Question

Solve for $x$ using the quadratic formula. Write your answer correct to two significant figures. $(x-1{)}^2-3x+4=0$

Answer 1

Given: $(x-1{)}^2-3x+4=0$
$\Rightarrow x^2+1-2x-3x+4=0$
$\Rightarrow x^2-5x+5=0$
Comparing $x^2-5x+5=0$ with $a{x}^2+bx+c=0$, we get $a=1,b=-5,c=5$.
$\therefore x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$x=\frac{(-5)\pm \sqrt{(-5{)}^2-4(1\left)\right(5)}}{2\times 1}$
$=\frac{5\pm \sqrt{25-20}}{2}=\frac{5\pm \sqrt{5}}{2}$
$=\frac{5\pm 2.236}{2}=\frac{5+2.236}{2}$ and $\frac{5-2.236}{2}$
$=\frac{7.236}{2}$ and $\frac{2.764}{2}$
$=3.62$ and $1.38$