# Thread: Why is discriminant taken to be zero when finding the range of y?

1. ## Why is discriminant taken to be zero when finding the range of y?

So, I was given a question which went like this

So, my teacher said that I needed to make a quadratic in x by taking all the terms in one side and then I had to make the discriminant greater than or equal to zero since x is real. But, discriminant is greater than or equal to zero when roots are real and distinct. So, I am unable to relate to that theory of quadratic equation.

2. ## Re: Why is discriminant taken to be zero when finding the range of y?

Originally Posted by abhistud
So, I was given a question which went like this

I'm not sure what you are talking about. Do you mean a fraction in which both numerator and denominator are quadratic in x? Could you give a specific example?

So, my teacher said that I needed to make a quadratic in x by taking all the terms in one side and then I had to make the discriminant greater than or equal to zero since x is real. But, discriminant is greater than or equal to zero when roots are real and distinct. So, I am unable to relate to that theory of quadratic equation.
No, if the discriminant is greater than 0, the roots are real and distinct. When the discriminant is 0 the roots are real but not distinct. There is one real root.

3. ## Re: Why is discriminant taken to be zero when finding the range of y?

Yeah!
Say,
Find the range of
$\frac{x^2+2x-2}{x^2+2x+1}$

4. ## Re: Why is discriminant taken to be zero when finding the range of y?

What I wanted to say was that the roots may be real and distinct or they might be equal.

5. ## Re: Why is discriminant taken to be zero when finding the range of y?

The domain of the function, the values that $\displaystyle x$ can take, are real. This means that the discriminant must be non-negative because a negative discriminant gives complex values of $\displaystyle x$.

6. ## Re: Why is discriminant taken to be zero when finding the range of y?

Originally Posted by abhistud
Yeah!
Say,
Find the range of
$\frac{x^2+2x-2}{x^2+2x+1}$
This is quite different from the question you first asked. The first thing I would do the indicated division to write this as $\displaystyle f(x)= 1- \frac{3}{x^+ 2x+ 1}$. To find the range of that first find the range of the denominator. And that has nothing to do with finding a "discriminant". You should be able to immediately see that $\displaystyle x^2+ 2x+ 1= (x+ 2)^2$. This function is $\displaystyle f(x)= 1- \frac{3}{(x+ 1)^2}$. That fraction takes on all values except 0 (a fraction is never 0 unless the numerator is 0) so f takes on all values except 1.

I wouldn't worry about a discriminant

7. ## Re: Why is discriminant taken to be zero when finding the range of y?

$\displaystyle y=\frac{x^2+2x-2}{x^2+2x+1}$

gives a second degree equation in $\displaystyle x$

$\displaystyle (y-1)x^2+2(y-1)x+(y+2)=0$

$\displaystyle y$ is in the range if and only if this equation has a real solution

therefore the discriminant $\displaystyle 12-12y>0$ so the range is $\displaystyle y<1$

8. ## Re: Why is discriminant taken to be zero when finding the range of y?

Originally Posted by HallsofIvy
This is quite different from the question you first asked. The first thing I would do the indicated division to write this as $\displaystyle f(x)= 1- \frac{3}{x^+ 2x+ 1}$. To find the range of that first find the range of the denominator. And that has nothing to do with finding a "discriminant". You should be able to immediately see that $\displaystyle x^2+ 2x+ 1= (x+ 2)^2$. This function is $\displaystyle f(x)= 1- \frac{3}{(x+ 1)^2}$. That fraction takes on all values except 0 (a fraction is never 0 unless the numerator is 0) so f takes on all values except 1.

I wouldn't worry about a discriminant
HallsOfIvy has had a nightmare with this answet. Most uncharacteristic. He's right about the discriminant being unnecessary, but it is still a perfectly valid way to approach the question. See Idea's post.