SOLUTION: Find the product of zeros of -2x2 +kx+6 .

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Question 757768: Find the product of zeros of -2x2 +kx+6 .
Found 3 solutions by josgarithmetic, Edwin McCravy, solver91311:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
-2x%5E2%2Bkx%2B6=0

Using r and s for the zeros,


s=%28-k-sqrt%28k%5E2-4%28-2%296%29%29%2F%28-4%29=%28-k-sqrt%28k%5E2%2B48%29%29%2F%28-4%29

Better simplifying, multiplying each by %28-1%29%2F%28-1%29:
r=%28k-sqrt%28k%5E2%2B48%29%29%2F4
s=%28k%2Bsqrt%28k%5E2%2B48%29%29%2F4

The product:

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
-2x² + kx + 6

If r1 and r2 are the zeros of

ax² + bx + c

then sum of zeros = r1+r2 = -b%2Fa
and product of zeros = r1r·sub>2 = c%2Fa

For your problem:

-2x² + kx + 6

a = -2, b = k, c = 6

and product of zeros = c%2Fa = 6%2F%28-2%29 = -3

Edwin

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!




Use the quadratic formula:





Which you should recognize as a conjugate pair the product of which is the difference of two squares:



John

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