# How do you solve for x ? (x-2)(x-3)=34/33^2

Jun 24, 2016

$\frac{100}{33} \text{ and } \frac{65}{33}$

#### Explanation:

$\left(x - 2\right) \left(x - 3\right) = \frac{34}{33} ^ 2$
$\implies \left(33 x - 66\right) \left(33 x - 99\right) = 34$

being hopeful that $\left(33 x - 66\right) \mathmr{and} \left(33 x - 99\right)$ are integers , at this point, I note that this equation is basically a factorization of 34 such that $a . b = 34$ and $a - b = 33$

Naturally, the factors are 34 and 1 or (-1) and (-34).

There are two options :

Case I : $a = 34 \mathmr{and} b = 1 \setminus \implies x = \frac{100}{33}$

Case II : $a = - 1 \mathmr{and} b = - 34 \setminus \implies x = \frac{65}{33}$

Jun 25, 2016

$x = \frac{65}{33} \mathmr{and} \frac{100}{33}$

#### Explanation:

Let x-3 =a then x-2==a+1

$\left(x - 2\right) \left(x - 3\right) = \frac{33 + 1}{33} ^ 2$

$\implies \left(a + 1\right) a = \frac{33}{33} ^ 2 + \frac{1}{33} ^ 2$

$\implies {a}^{2} + a - \frac{1}{33} ^ 2 - \frac{1}{33} = 0$

$\implies \left(a + \frac{1}{33}\right) \left(a - \frac{1}{33}\right) + 1 \left(a - \frac{1}{33}\right) = 0$

$\implies \left(a - \frac{1}{33}\right) \left(a + \frac{1}{33} + 1\right) = 0$

$\implies \left(a - \frac{1}{33}\right) \left(a + \frac{34}{33}\right) = 0$

$a = \frac{1}{33} \mathmr{and} a = - \frac{34}{33}$

when $a = \frac{1}{33}$

then $x - 3 = \frac{1}{33}$

$x = 3 + \frac{1}{33} = \frac{100}{33}$

when $a = - \frac{34}{33}$

then $x - 3 = - \frac{34}{33}$

$x = 3 - \frac{34}{33} = \frac{99 - 34}{33} = \frac{65}{33}$