[SOLVED] Inequality. If x is real, why does it mean this equation has real roots?

**tleave2000** · April 7th 2009, 12:07 AM

Hiya. There's an example in a book that I can follow the algebra of, but I don't quite understand the reasoning of one of the steps. The underlined part is the one and only part I don't understand. It would be great if someone could explain it some more.

Here's what the book has:
Q: If x is real, find the possible values of the function: $\frac{x^2}{x+1}$
A: Use $y = \frac{x^2}{x+1} \Rightarrow x^2-yx-y=0$
Since x is real, the roots of this equation are real, so $b^2-4ac \ge 0$

Then there's a bit of algebra to get: $y\le-4$ or $y\ge0$

The algebra's fine and I get that if the roots are real, then $b^2-4ac \ge 0$ , but I don't see why x being real means that that equation must have real roots.

If anyone could help me get some more understanding of this, intuitive or otherwise, it'd be greatly appreciated. Thanks.

**Kiwi_Dave** · April 7th 2009, 01:15 AM

By roots I assume they mean "values of y". Since x is real there exist integers m and n such that x=n/m.

$y = \frac{x^2}{x+1} = \frac{\frac{n^2}{m^2}}{\frac n m + 1}=\frac{n^2}{nm+m^2}=\frac a b$ where $a=n^2$ and $b=nm+m^2$ and a,b are both integers.

Therefore, y must be real.

**tleave2000** · April 7th 2009, 02:01 AM

Originally Posted by Kiwi_Dave

Since x is real there exist integers m and n such that x=n/m.

I'm not sure but I think that would only be true if the statement was "Since x is rational...".

But x being real doesn't necessarily mean it can be represented by a fraction. Because x could be irrational at the same time as being real. If someone could confirm or deny that, it'd be appreciated. Thanks for the input either way, it's interesting to see people's approach to the problem. Would love to see more.

**Kiwi_Dave** · April 7th 2009, 02:18 AM

You're absolutely right.

**stapel** · April 7th 2009, 03:43 AM

Originally Posted by tleave2000

Q: If x is real, find the possible values of the function: $\frac{x^2}{x+1}$
A: Use $y = \frac{x^2}{x+1} \Rightarrow x^2-yx-y=0$
Since x is real, the roots of this equation are real, so $b^2-4ac \ge 0$ ....

You have defined x to be a real number. To solve the quadratic equation, you will apply the Quadratic Formula. This will give you an expression for x (in terms of y).

Since x is defined as being real, then you cannot have a negative inside the square root in the Formula. Therefore the "b^2 - 4ac" part has to be non-negative.

**tleave2000** · April 7th 2009, 10:04 AM

Thanks that's good. Although I think it might have been a restatement of the part I had a problem with rather than quite explaining it, it might have helped me to specify my question better:

Why does x being real force the discriminant of $x^2-yx-y=0$ to be positive?
The domain (x) has been real for all quadratic equations that I've dealt with so far, but I've found plenty of ones with no real roots. So why does this one have to have real roots, just because x is real?

**stapel** · April 7th 2009, 10:42 AM

Originally Posted by tleave2000

Why does x being real force the discriminant of $x^2-yx-y=0$ to be positive?

If the discriminant is negative, will the square root evaluate to a real number?

Since x is real, and since $x\,=\, \frac{-b\, \pm\, \sqrt{b^2\, -\, 4ac}}{2a}$ , can you have a value in the square root that is negative?

**Reckoner** · April 7th 2009, 10:44 AM

Originally Posted by tleave2000

Why does x being real force the discriminant of $x^2-yx-y=0$ to be positive?
The domain (x) has been real for all quadratic equations that I've dealt with so far, but I've found plenty of ones with no real roots. So why does this one have to have real roots, just because x is real?

$x$ is a root of the equation and we have defined $x$ to be real. So any roots of the equation must be real. We do not consider values of $y$ that produce non-real roots, for then $x$ would not be real. Consequently, we take $b^2-4ac\geq0.$

**tleave2000** · April 7th 2009, 12:15 PM

Originally Posted by Reckoner

$x$ is a root of the equation and we have defined $x$ to be real.

Ok I think I get it now, thanks : ).

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