Question

Plz answer my question Find the roots by quadratic formula.

Answer 1

Answer:

The 2 roots are $\sqrt{2} \ and \ -3\sqrt{2}$.

Step-by-step explanation:

Given:

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

We need to find the roots of quadratic equation using quadratic formula.

For quadratic expression $ax^2+bx+c=0$

Quadratic formula $x= \frac{-b\±\sqrt{b^2-4ac}}{2a}$

In Given expression the values of a,b,c are;

$a=1, b=2\sqrt{2},c=-6$

We will first find $\sqrt{b^2-4ac}$ by substituting the given values;

$\sqrt{b^2-4ac}$ = $\sqrt{(2\sqrt{2})^2-4\times1\times(-6) } =\sqrt{(4\times2)+24} =\sqrt{8+24}=\sqrt{32}= \sqrt{16\times2}=\sqrt{(4)^2\times2}=4\sqrt{2}$

Now Substituting the above value in quadratic formula we get;

$x=\frac{-b\±\sqrt{b^2-4ac}}{2a}\\ \\x=\frac{-2\sqrt{2} \±4\sqrt{2} }{2\times1}\\\\x= \frac{2(-\sqrt{2} \±2\sqrt{2})}{2}\\\\x=-\sqrt{2} \±2\sqrt{2}$

Now we will find 2 different roots as;

$x_1=-\sqrt{2} +2\sqrt{2}=\sqrt{2}\\x_2=-\sqrt{2} -2\sqrt{2}=-3\sqrt{2}$

Hence the 2 roots are $\sqrt{2} \ and \ -3\sqrt{2}$.