Question 788911
THE FIFT GRADER APPROACH:
If the length and the width of the rectangle are each increases by 2 in., in the larger final rectangle the length will still be 4 in. longer than the width.
So we are looking for two numbers (measurements in inches) that multiply to yield 60 (square inches) for the area of the rectangle.
{{{6*10=60}}}, so the larger final rectangle is 6 inches wide by 10 inches long.
Then, the original smaller rectangle (before increasing each side by 2 inches) measured {{{6in-2in=highlight(4in)}}} by {{{10in-2in=highlight(8in)}}}.
 
MAKING IT LOOK MORE DIFFICULT WITH ALGEBRA:
{{{x}}}= width of the original rectangle (in inches)
{{{x+4}}}= length of the original rectangle (in inches)
As we increase each side by 2 inches, we get
{{{x+2}}}= width of the new, larger rectangle (in inches)
{{{x+4+2=x+6}}}= length of the new, larger rectangle (in inches)
{{{(x+2)(x+6)}}}= area of the new, larger rectangle (in square inches)
So, {{{(x+2)(x+6)=60}}} is our equation.
Simplifying:
{{{(x+2)(x+6)=60}}}
{{{x^2+6x+2x+12=60}}}
{{{x^2+8x+12=60}}}
{{{x^2+8x+12-60=0}}}
{{{x^2+8x-48=0}}}
Now we need to solve that quadratic equation and I can see 3 options.
 
Solving by factoring:
{{{x^2+8x-48=0}}}
{{{(x+12)(x-4)=0}}}
The solutions to that equation are
{{{x=-12}}} (from({{{x+12=0}}}) and
and {{{x=4}}} (from({{{x-4=0}}}).
We discardthe solution {{{x=-12}}} because the lwidth of a recatngle (in inches) must be a positive number.
{{{x=highlight(4)}}} is the width of the original rectangle (in inches)
{{{x+4=4+4=highlight(8)}}} is the length of the original rectangle (in inches).
 
Completing the square:
{{{x^2+8x-48=0}}}
{{{x^2+8x=48}}}
{{{x^2+8x+16=48+16}}}
{{{(x+4)^2=64}}}
{{{x+4=highlight(8)}}} is the length of the original rectangle in inches.
So, {{{x=8-4=highlight(4)}}} is the width of the original rectangle in inches.
 
Using the quadratic formula:
The solutions to an equation of the form {{{ax^2+bx+c=0}}} can be calculated as
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In the case of {{{x^2+8x-48=0}}},
{{{a=1}}}, {{{b=8}}}, and {{{c=-48}}}, so
{{{x = (-8 +- sqrt(8^2-4*1*(-48)))/(2*1) = (-8 +- sqrt(64+192))/2 = (-8 +- sqrt(256))/2 = (-8 +- 16)/2 }}}
One solution to that equation is
{{{x = (-8 - 16)/2 =-24/2=-12}}},
but that negative number cannot be the width of the original rectangle in inches.
The other solution is
{{{x = (-8 + 16)/2 =8/2=highlight(4)}}} and that is the width of the original rectangle in inches.
Then, {{{x=4=4+4=highlight(8)}}} is the length of the original rectangle in inches.