Question

find the roots of quadratic equation x^2- 4 √2x + 6 =0

Answer 1

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Answer 2

$3\sqrt{2}$ and $\sqrt{2}$ are the roots of the equation $x^2-4\sqrt{2}x+6=0$.

Given:

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

To Find:

The roots of the equation.

Solution:

The roots of the quadratic equation $ax^2+bx+c=0$ are given by,

$x_1=\frac{-b+\sqrt{b^2-4ac} }{2a}$ and $x_2=\frac{-b-\sqrt{b^2-4ac} }{2a}$.

In the equation, $x^2-4\sqrt{2}x+6=0$,  $a=1,\, b=-4\sqrt{2},\, c=6$.

Substitute $a=1,\, b=-4\sqrt{2},\, c=6$ into $x_1=\frac{-b+\sqrt{b^2-4ac} }{2a}$ and find the first root.

$x_1=\frac{-(-4\sqrt{2})+\sqrt{(-4\sqrt{2})^2-4\cdot 1\cdot 6 } }{2\cdot 1} \\=\frac{4\sqrt{2}+\sqrt{32-24 } }{2} \\=\frac{4\sqrt{2}+\sqrt{8} }{2} \\=\frac{4\sqrt{2}+2\sqrt{2} }{2} \\=2\sqrt{2}+\sqrt{2}\\ =3\sqrt{2}$

Substitute $a=1,\, b=-4\sqrt{2},\, c=6$ into $x_2=\frac{-b-\sqrt{b^2-4ac} }{2a}$  and find the second root.

$x_2=\frac{-(-4\sqrt{2})-\sqrt{(-4\sqrt{2})^2-4\cdot 1\cdot 6 } }{2\cdot 1} \\=\frac{4\sqrt{2}-\sqrt{32-24 } }{2} \\=\frac{4\sqrt{2}-\sqrt{8} }{2} \\=\frac{4\sqrt{2}-2\sqrt{2} }{2} \\=2\sqrt{2}-\sqrt{2}\\ =\sqrt{2}$

If the obtained roots are substituted into the given equation one by one then they will satisfy the equation which means the left-hand side will be equal to the right-hand side.

Thus, the roots of the equation $x^2-4\sqrt{2}x+6=0$ are $3\sqrt{2}$ and $\sqrt{2}$.