1. ## b^2-4ac>0

Show that the sum of the two solutions of this equaion is -(b/a)

show that the product of the two solutions of this equation is c/a

2. A quick look at the Quadratic Formula gives both answers. Write the two solutions separately and you will see it.

3. $\displaystyle a \ne 0$
$\displaystyle ax^2 + bx + c = (x - r_1 )(x - r_2 ) = x^2 - \left( {r_1 + r_2 } \right) + r_1 r_2$
$\displaystyle \frac{b}{a} = - \left( {r_1 + r_2 } \right),\;\frac{c}{a} = \left( {r_1 r_2 } \right)$

4. could you explain it more please?

5. Let's define $\displaystyle ax^2+bx+c=0,\,\forall a,b,c\in\mathbb R,\,a\ne0$

We all know that the following formula solves the given equation

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

We have two solutions here, so we can set

$\displaystyle x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}\,\wedge\,x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$

Add them, and multiply them, what do you get?

6. multiply?

7. It's the same as what Plato has done, really - but I'll try to explain it a bit more. (I'm new here, and I don't mean to 'trump' anyone else's answer - so if this is unacceptable, please let me know. Thanks. )

If the roots of $\displaystyle ax^2 + bx + c = 0$ are $\displaystyle \alpha$, $\displaystyle \beta$

$\displaystyle x = \alpha$ or $\displaystyle \beta$

Think about what you're doing when you solve a quadratic, and then just do the reverse....

$\displaystyle (x-\alpha)(x-\beta)=0$

$\displaystyle x^2- \alpha x- \beta x+ \alpha\beta=0$

$\displaystyle x^2-x(\alpha+\beta)+\alpha\beta=0$

(A) $\displaystyle ax^2+bx+c=0$
and (B) $\displaystyle x^2-x(\alpha+\beta)+\alpha\beta=0$
must be the same equation.

Divide equation (A) by a, so it becomes $\displaystyle x^2 + \frac{b}{a}x + \frac{c}{a}=0$

Then compare coefficients:

x coeff $\displaystyle \rightarrow \alpha+\beta=-\frac{b}{a}$

constant $\displaystyle \rightarrow \alpha\beta=\frac{c}{a}$

8. Original Problem Statements (Emphasis Added)

"Show that the sum of the two solutions of this equaion is -(b/a)"

"show that the product of the two solutions of this equation is c/a"

"Sum" is addition. "Product" is multiplication.

Original Reply: "A quick look at the Quadratic Formula gives both answers. Write the two solutions separately and you will see it."

Have you done this, yet?

Write them down. Add them. What is the result?
Write them down again. Myltiply them. What is the result?

Note: $\displaystyle (p+q)(p-q) = p^{2} - q^{2}$