Answer:
The 2 roots are [tex]\sqrt{2} \ and \ -3\sqrt{2}[/tex].
Step-by-step explanation:
Given:
[tex]x^2+2\sqrt{2}x-6=0[/tex]
We need to find the roots of quadratic equation using quadratic formula.
For quadratic expression [tex]ax^2+bx+c=0[/tex]
Quadratic formula [tex]x= \frac{-b\±\sqrt{b^2-4ac}}{2a}[/tex]
In Given expression the values of a,b,c are;
[tex]a=1, b=2\sqrt{2},c=-6[/tex]
We will first find [tex]\sqrt{b^2-4ac}[/tex] by substituting the given values;
[tex]\sqrt{b^2-4ac}[/tex] = [tex]\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}[/tex]
Now Substituting the above value in quadratic formula we get;
[tex]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}[/tex]
Now we will find 2 different roots as;
[tex]x_1=-\sqrt{2} +2\sqrt{2}=\sqrt{2}\\x_2=-\sqrt{2} -2\sqrt{2}=-3\sqrt{2}[/tex]
Hence the 2 roots are [tex]\sqrt{2} \ and \ -3\sqrt{2}[/tex].
