The3Mathketeers

There seem to be a lot of questions about families of quadratics and their respective number of roots!

I will write the given quadratic equation in standard form.

\(3x^2 + 7x + c = 15x - 10 \\ 3x^2 - 8x + (c + 10) = 0\)

In order for this family of quadratics to have 2 real roots, the discriminant of the quadratic must be greater than zero.

\(a = 3; b = -8; c = c + 10 \\ D = b^2 - 4ac \\ D = (-8)^2 - 4 * 3 * (c + 10) \\ D = 64 - 12(c + 10) \\ D = -12c -56\)

Now that I have simplified the discriminant as much as possible, I will determine what values of c make the discriminant greater than 0.

\(D > 0 \\ -12c - 56 > 0 \\ -12c > 56 \\ c < -\frac{14}{3}\)

There are no positive integers c such that the given quadratic has two real roots! Therefore, I do not think it makes sense to take the product!

I suppose I will ignore the fact that the given diagram does not match the description given of this arbitrary rational function. The written words state that x = - 2 and y = 2 act as the only asymptotes, yet the diagram shows that x = 2 and y = 1 are the asymptotes. 

Despite this, I noticed that there is a hole at approximately (3, 2) of the given diagram. A hole indicates that the numerator and denominator share a common factor at x = 3. However, the question is asking for the value of \(\frac{p(3)}{q(3)}\), which is the x-value of the hole. The denominator equals zero in this case, so \(\frac{p(3)}{q(3)}\) is undefined.

Unfortunately, I am not aware of a more efficient full-proof primality test that does not involve testing all the prime factors not exceeding the square root of the number to test. However, there exists probabilistic primality tests that are computationally much faster to perform. This essentially means that the outcome of these tests have a high probability of giving the correct answer, but they can be wrong. I have not read too much about these probabilistic primality tests, but a lot of them are likely above my mathematical chops.

If you are interested in generating many prime numbers at once, then you can explore the Sieve of Erathosenes. I have used this before on a previous college computer science project, and I think the algorithm is understandable to most laypeople.

First, I will label one point in the diagram to ease the solving process. Let S be the point where \(\overline{\text{AB}}\) intersects the circle with center O. Because \(\overline{\text{AB}}\) intersects at one point, that makes the segment a tangent segment to the circle with center O by definition.

Now, consider \(\overline{\text{BR}} \text{ and } \overline{\text{BS}}\). Both of these segments form a tangent segment to the circle with center O. We can use a variety of methods to prove that \(\triangle \text{OBR} \cong \triangle \text{OBS}\) because of Hypotenuse-Leg Congruency Theorem. Because of the presence of congruent triangles, we can conclude that \(m \angle \text{ROB} = m\angle \text{BOS}\). Similarly, we can conclude that \(m \angle \text{SOA} = m\angle \text{OAT}\).

By the Angle Addition Postulate, we know that \(m \angle \text{BOS} + m \angle \text{SOA} = 40^{\circ}\)We can use this information to find the measure of the central angle \(m\angle \text{ROT}\).

\(m \angle \text{ROT} = m \angle \text{ROB} + m \angle \text{BOS} + m \angle \text{SOA} + m \angle \text{AOT} \\ m \angle \text{ROT} = m \angle \text{BOS} + m \angle \text{BOS} + m\angle \text{SOA} + m \angle \text{SOA} \\ m \angle \text{ROT} = (m \angle \text{BOS} + m \angle \text{SOA}) + (m \angle \text{BOS} + m\angle \text{SOA}) \\ m \angle \text{ROT} = 40^{\circ} + 40^{\circ} \\ m \angle \text{ROT} = 80^{\circ}\)

Now, consider the figure ROTP. This forms a quadrilateral. The sum of the interior angles of any quadrilateral is 360 degrees. We actually know three of the angles already. \(m \angle \text{PRO} = m \angle \text{OTP} = 90^{\circ}\) because the corresponding angle between tangent segments and the radius of a circle are always perpendicular. Now, we can find the \(m \angle \text{RPA}\)

\(m \angle \text{RPA} + m \angle \text{PRO} + m \angle \text{ROT} + m \angle \text{OTP} = 360^{\circ} \\ m \angle \text{RPA} + 90^\circ + 80^\circ + 90^\circ = 360^\circ \\ m \angle \text{RPA} = 100^{\circ}\)

Of course, \(m \angle \text{BPA} = m \angle \text{RPA} = 100^{\circ}\).

