SOLUTION: A number is 1 more than twice another. Their squares differ by 176. What are the numbers?

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Question 752011: A number is 1 more than twice another. Their squares differ by 176. What are the numbers?
Answer by tommyt3rd(5050) About Me  (Show Source):
You can put this solution on YOUR website!
We'll call them x and y

y=2x%2B1
x%5E2-y%5E2=176

then



x%5E2-%282x%2B1%29%5E2=176
x%5E2-%284x%5E2%2B4x%2B1%29=176
-3x%5E2-4x-177=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 3x%5E2%2B4x%2B177+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%284%29%5E2-4%2A3%2A177=-2108.

The discriminant -2108 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -2108 is + or - sqrt%28+2108%29+=+45.912961132996.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B4%2Ax%2B177+%29






so we see that no pair of real numbers wtii work but


%0D%0A%281%2F3%29%2A%28-2+%2B-+i+%2Asqrt%28527%29%29%0D%0A

does work

:)