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

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Answer 1 · 2017-01-12T07:10:36

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

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 · 2017-01-12T10:18:04

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

#(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#