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what are the zeros of the function f(x)=(x+2)^2-25

Sagot :

Panoyin
f(x) = (x + 2)² - 25
f(x) = (x + 2)(x + 2) - 25
f(x) = x(x + 2) + 2(x + 2) - 25
f(x) = x(x) + x(2) + 2(x) + 2(2) - 25
f(x) = x² + 2x + 2x + 4 - 25
f(x) = x² + 4x + 4 - 25
f(x) = x² + 4x - 21
   0 = x² + 4x - 21
   x = -(4) ± √((4)² - 4(1)(-21))
                       2(1)
   x = -4 ± √(16 + 84)
                    2
   x = -4 ± √(100)
                 2
   x = -4 ± 10
              2
   x = -2 ± 5
   x = -2 + 5    U    x = -2 - 5
   x = 3    U    x = -7
The solution set is {3, -7}.
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