Question

Solve:
$(x-3)(x-4)=\frac{34}{{33}^2}$

Answer 1

We have
$(x-3)(x-4)=\frac{34}{{33}^2}$

$\Rightarrow x^2-7x+12-\frac{34}{{33}^2}=0$

$\Rightarrow x^2-7x+\frac{13034}{{33}^2}=0$

$\Rightarrow x^2-7x+\frac{98}{33}\times \frac{133}{33}=0$

$\Rightarrow x^2-\frac{231}{33}x+\frac{98}{33}\times \frac{133}{33}=0$

$\Rightarrow x^2-(\frac{98}{33}+\frac{133}{33})x+\frac{98}{33}\times \frac{133}{33}=0$

$\Rightarrow x^2-\frac{98}{33}x-\frac{133}{33}x+\frac{98}{33}\times \frac{133}{33}=0$

$\Rightarrow (x^2-\frac{98}{33}x)-(\frac{133}{33}x+\frac{98}{33}\times \frac{133}{33})=0$

$\Rightarrow x(x-\frac{98}{33})-\frac{133}{33}(x-\frac{98}{33})=0$

$\Rightarrow (x-\frac{98}{33})(x-\frac{133}{33})=0$

$\Rightarrow x=\frac{98}{33}orx=\frac{133}{33}$