n, n+1 - two consecutive integers
n(n + 1) = 50 <em>use distributive property</em>
n² + n = 50 <em>subtract 50 from both sides</em>
n² + n - 50 = 0
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ax² + bx + c =0
if b² - 4ac > 0 then we have two solutions:
[-b - √(b² - 4ac)]/2a and [-b - √(b² + 4ac)]/2a
if b² - 4ac = 0 then we have one solution -b/2a
if b² - 4ac < 0 then no real solution
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n² + n - 50 = 0
a = 1, b = 1, c = -50
b² - 4ac = 1² - 4(1)(-50) = 1 + 200 = 201 > 0 → two solutions
√(b² - 4ac) = √(201) - it's the irrational number
Answer: There are no two consecutive integers whose product is 50.
![\bf \stackrel{f(x)}{0}=-4cos\left(x-\frac{\pi }{2} \right)\implies 0=cos\left(x-\frac{\pi }{2} \right) \\\\\\ cos^{-1}(0)=cos^{-1}\left[ cos\left(x-\frac{\pi }{2} \right) \right]\implies cos^{-1}(0)=x-\cfrac{\pi }{2} \\\\\\ x-\cfrac{\pi }{2}= \begin{cases} \frac{\pi }{2}\\\\ \frac{3\pi }{2} \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bf%28x%29%7D%7B0%7D%3D-4cos%5Cleft%28x-%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%20%5Cright%29%5Cimplies%200%3Dcos%5Cleft%28x-%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%20%5Cright%29%0A%5C%5C%5C%5C%5C%5C%0Acos%5E%7B-1%7D%280%29%3Dcos%5E%7B-1%7D%5Cleft%5B%20cos%5Cleft%28x-%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%20%5Cright%29%20%5Cright%5D%5Cimplies%20cos%5E%7B-1%7D%280%29%3Dx-%5Ccfrac%7B%5Cpi%20%7D%7B2%7D%0A%5C%5C%5C%5C%5C%5C%0Ax-%5Ccfrac%7B%5Cpi%20%7D%7B2%7D%3D%0A%5Cbegin%7Bcases%7D%0A%5Cfrac%7B%5Cpi%20%7D%7B2%7D%5C%5C%5C%5C%0A%5Cfrac%7B3%5Cpi%20%7D%7B2%7D%0A%5Cend%7Bcases%7D)
