Question
Solve the following equation by factorization $\sqrt{3} x^2+10 x+7 \sqrt{3}=0$
✓
Answer
$
\begin{aligned}
& \sqrt{3} x^2+10 x+7 \sqrt{3}=0 \\
& {[\because \sqrt{3} \times 7 \sqrt{3}=7 \times 3=21]} \\
& \Rightarrow \sqrt{3} x(x+\sqrt{3}+7(x+\sqrt{3}=0 \\
& \Rightarrow(x+\sqrt{3})(\sqrt{3} x+7)=0
\end{aligned}
$
Either $x+\sqrt{3}=0$,
then $x=-\sqrt{3}$
or
$
\sqrt{3} x+7=0
$
then $\sqrt{3} x=-7$
$
\begin{aligned}
& \Rightarrow x =\frac{-7}{\sqrt{3}} \\
& \Rightarrow x =\frac{-7 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\
& =\frac{-7 \sqrt{3}}{3}
\end{aligned}
$
Hence $x =-\sqrt{3},-\frac{7 \sqrt{3}}{3}$.
\begin{aligned}
& \sqrt{3} x^2+10 x+7 \sqrt{3}=0 \\
& {[\because \sqrt{3} \times 7 \sqrt{3}=7 \times 3=21]} \\
& \Rightarrow \sqrt{3} x(x+\sqrt{3}+7(x+\sqrt{3}=0 \\
& \Rightarrow(x+\sqrt{3})(\sqrt{3} x+7)=0
\end{aligned}
$
Either $x+\sqrt{3}=0$,
then $x=-\sqrt{3}$
or
$
\sqrt{3} x+7=0
$
then $\sqrt{3} x=-7$
$
\begin{aligned}
& \Rightarrow x =\frac{-7}{\sqrt{3}} \\
& \Rightarrow x =\frac{-7 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\
& =\frac{-7 \sqrt{3}}{3}
\end{aligned}
$
Hence $x =-\sqrt{3},-\frac{7 \sqrt{3}}{3}$.
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| Class-interval | 0 - 20 | 20 - 40 | 40 - 60 | 60 - 80 |
| Frequency | 10 | 8 | 15 | 7 |