## anonymous one year ago What is the greatest integer c for which the quadratic polynomial 5x^2 + 11x +c has two distinct rational roots?

1. anonymous

Well, if you consider a quadratic $$ax^{2} + bx + c$$ and look at its discriminant $$b^{2} - 4ac$$, then we can have 3 different cases depending on the value of that discriminant. $$b^{2} - 4ac > 0$$ gives 2 distinct real zeros $$b^{2} - 4ac = 0$$ gives a repeated real root $$b^{2} - 4ac < 0$$ gives conjugate complex roots. So for your question, we want $$b^{2} - 4ac$$ to be greater than 0. But we also want it to be rational, which means we don't want silly square roots or anything. When you look at the quadratic formula, you have that $$\sqrt{b^{2}-4ac}$$. We will only get rational roots if we can make a perfect square. So let's look at the quadratic formula with our values of a = 5, b = 11, and c. $\frac{ -11 \pm \sqrt{121 - 20c} }{ 10 }$ We will get rational values as long as we can make $$121- 20c$$ be a perfect square root. So we are only allowed integer values of c and we need $$121 - 20c >0$$. So the largest integer value that will keep this expression positive is $$c = 6$$. Conveniently, if c = 6, we get $$121-20(6) = 121 - 120 = 1$$. And 1 is a perfect square, so that will give us 2 distinct rational roots. So the value you're looking for is 6. So to recap, the idea is that square root in the quadratic formula controls what kind of roots we have. That discriminant, which is the part under the square root in the quadratic formula, needs to be positve in order to get 2 distinct roots. Additionally, we need rational roots, which means we can't get odd square roots. The way we avoid that is making sure that $$b^{2} - 4ac$$ is a perfect square. With our particular values of a and b, we find that we need $$121-20c$$ to be positive and a perfect square. Since c must be an integer, we determine that the largest value that can meet our conditions is c = 6, which happens to cause that expression to be a perfect square. Hopefully rhat makes somewhat sense, lol.

2. anonymous

Yes, that makes perfect sense. This is fantastic, thank you!

3. anonymous

You're welcome :)