SOLUTION: Hi , I was hoping you could help me with two more question. I have tried to find a formula but I can't and I'm wasting too much time. My class is tomorrow evening. 1. Deter

Question 134085: Hi ,

I was hoping you could help me with two more question. I have tried to find a formula but I can't and I'm wasting too much time. My class is tomorrow evening.

1. Determine two numbers whose sum is 15 and the sum of its square roots is 137. I know the answer is 11 and 4 but what formula do I use.

7. The perimeter of a rectangle is 20 inches and its area is 24 inches squared. Determine the lengths of its sides. I know that the sides are 6" and 4" but I need to show my work. I just used logic to find the numbers.

8. The perimeter of a rectangle es 24cm and its area is 32 cm squared. Find the lengths of its sides. Again I know that the answer is 8cm and 4cm but what is the formula?

I really appreciate the help

Thanks,
Anne

Found 2 solutions by solver91311, Ganesha:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
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:
Second number:
Square of the first number:
Square of the second number:
Sum of the numbers:
Sum of the squares:

We are given:

Solve the first equation for either of the variables:

Substitute this value into the second equation:

Expand and simplify:

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

Factor the trinomial:

and , so:

, so:

or

Check the answer:

=================================
Perimeter =
Area =

Solve the perimeter equation for either variable:

Substitute in the Area equation:

Factor:

or

Traditionally, the width is shorter than the length, so the width is 4, so =>=>

===============================
Do problem #8 the same way I just showed you for problem #7
Answer by Ganesha(13) (Show Source): You can put this solution on YOUR website!
1
x+y=15 (1)and x^2+y^2 =137
x+y)^2 = 15^=225. 2xy = 225-137= 88
x-y)^2 =137-88 = 49, Hence
x-y=7 (2). from (1) and (2) 2x=22 x=11 2y=8 y=4
The other 2 are similar please do it similarly.
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