SOLUTION: Not sure what is being asked. This is a homework assignment. y=x^2 - 5x + 6 and y=-x^2 + 2x-1 Thanks.

Algebra ->  Radicals -> SOLUTION: Not sure what is being asked. This is a homework assignment. y=x^2 - 5x + 6 and y=-x^2 + 2x-1 Thanks.       Log On


   



Question 171705: Not sure what is being asked. This is a homework assignment.
y=x^2 - 5x + 6
and y=-x^2 + 2x-1
Thanks.

Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
y=x^2 - 5x + 6
and y=-x^2 + 2x-1
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Let's assume the two quations are a system.
Substitute to get:
x^2 -5x + 6 = -x^2 + 2x -1
2x^2 - 7x +7= 0
Use the Quadratic formula to get:
x = [7 +- sqrt(49 -4*2*7)]/4
x = [7 +- sqrt(-7)]/4
x = [7 +- isqrt(7)]/4
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Hmmmmm?
If they are a system of equations they do not have a Real Number solution.
=======================
Cheers,
Stan H.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
It looks like they want you to factor the equations
y+=+x%5E2+-+5x+%2B+6
Look at the constant term, 6
Notice that %28-2%29%2A%28-3%29+=+6 and -2+%2B+%28-3%29+=+-5
y+=+%28x+-+2%29%28x+-+3%29 answer
----------------------
y+=+x%5E2+%2B+2x+-+1
A good method is the "complete the square" method
Set y+=+0 to find the roots
Take 1/2 of the coefficient of x, square it
and add it to both sides
First, though, add 1 to both sides
1+=+x%5E2+%2B+2x
1+%2B+%282%2F2%29%5E2+=+x%5E2+%2B+2x+%2B+%282%2F2%29%5E2
2+=+x%5E2+%2B+2x+%2B+1
2+=+%28x+%2B+1%29%5E2
Take the square root of both sides
sqrt%282%29+=+x+%2B+1
The roots of the equation are:
x+=+-1+%2B+sqrt%282%29 and
x+=+-1+-+sqrt%282%29
The factors of the equation are
%28x+%2B+1+-+sqrt%282%29%29%28x+%2B+1+%2B+sqrt%282%29%29
Note that the factors are just x+-+r%5B1%5D and x+-+r%5B2%5D
where r%5B1%5D and r%5B2%5D are the roots