Question
Solve using the quadratic formula $x^2 – 4x + 1 = 0$
✓
Answer
$x^2-4 x+1=0$
$a=1, b=-4, c=1$
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(-4) \pm \sqrt{(-4)^2-4 \times 1 \times 1}}{2 \times 1} $
$=\frac{4 \pm \sqrt{16-4}}{2} $
$ =\frac{4+\sqrt{12}}{2}$
Taking $(+)$
$=\frac{4+2 \sqrt{3}}{2} $
$=2+\sqrt{3}$
$\therefore x = 2 + 1·732$
$= 3·732$
Taking $(-)$
or $x = \frac{4-\sqrt{12}}{2}$
$=\frac{4-2 \sqrt{3}}{2}$
$=2-\sqrt{3}$
$\therefore x = 2 - 1.732$
$= 0.268$
Hence, $x=2+\sqrt{3}$ and $2-\sqrt{3}$
or $3.732$ and $0.268$
$a=1, b=-4, c=1$
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(-4) \pm \sqrt{(-4)^2-4 \times 1 \times 1}}{2 \times 1} $
$=\frac{4 \pm \sqrt{16-4}}{2} $
$ =\frac{4+\sqrt{12}}{2}$
Taking $(+)$
$=\frac{4+2 \sqrt{3}}{2} $
$=2+\sqrt{3}$
$\therefore x = 2 + 1·732$
$= 3·732$
Taking $(-)$
or $x = \frac{4-\sqrt{12}}{2}$
$=\frac{4-2 \sqrt{3}}{2}$
$=2-\sqrt{3}$
$\therefore x = 2 - 1.732$
$= 0.268$
Hence, $x=2+\sqrt{3}$ and $2-\sqrt{3}$
or $3.732$ and $0.268$
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