2} ) {}^{2} = 8 - 6 \\ = > x - 2 \sqrt{2} = ( + - ) \sqrt{2} \\ = > x = 2 \sqrt{2} ( + - ) \sqrt{2} \\ = > x = 2 \sqrt{2} + \sqrt{2} \: or \: 2 \sqrt{2} - \sqrt{2} \\ = > x = 3 \sqrt{2} \: or \: \sqrt{2} \: ans$

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Answer 2

The zeros of the given equation are $x=3\sqrt{2},\sqrt{2}$.

Step-by-step explanation:

Quadratic formula : If a quadratic equation is $ax^2+bx+c=0$, then the root of quadratic equations are

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

The given equation is

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

Here, a=1, b=-4√2 and c=6.

$x=\dfrac{-(-4\sqrt{2})\pm \sqrt{(-4\sqrt{2})^2-4(1)(6)}}{2(1)}$

$x=\dfrac{4\sqrt{2}\pm \sqrt{32-24}}{2}$

$x=\dfrac{4\sqrt{2}\pm \sqrt{8}}{2}$

$x=\dfrac{4\sqrt{2}\pm 2\sqrt{2}}{2}$

$x=\dfrac{4\sqrt{2}+2\sqrt{2}}{2},\dfrac{4\sqrt{2}-2\sqrt{2}}{2}$

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

Therefore, the zeros of the given equation are $x=3\sqrt{2},\sqrt{2}$.

#Learn more

If one root of quadratic equations x^2 +6x+k=0 is h + 2√6 .Find h and k .....Give step by step explanation plzz

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