Question

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

Answer 1

`x=11/2+-13/24` i.e. `4.9583` or `6.0417`

Explanation:

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

`w(w+1)=a^2` or `w^2+w-a^2=0`

and hence using quadratic formula `(-b+-sqrt(b^2-4ac))/(2a)`

`w=(-1+-sqrt(1+4xx1xxa^2))/2`

= `-1/2+-sqrt(1+4a^2)/2`

As `a=5/24`,

`sqrt(1+4a^2)=sqrt(1+4xx25/24^2)`

= `sqrt(1+25/12^2)=sqrt((144+25)/12^2)=13/12`

and `w=-1/2+-13/24`

and `x=w+6=11/2+-13/24`

or `x=5.5+-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]}

Answer 2

`x=4 23/24 and 6 1/24`

Explanation:

`(x-5)(x-6)=25/24^2`

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

Now the equation becomes

`(a+1)a=(24+1)/24^2=1/24(1+1/24)`

`=>a^2+a-1/24(1+1/24)=0`

`=>a^2+(1+1/24)a-a/24-1/24(1+1/24)=0`

`=>a(a+1+1/24)-1/24(a+1+1/24)=0`

`=>(a+1+1/24)(a-1/24)=0`

`=>(a+25/24)(a-1/24)=0`

So `a=-25/24 and a=1/24`

Now
when `a=-25/24`

`x=a+6=-25/24+6=4 23/24`

Again

when `a=1/24`

`x=a+6=1/24+6=6 1/24`