Find the roots of the quadratic equation 2 6 x x .
Step-by-Step Solution
Key Concept: Use the quadratic formula $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ for a quadratic equation $ax^{2}+bx+c=0$.
1. Write the given equation in standard form $ax^{2}+bx+c=0$:\\
$$2x^{2}-6x+2=0$$\
Hence, $a=2$, $b=-6$, $c=2$.\\
2. Compute the discriminant $D=b^{2}-4ac$:\\
$$D = (-6)^{2}-4\times2\times2 = 36-16 = 20.$$\\
3. Since $D>0$, the equation has two distinct real roots.\\
4. Apply the quadratic formula:\\
$$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-6) \pm \sqrt{20}}{2\times2} = \frac{6 \pm \sqrt{20}}{4}.$$\\
5. Simplify $\sqrt{20}=\sqrt{4\times5}=2\sqrt{5}$ and reduce the fraction:\\
$$x = \frac{6 \pm 2\sqrt{5}}{4}=\frac{2(3 \pm \sqrt{5})}{4}=\frac{3 \pm \sqrt{5}}{2}.$$\\
6. Hence the two roots are:\\
$$x_{1}=\frac{3+\sqrt{5}}{2}, \qquad x_{2}=\frac{3-\sqrt{5}}{2}.$$
Correct Answer: $x = \frac{3+\sqrt{5}}{2}\;\text{or}\;x = \frac{3-\sqrt{5}}{2}$