Relation between coefficients and zeros of a quadratic polynomial

[physics kiddy]

Homework Statement

For any quadratic polynomial ax²+bx+c having zeros β and α
Prove that β + α = -b/a and αβ = c/a.

Homework Equations

The Attempt at a Solution

I have found a method myself to prove α+ β = -b/a. However, I could not prove αβ = c/a.

It goes like this.

If α and β are the zeros of the given polynomial.

a(α)²+b(α) + c = 0 ...... (i)

Also,

a(β)²+b(β) + c = 0 ......(ii)

Comparing (i) and (ii)

a(α)²+b(α) + c = a(β)²+b(β) + c

=> aα²-aβ² = bβ - bα
=>a(α²-β²) = -b(α-β)
=>α²-β²/α-β = -b/a
=> α+β = -b/a

Please help me prove αβ = c/a using the same method.

 

[SammyS]

Homework Statement

For any quadratic polynomial ax²+bx+c having zeros β and α
Prove that β + α = -b/a and αβ = c/a.

Homework Equations

The Attempt at a Solution

I have found a method myself to prove α+ β = -b/a. However, I could not prove αβ = c/a.

It goes like this.

If α and β are the zeros of the given polynomial.

a(α)²+b(α) + c = 0 ...... (i)

Also,

a(β)²+b(β) + c = 0 ......(ii)

Comparing (i) and (ii)

a(α)²+b(α) + c = a(β)²+b(β) + c

=> aα²-aβ² = bβ - bα
=>a(α²-β²) = -b(α-β)
=>α²-β²/α-β = -b/a
=> α+β = -b/a

Please help me prove αβ = c/a using the same method.

Use the quadratic formula for the two zeros of the quadratic polynomial.

 

[Hurkyl]

Please help me prove αβ = c/a using the same method.

Do you have reason to think the same method will work?

Use the quadratic formula for the two zeros of the quadratic polynomial.

It would be easier to just write down the factorization of the polynomial, I think.

 

[NascentOxygen]

Given: aα²+bα + c = 0 ...... (i)
it factorizes: aα(α + b/a) + c = 0

Substitute for the bold α.

 

[physics kiddy]

Thank you very very very very much. I can't explain how happy I am to get the answer. Thanks again.

 

[SammyS]

Use the quadratic formula for the two zeros of the quadratic polynomial.

It would be easier to just write down the factorization of the polynomial, I think.

Yes, that's correct supposing that physics kiddy knows:

If α and β are the solutions to [itex]\text{a}x^2+bx+c=0\,,[/itex] then [itex]\text{a}x^2+bx+c=\text{a}(x-\alpha)(x-\beta)\ .[/itex]​

Of course, I agree that is a very handy thing to know!