Question

Solve the following equations:
$\frac{3x-2}{2}+\sqrt{x^2-5x+3}=\frac{(x+1{)}^2}{3}$.

Answer 1

$\frac{3x-2}{2}+\sqrt{{2x}^2-5x+3}=\frac{{(x+1)}^2}{3}$

$\sqrt{{2x}^2-5x+3}=\frac{{(x+1)}^2}{3}-\frac{3x-2}{2}$

$\sqrt{{2x}^2-5x+3}=\frac{{2(x+1)}^2-3(3x-2)}{6}$

$6\sqrt{{2x}^2-5x+3}=2(x^2+2x+1)-9x+6$

$6\sqrt{{2x}^2-5x+3}=2x^2+4x+2-9x+6$

$6\sqrt{{2x}^2-5x+3}=2x^2-5x+8$

$6\sqrt{{2x}^2-5x+3}=2x^2-5x+3+5$

$6\sqrt{{2x}^2-5x+3}={\left(\sqrt{{2x}^2-5x+3}\right)}^2+5$

Put $\sqrt{{2x}^2-5x+3}=t$, we get

$6t=t^2+5t^2-6t+5=0$

$t^2-5t-t+5=0$

$t(t-5)-1(t-5)=0$

$(t-1)(t-5)=0$

$t=1,5$

$\therefore \sqrt{{2x}^2-5x+3}=t$

${2x}^2-5x+3=t^2$

${2x}^2-5x+3=1^2$ ...... $[\because t=1]$

${2x}^2-5x+2=\mathrm{0.......}\left(i\right)$

${2x}^2-4x-x+2=0$

$2x(x-2)-1(x-2)=0$

$(2x-1)(x-2)=0$

$x=\frac{1}{2},2$

Also ${2x}^2-5x+3=5^2$

${2x}^2-5x-22=\mathrm{0.......}\left(ii\right)$

Using quadratic formula

$x=\frac{5\pm \sqrt{25-4\left(2\right)(-22)}}{2\left(2\right)}=\frac{5\pm \sqrt{201}}{4}$

So the values of $x$ are $\frac{5\pm \sqrt{201}}{4},\frac{1}{2}$ and $2$