Question
Solve the following equations by using the method of completing the square:
$x^2 - 6x + 3 = 0$
$x^2 - 6x + 3 = 0$
✓
Answer
$x^2 - 6x + 3 = 0$
$\Rightarrow x^2 - 6x = -3$
$\Rightarrow x^2 - 2 \times x \times 3 + 3^2 = -3 + 3^2$ (Adding $3^2$ on both sides)
$\Rightarrow (x - 3)^2= -3 + 9 = 6$
$\Rightarrow\text{x}-3=\pm\sqrt6$ (Taking square root on both sides)
$\Rightarrow\text{x}-3=\sqrt6$ or $\text{x}-3=-\sqrt6$
$\Rightarrow\text{x}=3+\sqrt6$ or $\text{x}=3-\sqrt6$
Hence, $3+\sqrt6$ and $3-\sqrt6$ are the roots of the given equation.
$\Rightarrow x^2 - 6x = -3$
$\Rightarrow x^2 - 2 \times x \times 3 + 3^2 = -3 + 3^2$ (Adding $3^2$ on both sides)
$\Rightarrow (x - 3)^2= -3 + 9 = 6$
$\Rightarrow\text{x}-3=\pm\sqrt6$ (Taking square root on both sides)
$\Rightarrow\text{x}-3=\sqrt6$ or $\text{x}-3=-\sqrt6$
$\Rightarrow\text{x}=3+\sqrt6$ or $\text{x}=3-\sqrt6$
Hence, $3+\sqrt6$ and $3-\sqrt6$ are the roots of the given equation.
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