Find the sum of all integral values of $c$ with $c\le 25$ and $c\ge1$ for which the equation $y=x^2-17x-c$ has two integral roots.

bingboy Aug 25, 2023

#1**+1 **

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. 17^{2} = 289 and the lower bound is 293. This means that the radicand must have a perfect square greater than 17. 20^{2} = 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 = 18^{2} or 289 + 4c = 19^{2}. We can find the corresponding value for c and determine whether or not this yields integer solutions.

\(289 + 4c = 18^2 \\ 289 + 4c = 324 \\ 4c = 35 \\ c = \frac{35}{4} \not\in \mathbb{Z}\) | \(289 + 4c = 19^2 \\ 289 + 4c = 361 \\ 4c = 72 \\ c = 18 \in \mathbb{Z}\) |

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.

The3Mathketeers Aug 25, 2023

#1**+1 **

Best Answer

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. 17^{2} = 289 and the lower bound is 293. This means that the radicand must have a perfect square greater than 17. 20^{2} = 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 = 18^{2} or 289 + 4c = 19^{2}. We can find the corresponding value for c and determine whether or not this yields integer solutions.

\(289 + 4c = 18^2 \\ 289 + 4c = 324 \\ 4c = 35 \\ c = \frac{35}{4} \not\in \mathbb{Z}\) | \(289 + 4c = 19^2 \\ 289 + 4c = 361 \\ 4c = 72 \\ c = 18 \in \mathbb{Z}\) |

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.

The3Mathketeers Aug 25, 2023