How do you solve for x? $(x - 5) (x - 6) = \frac{25}{24^{2}}$ $(x-5)(x-6)=25/24^2$

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How do you solve for x? $(x - 5) (x - 6) = \frac{25}{24^{2}}$ $(x-5)(x-6)=25/24^2$

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Answer 1 · 2017-01-12T07:10:36

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

Explanation:

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

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

and hence using quadratic formula $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b+-sqrt(b^2-4ac))/(2a)$

$w = \frac{- 1 \pm \sqrt{1 + 4 \times 1 \times a^{2}}}{2}$ $w=(-1+-sqrt(1+4xx1xxa^2))/2$

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

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

$\sqrt{1 + 4 a^{2}} = \sqrt{1 + 4 \times \frac{25}{24^{2}}}$ $sqrt(1+4a^2)=sqrt(1+4xx25/24^2)$

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

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

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

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

Answer 2 · 2017-01-12T10:18:04

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

Explanation:

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

Let $(x - 6) = a; t h e n (x - 5) = a + 1$ $(x-6)=a;" "then" "(x-5)=a+1$

Now the equation becomes

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

$\Rightarrow a^{2} + a - \frac{1}{24} (1 + \frac{1}{24}) = 0$ $=>a^2+a-1/24(1+1/24)=0$

$\Rightarrow a^{2} + (1 + \frac{1}{24}) a - \frac{a}{24} - \frac{1}{24} (1 + \frac{1}{24}) = 0$ $=>a^2+(1+1/24)a-a/24-1/24(1+1/24)=0$

$\Rightarrow a (a + 1 + \frac{1}{24}) - \frac{1}{24} (a + 1 + \frac{1}{24}) = 0$ $=>a(a+1+1/24)-1/24(a+1+1/24)=0$

$\Rightarrow (a + 1 + \frac{1}{24}) (a - \frac{1}{24}) = 0$ $=>(a+1+1/24)(a-1/24)=0$

$\Rightarrow (a + \frac{25}{24}) (a - \frac{1}{24}) = 0$ $=>(a+25/24)(a-1/24)=0$

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

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

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

Again

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

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

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