How do we find the sum of the roots in a quadratic equation?

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SUMMARY

The sum of the roots of the quadratic equation x² + px + q = 0 is definitively -p. This conclusion is derived from the quadratic formula, where the sum S of the roots is expressed as S = -b/a. By substituting b with p and a with 1, the result simplifies to -p. The discussion emphasizes the importance of understanding the relationship between the coefficients and the roots of quadratic equations.

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mathdad
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Show that the sum of the roots of the equation

x^2 + px + q = 0 is -p.

I need help with the set up.

Is the discriminant involved here?
 
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What do you get when you expand (x - a)(x - b)?
 
greg1313 said:
What do you get when you expand (x - a)(x - b)?

(x - a)(x - b)

x^2 - bx - ax + ab

After factoring by grouping, I found the roots to be x = a and x = b.

What is next?
 
Here's another approach:

Suppose we have:

$$ax^2+bx+c=0$$

Them by the quadratic formula, we have that the sum $S$ of the roots is given by:

$$S=\frac{-b+\sqrt{b^2-4ac}}{2a}+\frac{-b-\sqrt{b^2-4ac}}{2a}=-\frac{b}{a}$$

Use this formula on the given quadratic...what do you find?
 
MarkFL said:
Here's another approach:

Suppose we have:

$$ax^2+bx+c=0$$

Them by the quadratic formula, we have that the sum $S$ of the roots is given by:

$$S=\frac{-b+\sqrt{b^2-4ac}}{2a}+\frac{-b-\sqrt{b^2-4ac}}{2a}=-\frac{b}{a}$$

Use this formula on the given quadratic...what do you find?

Great job!

Ok. You said use -b/a.

Let b = p

Let a = 1

We get -p/1 = -p.

I got it!
 

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