I apologize, but I have to amend my answer because I made an oversight that I cannot leave uncorrected.To minimize \((t - p)(t - q)(t - r)(t - s)\) where \(p, q, r, \text{ and } s\) are all distinct integers, I previously suggested \(1 * 2 * 3 * 4\) as a product of four integers that minimizes the product but ensuring the result is positive, but I must not forget about the negative integers! With the negative numbers, the minimized product would be \(1 * -1 * 2 * -2\). Notice that since 2 of the integers are negative that the end result will indeed be positive. With this correction, \(g(x) = (x - 1)(x + 1)(x - 2)(x + 2)\), where t is minimized when t = 0, so \(g(t) = g(0) = -1 * 1 * -2 * 2 = 4\), which is significantly less than 24 from my previous answer!

Now, perform the transformation as I described in my previous answer. That would result in \(f(x) = (x - 1)(x + 1)(x - 2)(x + 2) + 1\), so \(f(t) = f(0) = -1 * 1 * -2 * 2 + 1 = 5\).

One of the equations is y = 7 and the other equation is y = 2x² + 8x + 4 - 3x + 3. Since both equations are solved for y, you can set each equation equal to each other and solve for the corresponding x-variable.

\(y = 7; y = 2x^2 + 8x + 4 - 3x + 3\\ 2x^2 + 8x + 4 - 3x + 3 = 7 \\ 2x^2 + 5x + 7 = 7\\ 2x^2 + 5x = 0 \\ x(2x + 5) = 0 \\ x = 0 \text{ or } 2x + 5 = 0 \\ x = 0 \text{ or } x = -\frac{5}{2}\)

As you mentioned, when x = 0, y = 7 as we found. This means that the coordinates of the endpoints are \((0, 7) \text{ and } \left(-\frac{5}{2}, 7\right)\). Do you know how to carry on from here?

This quadratic is in the form of \(y = x^2 - 17x - c\). We can find the solutions of a quadratic equation by utilizing the quadratic formula.

\(x_{1, 2} = \frac{-(-17) \pm \sqrt{(-17)^2 - 4 * -c}}{2*1} \\ x_{1, 2} = \frac{17 \pm \sqrt{289 + 4c}}{2}\)

All solutions are of this particular form. In order for this family of quadratics to have any chance of having integer roots, the radicand \(289 + 4c\) must be a perfect square. We can combine this with the constraint information \(1 < c < 25\) to narrow the options significantly.

\(1 < c < 25 \\ 4 < 4c < 100 \\ 293 < 289 + 4c < 389\)

In other words, any candidate perfect square must lie between 293 and 389. 17² = 289 and the lower bound is 293. This means that the radicand must have a perfect square greater than 17. 20² = 400 and the upper bound is 389. This means that the radicand must have a perfect square less than 20. This restricts the options considerably to either 289 + 4c = 18² or 289 + 4c = 19². We can find the corresponding value for c and determine whether or not this yields integer solutions.

We have now determined that, when c = 18, the radicand it a perfect square! Now, we should check that it generates integer solutions.

\(\text{Let } c = 18: \\ \Delta = 289 + 4c = 19^2 \\ x_{1, 2} = \frac{17 \pm \sqrt{19^2}}{2} \\ x_1 = \frac{17 + 19}{2} \text{ or } x_2 = \frac{17 - 19}{2}\)

At this point, it is clear that the solutions are integers because the numerator is even.

When 1 < c < 25, c = 18 is the only value that generated integer solutions of the given family of quadratics, so the sum of all possible integer values for c is 18.

If, for some reason, you are convinced that the question is worded correctly despite the evidence otherwise, you can still find a value for n that will satisfy this problem. I will pick up where Cphill left off.

\( x = 2 + \sqrt{\frac{13}{10}} \\ 2 + \sqrt{\frac{13}{10}} = \sqrt{n} + 3 \\ \sqrt{n} = -1 + \sqrt{\frac{13}{10}} \\ n = 1 - 2\sqrt{\frac{13}{10}} + \frac{13}{10} \\ n = \frac{23}{10} - 2\sqrt{\frac{13}{10}}\)

However, this is quite an odd answer given the question.