Question
Solve the following equations by using the method of completing the square:
$3x^2 - x - 2 = 0$
$3x^2 - x - 2 = 0$
✓
Answer
$3x^2 - x - 2 = 0$
$\Rightarrow 9x^2 - 3x - 6 = 0$ (Multiplying both sides by $3$)
$\Rightarrow 9x^2 - 3x = 6$
$\Rightarrow(\text{3x})^2-2\times\text{3x}\times\frac{1}{2}+\Big(\frac{1}{2}\Big)^2\\=6+\Big(\frac{1}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{1}2{}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{3x}-\frac{1}{2}\Big)^2$
$=6+\frac{1}{4}$
$=\frac{25}{4}=\Big(\frac{5}{2}\Big)^2$
$\Rightarrow\text{3x}-\frac{1}{2}=\pm\frac{5}{2}$ (Taking square root on both sides)
$\Rightarrow\text{3x}-\frac{1}{2}=\frac{5}{2}$ or $\text{3x}-\frac{1}{2}=-\frac{5}{2}$
$\Rightarrow\text{3x}=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3$ or $\text{3x}=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=-2$
$\Rightarrow x = 1$ or $\text{x}=-\frac{2}3{}$
Hence 1 and $-\frac{2}{3}$ are the roots of the given equation.
$\Rightarrow 9x^2 - 3x - 6 = 0$ (Multiplying both sides by $3$)
$\Rightarrow 9x^2 - 3x = 6$
$\Rightarrow(\text{3x})^2-2\times\text{3x}\times\frac{1}{2}+\Big(\frac{1}{2}\Big)^2\\=6+\Big(\frac{1}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{1}2{}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{3x}-\frac{1}{2}\Big)^2$
$=6+\frac{1}{4}$
$=\frac{25}{4}=\Big(\frac{5}{2}\Big)^2$
$\Rightarrow\text{3x}-\frac{1}{2}=\pm\frac{5}{2}$ (Taking square root on both sides)
$\Rightarrow\text{3x}-\frac{1}{2}=\frac{5}{2}$ or $\text{3x}-\frac{1}{2}=-\frac{5}{2}$
$\Rightarrow\text{3x}=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3$ or $\text{3x}=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=-2$
$\Rightarrow x = 1$ or $\text{x}=-\frac{2}3{}$
Hence 1 and $-\frac{2}{3}$ are the roots of the given equation.
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