Finding the largest $c$ such that $4.7(x-c) = 9-\frac12x^2$ has a real solution $x$

Question

Suppose I'm given the following:

$$4.7(x-c) = 9-\frac12x^2$$

How would I go about determining the largest $c$ such that there exists a solution $x\in\mathbb{R}$?

Answer 1

There are two ways:

• compute the discriminant of the quadratic trinomial and find the largest $c$ that makes it non-negative;

• express $c$ as a function of $x$, and find the global maximum.

There is no largest $c$.

Answer 2

Use the determinant, since a cuadratic equation $ax^2+bx+c=0$ has two exact solutions of the form $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

So, for the solution to be real, $b^2-4ac\geq0$ must be true. Applying this method to your question

$$4.7x-4.7c=9-\frac{1}{2}x^2\\ \frac{1}{2}x^2+4.7x-4.7c-9=0\\ \underbrace{\frac{1}{2}}_a x^2+\underbrace{4.7}_bx\underbrace{-4.7c-9}_c=0$$ So $$4.7^2-4\left(\frac{1}{2}\right)(-4.7c-9)\geq0\\ (-2)(-4.7c-9)\geq-4.7^2\\ 9.14c+18\geq-4.7^2\\ c\geq\frac{-4.7^2-18}{9.14}$$

From here try determining the largest possible value of c.

Answer 3

If we solve for x \begin{equation} 4.7(x-c) = 9-(x^2)/2\implies x = \frac{\pm\sqrt{940 c + 4009} - 47}{10}\\ \implies x\in\mathbb{R} \iff c > -\frac{4009}{940}\approx -4.264893617\qquad f(-4.264893617)=-4.7 \end{equation}

This means that there is no upper limit to the value of c, only a lower limit.

If we solve for c

$$c = \frac{5 x^2 + 47x - 90}{47}\qquad f(-4.7\approx -4.264893617$$

All other valid values of c and x are larger.

Answer 4

The equation $ \ 4.7(x-c) \ = \ 9 - \frac12x^2 \ $ can have its solutions (if they exist) interpreted as the intersections of the "downward opening" parabola $ \ y = 9 - \frac12x^2 \ $ with the line $ \ y \ = \ 4.7x - 4.7c \ \ . $ Since the line has a steep positive slope and the parabola is symmetrical about the $ \ y-$axis, there must be two intersections at least for all such lines with $ \ y-$intercepts less than or equal to $ \ 9 \ \ . $ This means that $ \ -4.7·c \ \le \ 9 \ \Rightarrow \ c \ \ge \ -\frac{9}{4.7} \ \ . $ So this already tells us that there is some minimum value for $ \ c \ \ , $ but no maximum.

Increasing $ \ c \ $ takes the line "downward and to the right" then, since that shifts its $ \ y-$intercept lower, so we wish to find the value of $ \ c \ $ at which the line becomes merely tangent to the parabola. (Beyond that, still smaller values of $ \ c \ $ correspond to lines which do not intersect the parabola at all.) If we rearrange the "intersection equation" and "complete the square", we obtain $$ \frac12x^2 \ + \ 4.7x \ - \ (4.7c \ + \ 9) \ \ = \ \ \frac12 · (x^2 \ + \ 9.4x \ + \ 4.7^2) \ - \ \left(4.7c \ + \ 9 \ + \ \frac{4.7^2}{2} \right) \ \ $$ $$ = \ \ \frac12 · (x \ + \ 4.7)^2 \ - \ 4.7·\left( c \ + \ \frac{9}{4.7} \ + \ \frac{4.7}{2} \right) \ \ = \ \ 0 \ \ . $$ This will only produce a single solution when the sum in parentheses equals zero, hence the line is tangent to the parabola for $$ c \ = \ -\frac{9}{4.7} \ - \ \frac{4.7}{2} \ \ = \ \ -\frac{18 \ + \ 4.7^2}{9.4} \ \ \approx \ \ -4.265 \ \ . $$ [This is of course equivalent to find the value of $ \ c \ $ for which the discriminant of the quadratic equation is zero.] Smaller values of $ \ c \ \ , $ as we've said, will leave no intersection between the curves, so we must have $ \ c \ \ge \ -\frac{18 \ + \ 4.7^2}{9.4} \ \ . $

We also see from this calculation that the point of tangency occurs at $ \ \left(-4.7 \ , \ 9 \ - \ \frac{[-4.7]^2}{2} \approx -2.045 \right) \ \ $ for the line $ \ y \ = \ 4.7x + \frac{18 \ + \ 4.7^2}{2} \ \approx \ 4.7x + 20.05 \ \ . $