# How do you solve for x? (x-5)(x-6)=25/24^2

Jan 12, 2017

$x = \frac{11}{2} \pm \frac{13}{24}$ i.e. $4.9583$ or $6.0417$

#### Explanation:

Let $x - 6 = w$ and then $x - 5 = w + 1$. Also let $a = \frac{5}{24}$ and then the equation $\left(x - 5\right) \left(x - 6\right) = \frac{25}{24} ^ 2$ can be written as

$w \left(w + 1\right) = {a}^{2}$ or ${w}^{2} + w - {a}^{2} = 0$

and hence using quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$w = \frac{- 1 \pm \sqrt{1 + 4 \times 1 \times {a}^{2}}}{2}$

= $- \frac{1}{2} \pm \frac{\sqrt{1 + 4 {a}^{2}}}{2}$

As $a = \frac{5}{24}$,

$\sqrt{1 + 4 {a}^{2}} = \sqrt{1 + 4 \times \frac{25}{24} ^ 2}$

= $\sqrt{1 + \frac{25}{12} ^ 2} = \sqrt{\frac{144 + 25}{12} ^ 2} = \frac{13}{12}$

and $w = - \frac{1}{2} \pm \frac{13}{24}$

and $x = w + 6 = \frac{11}{2} \pm \frac{13}{24}$

or $x = 5.5 \pm 0.5417$ i.e. $4.9583$ or $6.0417$
graph{(x-5)(x-6)-25/24^2 [4.859, 6.109, -0.255, 0.37]}

Jan 12, 2017

$x = 4 \frac{23}{24} \mathmr{and} 6 \frac{1}{24}$

#### Explanation:

$\left(x - 5\right) \left(x - 6\right) = \frac{25}{24} ^ 2$

Let (x-6)=a;" "then" "(x-5)=a+1

Now the equation becomes

$\left(a + 1\right) a = \frac{24 + 1}{24} ^ 2 = \frac{1}{24} \left(1 + \frac{1}{24}\right)$

$\implies {a}^{2} + a - \frac{1}{24} \left(1 + \frac{1}{24}\right) = 0$

$\implies {a}^{2} + \left(1 + \frac{1}{24}\right) a - \frac{a}{24} - \frac{1}{24} \left(1 + \frac{1}{24}\right) = 0$

$\implies a \left(a + 1 + \frac{1}{24}\right) - \frac{1}{24} \left(a + 1 + \frac{1}{24}\right) = 0$

$\implies \left(a + 1 + \frac{1}{24}\right) \left(a - \frac{1}{24}\right) = 0$

$\implies \left(a + \frac{25}{24}\right) \left(a - \frac{1}{24}\right) = 0$

So $a = - \frac{25}{24} \mathmr{and} a = \frac{1}{24}$

Now
when $a = - \frac{25}{24}$

$x = a + 6 = - \frac{25}{24} + 6 = 4 \frac{23}{24}$

Again

when $a = \frac{1}{24}$

$x = a + 6 = \frac{1}{24} + 6 = 6 \frac{1}{24}$