+24 1. Find the product of all positive integer values of $c$ such that $8x^2+15x+c=0$ has two real roots.
2. The equation $x^2+18x=27$ has two solutions. The positive solution has the form $\sqrt{a}-b$ for positive natural numbers $a$ and $b$. What is $a+b$
3. In the quadratic equation
x^2+(k-1/k)x-1=0. solve for $x$ in terms of $k$.
1. The quadratic has real roots when its discriminat is positive. So, we want 15^2 - 8c >= 0, or c <= 15^2/8 = 28. The answer is 28!.
2. By the quadratic formula, the positive root is 6*sqrt(3) - 9, so a + b = 6^2*3 + 9 = 117.
3. By the quaratic formula,
\(x = \frac{k - 1/k \pm \sqrt{(k - 1/k)^2 - 4}}{2} = \frac{k - 1/k \pm \sqrt{k^2 - 8 + 1/k^2}}{2}\)
