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For there to be two real solutions in a quadratic equation the discriminant, $b^2-4ac$, has to be positive, so $b^2-4ac > 0$. Rearranging the equation, I get $b^2 > 4ac$. Then $b > \sqrt{4ac}$. Since $a$ and $c$ are under the square root, this implies that $a$ and $c$ must be positive. Of the right side $\sqrt{4ac}$ is positive then so must be the left side.

Since $a,b,$ and $c$ are all positive, $bc < 0$ cannot hold true because one of them has to be negative.

What did I do wrong?

EDIT:I was asked to work backwards. that is why i started the proof at the end goal

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You made the mistake of taking the "positive square root", and hence your $b$ is positive :) In other words, your mistake is taking $\sqrt(b^2)=|b|$, which you claimed to be $b$. –  voldemort Apr 28 at 2:45
    
but if its -b > sqrt(4ac) that wouldn't be true because the right side is positive –  user137243 Apr 28 at 2:57
    
A couple of things. First, you began by assuming that the quadratic has two real solutions. This is what you are supposed to show, so you shouldn’t start drawing conclusions from that assumption. (Right away, you say that $b^2-4ac>0$, but your reason is that the quadratic has two real solutions, which you have not shown.) Although you shouldn’t be assuming $b^2>4ac$ to begin with, @voldemort is right that you can’t use that to conclude $b > \sqrt{4ac}$. $X>Y$ doesn’t imply that $\sqrt {X}>\sqrt{Y}$. (In particular, it’s an invalid step if $Y<0$.) –  Steve Kass Apr 28 at 3:04

6 Answers 6

$ab\cdot bc\lt0$, so $ac\lt0$

$b^2\gt 0\gt4ac$, we cannot conclude that

$$b > \sqrt{4ac}$$

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As we are interested only in the roots, multiplying the quadratic equation by $-1$, if necessary, we can assume $a>0$. Then given conditions force what the signs for $b$ and $c$ should be, and consequently you can determine the sign of the discriminant.

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Because $ab > 0$ and $bc < 0 $, then $(ab)(bc)<0 \rightarrow ac*b^2<0 \rightarrow ac<0.$

We KNOW that $ac<0$. Now let's look at the quadratic equation. $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Does this give you any ideas?

To elaborate, the equation $ax^2+bx+c=0$ will only have real solutions if the discriminant $\sqrt{b^2-4ac}$ is real. We know that it is real if $b^2-4ac$ is positive, which we know is positive because $-4ac>0$ and $b^2>0$

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What you did wrong was to start with the goal of the proof, and then follow it with implications. If you are trying to prove statement X, then the only informative logic is $\text{something} \rightarrow X$, not $X \rightarrow \text{something}$. An exception is when you try to make a statement of the form $X \rightarrow \text{false}$ in order to prove $\text{not } X$.

For example, suppose I try to prove that "y is divisible by 12", that is $12\mid y$. If something is divisible by 12, then it is also divisible by 4, so we could say $12\mid y \rightarrow 4 \mid y$, but that doesn't help because even if we prove that $4 \mid y$ then we still don't know that $12 \mid y$. On the other hand, if we could prove $24 \mid y$, that would help because $24 \mid y \rightarrow 12 \mid y$.

I suggest approaching your problem another way, consider well known the function:

$$\text{sgn}(x) = \begin{cases} -1 & \text{ for } x < 0 \\ 0 & \text{ for } x = 0 \\ 1 & \text{ for } x > 0 \end{cases}$$

sgn just tells us if x is positive, negative, or zero.

We know a,b, and c aren't zero.

If $ab > 0$, then $\text{sgn}(a) = \text{sgn}(b)$. If $bc < 0$, then $\text{sgn}(b) = -\text{sgn}(c)$. Together we get that $\text{sgn}(a) = -\text{sgn}(c)$. That implies that $ac < 0$, which implies that $b^2 - 4ac > 0$.

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$ab$ is positive, $bc$ is negative. Therefore $ab \cdot bc$ is negative. This is $ab^2c$, and since $b^2$ is always positive $ac$ must be negative. The discriminant is $\sqrt{b^2 - 4ac}$, and $b^2$ and $-4ac$ are both positive so there are two real solutions.

You started off assuming what you wanted to prove, which doesn't really prove anything. Also, as @voldemort said, you did not consider the alternate case if $b$ is negative.

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Suppose not. Then $ax^2+bx+c=0$ has no real solutions. This implies that $b^2-4ac<0\implies $ either $a,c>0$ or $a,c<0$ because $4ac>b^2>0$.

Case $1$: $a,c>0$. Since by the hypothesis we have $ab>0$, and the fact that $a>0$, then we have $b>0.$ Also by the hypothesis we have $bc<0$. Since $c>0\implies b<0,$ we have a contradiction. i.e, $b>0$ and $b<0$.

Case $2$: $a,c<0$. Again by the hypothesis we have $ab>0\implies b<0$ since $a<0$. Also, $bc<0$ by hypothesis, implies that $b>0$. This is a contradiction as well.

Therefore, $ax^2+bx+c=0$ must have two real solutions.

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