How can I verify the statement A * B = c/a for quadratic equations?

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Discussion Overview

The discussion centers on verifying the statement A * B = c/a in the context of quadratic equations, specifically focusing on the roots of the equation ax^2 + bx + c = 0. Participants explore various methods to demonstrate this relationship, including algebraic manipulations and coefficient comparisons.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant proposes using the quadratic formula to express the roots A and B and suggests that multiplying these roots should yield c/a.
  • Another participant outlines a method involving the expansion of the quadratic equation and equating coefficients to derive the relationship AB = c/a.
  • Some participants reiterate the method of equating coefficients, seeking clarification on whether the approach aligns with earlier suggestions.
  • Hints are provided regarding algebraic identities, though their relevance to the verification process is questioned.
  • A later reply confirms the calculation of the product of the roots using the quadratic formula, asserting that it leads to the desired result.

Areas of Agreement / Disagreement

Participants express different methods for verifying the statement, and while some approaches are reiterated, there is no consensus on a single method being superior or definitive. The discussion remains exploratory with multiple perspectives presented.

Contextual Notes

Some participants' approaches depend on specific algebraic manipulations and assumptions about the coefficients of the quadratic equation. The discussion does not resolve potential ambiguities in the methods proposed.

mathdad
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Let A and B be roots of the quadratic equation
ax^2 + bx + c = 0. Verify the statement.

A * B = c/a

What are the steps to verify this statement?

I can let A = [-b + sqrt{b^2 - 4ac}]/2a and, of course, let
B = [-b - sqrt{b^2 - 4ac}]/2a. If I multiply the left side, the statement A * B becomes c/a, right?
 
Last edited:
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I think what I would do is write:

$$ax^2+bc+c=k(x-A)(x-B)=kx^2-k(A+B)x+kAB$$

Equating coefficients, we then find:

$$k=a$$

$$kAB=c\implies AB=\frac{c}{a}$$
 
MarkFL said:
I think what I would do is write:

$$ax^2+bc+c=k(x-A)(x-B)=kx^2-k(A+B)x+kAB$$

Equating coefficients, we then find:

$$k=a$$

$$kAB=c\implies AB=\frac{c}{a}$$

Ok but can it be done as expressed in my post?
 
Hint:

$$(a-b)(a+b)=a^2-b^2$$
 
greg1313 said:
Hint:

$$(a-b)(a+b)=a^2-b^2$$

Can you be more specific?
 
$$\frac{-b+\sqrt{b^2-4ac}}{2a}\cdot\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{b^2-(\sqrt{b^2-4ac})^2}{4a^2}=\frac{4ac}{4a^2}=\frac ca$$
 
greg1313 said:
$$\frac{-b+\sqrt{b^2-4ac}}{2a}\cdot\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{b^2-(\sqrt{b^2-4ac})^2}{4a^2}=\frac{4ac}{4a^2}=\frac ca$$

This is exactly what I thought should be done.
 

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