Question 978430
{{{(3x+1) (2x-1)=144}}}
{{{6x^2-3x+2x-1=144}}}
{{{6x^2-x-1=144}}}
{{{6x^2-x-1-144=0}}}
{{{6x^2-x-145=0}}}
We need to solve the quadratic equation above.
Quadratic equations can always be solved
by "completing the square"
and by applying the quadratic formula.
Some quadratic equations can also be solved by factoring,
and {{{6x^2-x-145=0}}} is one of those.
To factor, we look for factors of {{{6*145}}} .
No need to multiply, when we can find and recombine the prime factors:
{{{6*145=(2*3)*(5*29)=(2*3*5)*29=30*29}}} ,
{{{6x^2-x-145=6x^2-30x+29x-145=(6x^2-30x)+(29x-145)=6x(x-5)+29(x-5)=(6x+29)(x-5)}}}
So we can re-write {{{6x^2-x-145=0}}} as
{{{(6x+29)(x-5)=0}}}--->{{{system(6x+29=0,"or",x-5=0)}}} --->{{{system(x=-29/6,"or",x=5)}}} .
{{{x=5}}}--->{{{system(2x-1=2*5-1=10-1=9,3x+1=3*5+1=15+1=16)}}} ,
for the length and width of the rectangle to be 16 and 9cm respectively,
which makes perfect sense, so {{{highlight(x=5)}}} is a good answer.
On the other hand,
{{{x=-29/6}}}--->{{{system(2x-1=2*(-29/6)-1=-29/3-1=-32/3,3x+1=3*(-29/6)+1=-29/2+1=-27/2)}}} ,
yielding negative numbers for the length and width of the rectangle, does not make sense.
So, {{{x=-29/6}}} is a solution for the equation {{{(3x+1) (2x-1)=144}}} ,
but is not a solution to the word problem.