Question
Solve the following equation by using formula :
$2x^2 – 6x + 3 = 0$
$2x^2 – 6x + 3 = 0$
✓
Answer
$
\begin{aligned}
& 2 x^2-6 x+3=0 \\
& \text { Here } a=2, b=-6, c=3 \\
& \text { then } D=b^2-4 a c \\
& =(-6)^2-4 \times 2 \times 3 \\
& =36-24 \\
& =12
\end{aligned}
$
Now
$
\begin{aligned}
& x=\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{(-6) \pm \sqrt{12}}{2 \times 2} \\
& =\frac{6 \pm 2 \sqrt{3}}{4} \\
& \therefore x_1=\frac{6+2 \sqrt{3}}{4} \\
& =\frac{2(3+\sqrt{3})}{4} \\
& =\frac{3+\sqrt{3}}{2} \\
& x_2=\frac{6-2 \sqrt{3}}{4} \\
& =\frac{2(3-\sqrt{3})}{4} \\
& =\frac{3-\sqrt{3}}{2}
\end{aligned}
$
Hence $x=\frac{3+\sqrt{3}}{2}, \frac{3-\sqrt{3}}{2}$.
\begin{aligned}
& 2 x^2-6 x+3=0 \\
& \text { Here } a=2, b=-6, c=3 \\
& \text { then } D=b^2-4 a c \\
& =(-6)^2-4 \times 2 \times 3 \\
& =36-24 \\
& =12
\end{aligned}
$
Now
$
\begin{aligned}
& x=\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{(-6) \pm \sqrt{12}}{2 \times 2} \\
& =\frac{6 \pm 2 \sqrt{3}}{4} \\
& \therefore x_1=\frac{6+2 \sqrt{3}}{4} \\
& =\frac{2(3+\sqrt{3})}{4} \\
& =\frac{3+\sqrt{3}}{2} \\
& x_2=\frac{6-2 \sqrt{3}}{4} \\
& =\frac{2(3-\sqrt{3})}{4} \\
& =\frac{3-\sqrt{3}}{2}
\end{aligned}
$
Hence $x=\frac{3+\sqrt{3}}{2}, \frac{3-\sqrt{3}}{2}$.
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