Question
Solve the following equations by using the method of completing the square:
$3x^2 - 2x - 1 = 0$
$3x^2 - 2x - 1 = 0$
✓
Answer
$3x^2 - 2x - 1 = 0$
$\Rightarrow 9x^2 - 6x - 3 = 0$ (Multiplying both sides by 3)
$\Rightarrow 9x^2 - 6x = 3$
$\Rightarrow (3x)^2 - 2 \times 3x \times 1 + 1^2 = 3 + 1^2$ [Adding $1^2$ on both sides]
$\Rightarrow (3x - 1)^2 = 3 + 1 = 4 = (2)^2$
$\Rightarrow\text{3x}-1=\pm2$ (Taking square root on both sides)
⇒ 3x - 1 = 2 or 3x - 1 = -2
⇒ 3x = 3 or 3x = -1
⇒ x = 1 or $\text{x}=-\frac{1}{3}$
Hence, 1 and $-\frac{1}{3}$ are the roots of the given equation.
$\Rightarrow 9x^2 - 6x - 3 = 0$ (Multiplying both sides by 3)
$\Rightarrow 9x^2 - 6x = 3$
$\Rightarrow (3x)^2 - 2 \times 3x \times 1 + 1^2 = 3 + 1^2$ [Adding $1^2$ on both sides]
$\Rightarrow (3x - 1)^2 = 3 + 1 = 4 = (2)^2$
$\Rightarrow\text{3x}-1=\pm2$ (Taking square root on both sides)
⇒ 3x - 1 = 2 or 3x - 1 = -2
⇒ 3x = 3 or 3x = -1
⇒ x = 1 or $\text{x}=-\frac{1}{3}$
Hence, 1 and $-\frac{1}{3}$ are the roots of the given equation.
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