Question
Solve the following equation by using formula :$2 x^2+\sqrt{5}-5=0$
✓
Answer
$
\begin{aligned}
& 2 x^2+\sqrt{5}-5=0 \\
& \text { Here } a=2, b=\sqrt{5}, c=-5 \\
& D=b^2-4 a c \\
& =(\sqrt{5})^2-4 \times 2 \times(-5) \\
& =5+40 \\
& =45 \\
& \because x=\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{-\sqrt{5} \pm \sqrt{45}}{2 \times 2} \\
& =\frac{-\sqrt{5} \pm \sqrt{9 \times 5}}{4} \\
& =\frac{-\sqrt{5} \pm 3 \sqrt{5}}{4} \\
& \therefore x_1=\frac{-\sqrt{5}+3 \sqrt{5}}{4} \\
& =\frac{2 \sqrt{5}}{4} \\
& =\frac{\sqrt{5}}{2} \\
& x_2=\frac{-\sqrt{5}-3 \sqrt{5}}{4} \\
& =\frac{-4 \sqrt{5}}{4} \\
& =-\sqrt{5}
\end{aligned}
$
Hence $x =\frac{\sqrt{5}}{2},-\sqrt{5}$.
\begin{aligned}
& 2 x^2+\sqrt{5}-5=0 \\
& \text { Here } a=2, b=\sqrt{5}, c=-5 \\
& D=b^2-4 a c \\
& =(\sqrt{5})^2-4 \times 2 \times(-5) \\
& =5+40 \\
& =45 \\
& \because x=\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{-\sqrt{5} \pm \sqrt{45}}{2 \times 2} \\
& =\frac{-\sqrt{5} \pm \sqrt{9 \times 5}}{4} \\
& =\frac{-\sqrt{5} \pm 3 \sqrt{5}}{4} \\
& \therefore x_1=\frac{-\sqrt{5}+3 \sqrt{5}}{4} \\
& =\frac{2 \sqrt{5}}{4} \\
& =\frac{\sqrt{5}}{2} \\
& x_2=\frac{-\sqrt{5}-3 \sqrt{5}}{4} \\
& =\frac{-4 \sqrt{5}}{4} \\
& =-\sqrt{5}
\end{aligned}
$
Hence $x =\frac{\sqrt{5}}{2},-\sqrt{5}$.
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