Question 526943
EXTRA HELP:
The original problem and the extra example below start as geometry or word problems, but lead to equations involving rational expressions (quotients of polynomials). They require that you translate the words into an equation, and then solve the equation. However, you have to be careful and verify that the solutions you find are solutions to the original equation, and to the geometry or word problem you started from.
ORIGINAL PROBLEM
Let a and b be the two numbers. (No one said they were integers, or even rational numbers, so I'm just hoping for real numbers).
The sum of the numbers is 25 means that
{{{a+b=25}}}
The sum of their reciprocals is 1/4 means that
{{{1/a+1/b=(a+b)/ab=25/ab=1/4}}}
So
{{{ab=25(4)=100}}}
The numbers will be the solutions to the equation
{{{(x-a)(x-b)=x^2-2(a+b)x+ab=0}}}
Solving
{{{x^2-50x+100=0}}}
we find
{{{x=25+sqrt(525)}}} and {{{x=25-sqrt(525)}}}
Those are the two numbers that are solutions to the problem.
EXTRA EXAMPLE PROBLEM:
A rectangle has side lengths of 3/x+2 and 2/x-3. Its perimeter is 5 units. Find the dimensions of the rectangle. (That means finding x).
You know that the perimeter is calculated by adding the measures of all 4 sides, or adding the length and width and multiplying by 2, so the perimeter is
{{{2*(3/(x+2) + 2/(x-3))}}}
The equation is
{{{2*(3/(x+2) + 2/(x-3))=5}}}--->{{{6/(x+2) + 4/(x-3)=5}}}
To eliminate denominators we could multiply by (x+2)(x-3) to get
{{{6(x-3)+4(x+2)=5(x+2)(x-3)}}}-->{{{6x-18+4x+8=5(x^2-x-6)}}}-->{{{10x-10=5(x^2-x-6)}}}
At this point, I would divide both sides by 5, to get
{{{2x-2=x^2-x-6}}}-->{{{x^2-3x-4=0}}}
That has two solutions: x=4 and x=-1.
Those values work well in {{{2*(3/(x+2) + 2/(x-3))=5}}}
However, with x=-1, one of the lengths is
{{{ 2/(x-3)=2/(-1-3)=2/(-4)=-1/2}}}
and negative lengths are not good solutions, so the only solution that gives reasonable lengths is x=4.
Then the lengths of the sides of the rectangle are 1/2 and 4
{{{3/(x+2)=3/(4+2)=3/6=1/2}}} and
{{{ 2/(x-3)=2/(4-3)=4/1=4}}}