Question
Solve the following equations by using the method of completing the square:
$\sqrt2\text{x}^2-3\text{x}-2\sqrt2=0$
$\sqrt2\text{x}^2-3\text{x}-2\sqrt2=0$
✓
Answer
$\sqrt2\text{x}^2-3\text{x}-2\sqrt2=0$
$\Rightarrow2\text{x}^2-3\sqrt2\text{x}-4=0$ $\big($Multiplying both sides by $\sqrt2\big)$
$\Rightarrow\text{2x}^2-3\sqrt2\text{x}=4$
$\Rightarrow\big(\sqrt2\text{x}\big)^2-2\times\sqrt2\text{x}\times\frac{3}2{}+\Big(\frac{3}{2}\Big)^2\\=4+\Big(\frac{3}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{3}{2}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{2x}-\frac{3}{2}\Big)^2=4+\frac{9}{4}$
$=\frac{25}{4}=\Big(\frac{5}{2}\Big)^2$
$\Rightarrow\sqrt2\text{x}-\frac{3}{2}=\pm\frac{5}{2}$ (Taking square root on both sides)
$\Rightarrow\sqrt2\text{x}-\frac{3}{2}=\frac{5}{2}$ or $\sqrt2\text{x}-\frac{3}{2}=-\frac{5}{2}$
$\Rightarrow\sqrt2\text{x}=\frac{5}2{}+\frac{3}{2}=\frac{8}{2}=4$ or $\sqrt2\text{x}=-\frac{5}{2}+\frac{3}{2}=-\frac{2}{2}=-1$
$\Rightarrow\text{x}=\frac{4}{\sqrt2}=2\sqrt2$ or $\text{x}=-\frac{1}{\sqrt2}=-\frac{\sqrt2}{2}$
Hence, $2\sqrt2$ and $-\frac{\sqrt2}{2}$ are the roots of the given equation.
$\Rightarrow2\text{x}^2-3\sqrt2\text{x}-4=0$ $\big($Multiplying both sides by $\sqrt2\big)$
$\Rightarrow\text{2x}^2-3\sqrt2\text{x}=4$
$\Rightarrow\big(\sqrt2\text{x}\big)^2-2\times\sqrt2\text{x}\times\frac{3}2{}+\Big(\frac{3}{2}\Big)^2\\=4+\Big(\frac{3}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{3}{2}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{2x}-\frac{3}{2}\Big)^2=4+\frac{9}{4}$
$=\frac{25}{4}=\Big(\frac{5}{2}\Big)^2$
$\Rightarrow\sqrt2\text{x}-\frac{3}{2}=\pm\frac{5}{2}$ (Taking square root on both sides)
$\Rightarrow\sqrt2\text{x}-\frac{3}{2}=\frac{5}{2}$ or $\sqrt2\text{x}-\frac{3}{2}=-\frac{5}{2}$
$\Rightarrow\sqrt2\text{x}=\frac{5}2{}+\frac{3}{2}=\frac{8}{2}=4$ or $\sqrt2\text{x}=-\frac{5}{2}+\frac{3}{2}=-\frac{2}{2}=-1$
$\Rightarrow\text{x}=\frac{4}{\sqrt2}=2\sqrt2$ or $\text{x}=-\frac{1}{\sqrt2}=-\frac{\sqrt2}{2}$
Hence, $2\sqrt2$ and $-\frac{\sqrt2}{2}$ are the roots of the given equation.
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