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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 $k$.

 Dec 22, 2019
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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,

 

\(x = \frac{k - 1/k \pm \sqrt{(k - 1/k)^2 - 4}}{2} = \frac{k - 1/k \pm \sqrt{k^2 - 8 + 1/k^2}}{2}\)
 

 Dec 23, 2019

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