Find the Roots of the Quadratic Equation by Completing the Square: √2x² − 3x − 2√2 = 0
Question
Find the roots of the quadratic equation by the method of completing the square:
\[ \sqrt{2}x^2-3x-2\sqrt{2}=0 \]Solution
\[
\sqrt{2}x^2-3x=2\sqrt{2}
\]
Divide both sides by \(\sqrt{2}\):
\[
x^2-\frac{3}{\sqrt{2}}x=2
\]
Add the square of half the coefficient of \(x\) to both sides:
\[
x^2-\frac{3}{\sqrt{2}}x+\left(\frac{3}{2\sqrt{2}}\right)^2
=
2+\left(\frac{3}{2\sqrt{2}}\right)^2
\]
\[
\left(x-\frac{3}{2\sqrt{2}}\right)^2
=
2+\frac{9}{8}
=
\frac{25}{8}
\]
Taking square roots on both sides:
\[
x-\frac{3}{2\sqrt{2}}
=
\pm \sqrt{\frac{25}{8}}
=
\pm \frac{5}{2\sqrt{2}}
\]
\[
x
=
\frac{3}{2\sqrt{2}}
\pm
\frac{5}{2\sqrt{2}}
\]
Therefore,
\[
x=\frac{8}{2\sqrt{2}}
=\frac{4}{\sqrt{2}}
=2\sqrt{2}
\]
or
\[
x=\frac{-2}{2\sqrt{2}}
=-\frac{1}{\sqrt{2}}
=-\frac{\sqrt{2}}{2}
\]
Answer
\[
\boxed{x=2\sqrt{2}\quad \text{or}\quad x=-\frac{\sqrt{2}}{2}}
\]