# Thread: interesting quadratic equation problem

1. ## interesting quadratic equation problem

hi everyone.I'm just wondering ,if you have the roots of a quadratic equation is it possible to determine it's coefficients,i suspect i can determine b and c while a will have a range of possible values.What do you think?

2. If you have the equation
x^2 + 2x - 3 = 0 You can factor the quadratic to get
(x - 1)(x + 3) = 0 and hence the roots
x = 1 and x = -3
Conversely, if you have the roots x = 1 and x = -3
you can recover the quadratic equation by forming (x - 1)(x + 3) = 0.

The same is true if the roots are complex,

for example suppose that the roots are

5 + 3i and 5 - 3i then the quadratic equation is

(x - 5 - 3i)(x - 5 + 3i) = 0
which simplifies to

x^2 - 10x + 34 = 0

Cheers,
Adarsh

3. ## thanks for replying

thanks for replying.Using the method your using suggests a is one always.I'm trying to find a way of calculating at least one of the values of a that is not one.a is the first coefficient

4. So lets take it together

1)GIVEN: Roots are given

2) ALREADY FOUND : you found single b and c

3) NEXT FINDING: You found a range of "a" for a constant value of roots and other coefficients

If you agree to all the above steps than I dont agree with you
Now give a second thought and
tell me incase still you dont agree with it 5. Originally Posted by dynamo hi everyone.I'm just wondering ,if you have the roots of a quadratic equation is it possible to determine it's coefficients,i suspect i can determine b and c while a will have a range of possible values.What do you think?
You can find a monic polynomial with the given roots $\displaystyle P_1(x)=(x-r_1)(x-r_2)$, then all polynomials with these roots are of the form:

$\displaystyle P(x)=\alpha P_1(x)$

CB

6. ## thanks for replying

thanks for replying

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