1. ## [SOLVED] Algebra Proof

Consider the statement that may have occurred to early algebraists:
A quadratic equation ax^2 +bx+c = 0 has a solution in the set of real
numbers if and only if its discriminant is non-negative, i.e., b^2-4ac=>0.

Is this statement true or false? If true, give a proof and explain what a proof of a mathematical statement is. If false, give a counterexample and explain what a counterexample
to a statement is.

2. Originally Posted by ronaldo_07
Consider the statement that may have occurred to early algebraists:
A quadratic equation ax^2 +bx+c = 0 has a solution in the set of real
numbers if and only if its discriminant is non-negative, i.e., b^2-4ac=>0.

Is this statement true or false? If true, give a proof and explain what a proof of a mathematical statement is. If false, give a counterexample and explain what a counterexample
to a statement is.

Hint: look at the quadratic formula. note where the discriminant is. what would happen if it were less than zero?

when you come up with the reason we can talk about whether or not a counter-example exists and how to find one if it does

3. Originally Posted by ronaldo_07
Consider the statement that may have occurred to early algebraists: A quadratic equation ax^2 +bx+c = 0 has a solution in the set of real numbers if and only if its discriminant is non-negative, i.e., b^2-4ac=>0.
What do you know about a, b, & c?
Consider the quadratic equation $\displaystyle ix^2 - x + i = 0$.

4. Originally Posted by ronaldo_07
Consider the statement that may have occurred to early algebraists:
A quadratic equation ax^2 +bx+c = 0 has a solution in the set of real
numbers if and only if its discriminant is non-negative, i.e., b^2-4ac=>0.

Is this statement true or false? If true, give a proof and explain what a proof of a mathematical statement is. If false, give a counterexample and explain what a counterexample
to a statement is.

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

$\displaystyle = a(x^2+\frac{b}{a}x)+c = 0$

$\displaystyle = a(x^2+\frac{b}{a}x + (\frac{b}{2a})^2-(\frac{b}{2a})^2)+c = 0$

$\displaystyle = a((x+\frac{b}{2a})^2-(\frac{b}{2a})^2)+c = 0$

$\displaystyle = a(x+\frac{b}{2a})^2-a(\frac{b}{2a})^2+c = 0$

$\displaystyle = a(x+\frac{b}{2a})^2-(\frac{b^2}{4a})+c = 0$

$\displaystyle = a(x+\frac{b}{2a})^2 =(\frac{b^2}{4a})^2-c$

$\displaystyle = (x+\frac{b}{2a})^2=(\frac{b^2}{4a^2})-\frac{c}{a}$

$\displaystyle = (x+\frac{b}{2a})^2 =(\frac{b^2}{4a^2})-\frac{4ac}{4a^2}$

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

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

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

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

And you can see from this that there are only real solutions if $\displaystyle b^2-4ac \geq 0$.

5. Originally Posted by Mush
And you can see from this that there are only real solutions if $\displaystyle b^2-4ac \geq 0$.
Mush, have you considered my example?
$\displaystyle ix^2 - x + i = 0$

6. Originally Posted by Plato
Mush, have you considered my example?
$\displaystyle ix^2 - x + i = 0$
Ah. I didn't realise there wasn't a restriction for a, b and c to be real.

7. Thanks for the help I just used an example of a quadratic equation and showed that the negative can not be square rooted. is that enough for the proof?

8. ronaldo, a polynomial with two real roots implies that the discriminant is non-negative. However, a non-negative discriminant does not imply that the solutions are real. The proposition is then false. Plato gave the couterexample that proves this by contradiction:

Originally Posted by Plato
Mush, have you considered my example?
$\displaystyle ix^2 - x + i = 0$
$\displaystyle b^2-4ac = 1 - 4i^2 = 1 + 4 = 5$ but the solutions are not real.

9. what are these examples please explain

10. Originally Posted by ronaldo_07
what are these examples please explain
A simple counter-example has been given several times in this thread.