Question 134085
In the first place, you have a problem with the way you stated question number 1.  If two numbers have a sum of 15, they can't have a sum of their square ROOTS be a larger number, and certainly not 137.  Since you gave the answers as 11 and 4, then the question must have read:  '...the sum of their squares is 137.'


Presuming that to be true, here's how to solve it.

First number: {{{x}}}
Second number: {{{y}}}
Square of the first number: {{{x^2}}}
Square of the second number: {{{y^2}}}
Sum of the numbers: {{{x+y}}}
Sum of the squares: {{{x^2+y^2}}}


We are given:

{{{x+y=15}}}


{{{x^2+y^2=137}}}


Solve the first equation for either of the variables:


{{{y=15-x}}}


Substitute this value into the second equation:


{{{x^2+(15-x)^2=137}}}


Expand and simplify:


{{{x^2+225-30x+x^2=137}}}
{{{2x^2-30x+88=0}}}


The lead coefficient is an integer factor of all the coefficients, so divide:


{{{x^2-15x+44=0}}}


Factor the trinomial:


{{{-4 * -11 = 44}}} and {{{-4-11=-15}}}, so:


{{{(x-4)(x-11)=0}}}, so:


{{{x=4}}} or {{{x=11}}}


Check the answer:

{{{4+11=15}}}
{{{16 + 121 = 137}}}


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Perimeter = {{{2L + 2W=20}}}
Area = {{{LW=24}}}


Solve the perimeter equation for either variable:
{{{2L=20-2W}}}
{{{L=10-W}}}


Substitute in the Area equation:
{{{(10-W)W=24}}}
{{{10W-W^2-24=0}}}
{{{-W^2+10W-24=0}}}
{{{W^2-10W+24=0}}}


Factor:
{{{(W-6)(W-4)=0}}}


{{{W=6}}} or {{{W=4}}}


Traditionally, the width is shorter than the length, so the width is 4, so {{{2L+8=20}}}=>{{{2L=12}}}=>{{{L=6}}}


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Do problem #8 the same way I just showed you for problem #7