A quick look at the Quadratic Formula gives both answers. Write the two solutions separately and you will see it.

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- September 24th 2007, 07:45 AM #1

- September 24th 2007, 08:11 AM #2

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- September 24th 2007, 08:14 AM #3

- September 24th 2007, 04:56 PM #4

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- September 24th 2007, 05:02 PM #6

- September 24th 2007, 05:40 PM #7

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It's the same as what Plato has done, really - but I'll try to explain it a bit more. (I'm new here, and I don't mean to 'trump' anyone else's answer - so if this is unacceptable, please let me know. Thanks. )

If the roots of are ,

or

__Think about what you're doing when you solve a quadratic, and then just do the reverse....__

(A)

and (B)

must be the same equation.

__Divide equation (A) by a__, so it becomes

__Then compare coefficients:__

x coeff

constant

- September 25th 2007, 05:53 AM #8

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Original Problem Statements (Emphasis Added)

"Show that the**sum**of the two solutions of this equaion is -(b/a)"

"show that the**product**of the two solutions of this equation is c/a"

"Sum" is addition. "Product" is multiplication.

Original Reply: "A quick look at the Quadratic Formula gives**both**answers.**Write the two solutions separately**and you will see it."

Have you done this, yet?

Write them down. Add them. What is the result?

Write them down again. Myltiply them. What is the result?

Note: