Question 832595: Write a quadratic equation having the given numbers as solutions: -9 and -6.
Found 3 solutions by Alan3354, josgarithmetic, math-vortex:
Answer by Alan3354(69443)   (Show Source):
Answer by josgarithmetic(39626)  (Show Source):
Answer by math-vortex(648)  (Show Source):
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Hi, there--
THE PROBLEM:
Write a quadratic equation having the given numbers as solutions: -9 and -6.
A SOLUTION:
A quadratic equation is an equation with an x-squared term. It has at most two real roots. To find an
equation that goes with the roots -9 and -6, we will "work backwards."
If I had a quadratic equation and want to find its roots, I would factor the equation and solve each of the
factors. That's "working frontwards". I can also work backwards from the solutions.
The factors we find by working backwards always have the form (x - a).
Having a factor of (x - a) means the same thing as having a solution of x = a. In other words, if "x – a" is
a factor, then "x = a" is a solution, and vice versa. We use this fact to find quadratics from their roots.
We have a quadratic with two solutions: -9 and -6. This implies that x = -9 OR x = -6
If x = -9, then it came from the factor equation x + 9 = 0
If x = -6, then it came from the factor equation x + 6 = 0
This means that (x + 9) and (x + 6) are the factors of the quadratic
Remember, quadratic can have at most two solutions, so these are the only factors.
Therefore, the original quadratic in factored form was something like:
y = (x + 9)(x + 6)
You can leave this in factored form or multiply it out.
y = x^2 + 15x + 54
By the way, there are many, many other quadratic equations that have these two solutions. For
example, suppose you have the equation, y = 3(x + 9)(x + 6) We still have the same two solutions,
because the factor 3 does not yield a root.
You can read more about this here: http://www.purplemath.com/modules/fromzero.htm
Hope this helps! Feel free to email if you have any questions about the solution.
Good luck with your math,
Mrs. F
math.in.the.vortex@gmail.com
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