Question
Solve for x using the quadratic formula. Write your answer correct to two significant figures.
$(x – 1)^2 – 3x + 4 = 0$
$(x – 1)^2 – 3x + 4 = 0$
✓
Answer
$(x-1)^2-3 x+4=0 $
$ \Rightarrow x^2-2 x+1-3 x+4=0 $
$ \Rightarrow x^2-5 x+5=0$
Here $a = 1,b = -5$ and $c= 5$
$\therefore x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(-5) \pm \sqrt{-5}^2-4 \times 1 \times 5}{2 \times 1}$
$=\frac{5 \pm \sqrt{25-20}}{2}$
$\therefore x=\frac{5+2.24}{2} \text { or } x=\frac{5-2.24}{2}$
$\Rightarrow x=\frac{7.24}{2} \text { or } x=\frac{2.76}{2}$
$\Rightarrow x=3.6 \text { or } x=1.4$
$ \Rightarrow x^2-2 x+1-3 x+4=0 $
$ \Rightarrow x^2-5 x+5=0$
Here $a = 1,b = -5$ and $c= 5$
$\therefore x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(-5) \pm \sqrt{-5}^2-4 \times 1 \times 5}{2 \times 1}$
$=\frac{5 \pm \sqrt{25-20}}{2}$
$\therefore x=\frac{5+2.24}{2} \text { or } x=\frac{5-2.24}{2}$
$\Rightarrow x=\frac{7.24}{2} \text { or } x=\frac{2.76}{2}$
$\Rightarrow x=3.6 \text { or } x=1.4$
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