Question
Solve the following equations by using the method of completing the square:
$x^2 - 4x + 1 = 0$
$x^2 - 4x + 1 = 0$
✓
Answer
$x^2 - 4x + 1 = 0$
$\Rightarrow x^2 - 4x = -1$
$\Rightarrow x^2 - 2 \times x \times 2 + 2^2 = -1 + 2^2$ (Adding $2^2$ on both sides)
$\Rightarrow (x - 2)^2= -1 + 4 = 3$
$\Rightarrow\text{x}-2=\pm\sqrt3$ (Taking square root on both sides)
$\Rightarrow\text{x}-2=\sqrt3$ or $\text{x}-2=-\sqrt3$
$\Rightarrow\text{x}=2+\sqrt3$ or $\text{x}=2-\sqrt3$
Hence, $2+\sqrt3$ and $2-\sqrt3$ are the roots of the given equation.
$\Rightarrow x^2 - 4x = -1$
$\Rightarrow x^2 - 2 \times x \times 2 + 2^2 = -1 + 2^2$ (Adding $2^2$ on both sides)
$\Rightarrow (x - 2)^2= -1 + 4 = 3$
$\Rightarrow\text{x}-2=\pm\sqrt3$ (Taking square root on both sides)
$\Rightarrow\text{x}-2=\sqrt3$ or $\text{x}-2=-\sqrt3$
$\Rightarrow\text{x}=2+\sqrt3$ or $\text{x}=2-\sqrt3$
Hence, $2+\sqrt3$ and $2-\sqrt3$ are the roots of the given equation.
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