How is discriminant related to real $x$?

Question

Question in my text book

Solve for range of the function, $$y=\frac{x^2+4x-1}{3x^2+12x +20}$$

Text book says, cross multiply and express the obtained equation as a quadratic equation in $x$

So I get $ (3y-1)x^2 + (12y -4)x + (20y-1) = 0$

Now it says find discriminant $D$, so we have,

$$ D = -4 ( 3y-1)( 8y+3)$$

Now it says, set $D≥0$ as $x$ is real. Wait what?

Isn't $x \in \mathbb{R}$ the domain for a quadratic function? Meaning "$x$" is always real? What does a discriminant got anything to do with $x$ being real, when all discriminant tells us is whether or not the ROOTS are real? Help please.

To be more specific about my doubt, here's an edit.

EDIT : I'm confused, setting discriminant $≥0$ would tell whether or not roots are real, meaning whether the graph of the quadratic function cuts/touches X axis. Now tell me what does this got anything to do with range? As far as I know, quadratic fucntions that don't have real roots are also continous throughout the X axis, meaning there SHOULD always be a corresponding $y-value$

Answer 1

The domain of a quadratic polynomial with real coefficients need not be restricted to the reals.

The discriminant of a polynomial is a function of that polynomial's coefficients, and it can be used to deduce some properties of the roots (such as whether they are real). In the case of a quadratic, the roots will be real and distinct for $D>0$, real (but the same) if $D = 0$, but complex conjugates if $D <0$. That is to say, a negative discriminant implies that the roots of the quadratic will both be complex (i.e. there are no real roots).

What does a discriminant got anything to do with x being real, when all discriminant tells us is whether or not the ROOTS are real?

The $x$ in your above equation is the root(s), as you have set your expression equal to 0.

EDIT: I see your edit about the range, and will address it. Essentially, the problem asks for you to find the range of the function given the knowledge that its domain must be real. So, we can use the knowledge that the discriminant must be non-negative to solve for allowable y: $$(12y-4)^2 -4(3y-1)(20y+1) \geq 0$$ $$\implies -\frac{5}{8} \leq y < \frac{1}{3}$$

which is the range of your function.

EDIT 2: OP, note that the quadratic and therefore discriminant you calculated above is incorrect. Your "c" term should be $(20y+1)$, not $(20y-1)$.

Answer 2

The $y$-values making the discriminant non-negative are the $y$-values the function can produce (at least one real solution $x$ exists) . Therefore the set of $y$ making the discriminant non-negative is the range.

Answer 3

$$\frac{x^2+4x-1}{3x^2+12x+20}=y$$

After cross multiplying, we obtain

$$(3y-1)x^2+(12y-4)x+(20y+1)=0$$

If the discriminant is nonnegative, then we can find real $x$ as solution to the quadratic equation, and hence given a particular value of $y$ that makes the discriminant nonnegative, we can find $x$ that satisfy the original equation.

However, if the discriminant is negative, then $x$ that satisfies the quadratic equation is not real.

Answer 4

Here's an example that may clear up your confusion. Suppose we ask whether $1$ is in the range. We set the fraction equal to $1,$ cross multiply, and arrive at the equation $$ 2x^2+8x+19=0,$$ which has roots $$x=\frac{-8\pm \sqrt{64-152}}{4}$$
So, in order for $1$ to be in the range, $x$ must be non-real, and since $x$ is restricted to be real in the problem statement, $1$ is not in the range.

The method suggested in the book, is exactly this, with a general $y$ in place of $1$.

Answer 5

$$\frac{x^2+4x-1}{3x^2+12x +20} = \frac{x^2+4x + \frac{20}{3}}{3x^2+12x +20} - \frac{ \frac{23}{3}}{3x^2+12x +20} = \frac{1}{3}- \frac{ \frac{23}{3}}{3x^2+12x +20}$$ Meanwhile, $$ 3x^2 + 12 x + 20 = 3 (x+2)^2 + 8 $$ is always positive with a minimum value of $8$ when $x=-2,$ so the minimum value of the whole thing is $$ \frac{1}{3}- \frac{ \frac{23}{3}}{8} = \frac{1}{3}-\frac{23}{24}= -\frac{5}{8} $$ It has an upper bound of $\frac{1}{3}$ but never achieves the bound, it just gets very close when $|x|$ is large.