Easier way to find amount of solutions between a line and quadratic?

Question

Is there a better way of finding the number of solutions of the system: $y=(x-7)(3x+4)$ and $x=3y-1$ that doesn't involve calculus?

I know these are $2$ but that's due to substitution of one equation into the other and finding the discriminant of the resulting quadratic, but it's kind of tedious and this is suppose to be a question that can be done within a minute.

Can I just say that the vertex of the upward-opening parabola is below the $y$ -intercept of the line and this would result in $2$ intersections?

Answer 1

A simple idea. Notice that the point $(7, 8/3)$ lies inside the parabola (easy, since $y(7)=0<8/3$) and it is also a point on the line $x=3y-1$. Hence, a point of the line lies inside the parabola which means that the line will intersect the parabola in two points (unless the line is vertical which is not the case here).

Answer 2

It's true that if the parabola opens upward and any point on the parabola is directly below any point on the line, then the parabola and line intersect in two distinct points. But unless you already have this fact available, it merits justification in its own right, and that's not going to be easier than proving the original fact (one method to prove the latter fact, for example, is using the discriminant).

In any case, it's not too much work to prove the claim in this question using the discriminant: We can rewrite the line as $y = \frac{1}{3} x + \frac{1}{3}$ and the parabola as $y = 3 x^2 - 17 x - 28$. Thus, the $x$-coordinates of the intersection points are the solutions of $$3 x^2 - \frac{52}{3} x - \frac{85}{3} = 0 .$$ Computing the discriminant of the l.h.s. requires some arithmetic, but we only need its sign: In our case, $a > 0, c < 0$, so $b^2 - 4 a c > 0$ and hence there are two distinct solutions.

Alternatively, applying Descartes' Rule of Signs tells us more: That there is exactly one positive solution and one negative solution.

Answer 3

Step 1: Rewrite both equations into the form $y = f(x)$:

$$\begin{aligned} y &= 3x^2 - 17x - 28 \\ y &= \frac x3 + \frac13 \end{aligned}$$

Step 2: Subtract one equation from the other:

$$0 = 3x^2 - 17x - \frac x3 - 28 - \frac13$$

Step 3: Observe that the right-hand side is negative at $x = 0$ (since the constant term is negative) and tends to $+\infty$ as $x$ tends to either $+\infty$ or $-\infty$ (since the coefficient of the $x^2$ term is positive). Thus, there must be at least one positive and one negative solution. And two is the maximum number of solutions a (non-degenerate) quadratic equation can have, so there cannot be more.

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Ps. If the constant term had happened to be positive, we would've had to check whether the right hand side might still be negative somewhere else. A useful trick here is to find the minimum of the RHS from step 2 by taking its derivative with respect to $x$ and setting it to zero:

$$\frac{d}{dx} 3x^2 - 17x - \frac x3 - 28 - \frac13 = 2 \cdot 3x - 17 - \frac13 = 0 \\ \implies x = \frac{17 - \frac13}{6} = \frac{51 - 1}{18} = \frac{25}{9}$$

which you can then plug into the RHS to see if it's positive, zero or negative there (yielding 0, 1 or 2 solutions respectively). But depending on how fast you work and what tools you have, that may take a bit more than a minute.

Of course, if you expect that there are two solutions and just want to confirm it, you just need to find some value of $x$ for which the RHS is negative. So for a quick and dirty check, you could just pick some nice and round value of $x$ close to the minimum like, say, $x = 3 \approx \frac{25}{9}$, and plug that in. If you get a negative value, that's already enough to confirm the existence of two solutions.