Question

Solve each of the following quadratic equations:
(i) $\frac{x}{x-1}+\frac{x-1}{x}=4\frac{1}{4},x\ne 0,1$
(ii) $\frac{x-1}{2x-1}+\frac{2x-1}{x-1}=2,x\ne \frac{1}{2},1$

Answer 1

(i)$\frac{x}{x-1}+\frac{x-1}{x}=4\frac{1}{4}$

$\frac{x^2+(x-1{)}^2}{\left(x\right)(x-1)}=\frac{17}{4}$

$\frac{x^2+x^2+1-2x}{x^2-x}=\frac{17}{4}$

$17(x^2-x)=4(2x^2-2x+1)$

$17x^2-17x-8x^2+8x-4=0$

$9x^2-9x-4=0$

$9x^2-(12-3)x-4=0$

$9x^2-12x+3x-4=0$

$3x(3x-4)+1(3x-4)=0$

$(3x-4)(3x+1)=0$

$3x-4=0$

$x=\frac{4}{3}$ and $3x+1=0\Rightarrow x=\frac{-1}{3}$

(ii) $\frac{x-1}{2x-1}+\frac{2x-1}{x-1}=2$

$\frac{(x-1{)}^2+(2x-1{)}^2}{(2x-1)(x-1)}=2$

$\frac{x^2+1-2x+4x^2+1-4x}{2x^2-2x-x+1}=2$

$x^2+1-2x+4x^2+1-4x=4x^2-4x-2x+2$

$x^2+2=2$

$x^2=0$

$x=0$