Give the zeros (if any) of the quadratic function f(x)=3(x-4)(x+11).
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Answer:
10 terms
Step-by-step explanation:
equate the sum formula to 55 and solve for n
n(n + 1) = 55 ( multiply both sides by 2 to clear the fraction )
n(n + 1) = 110 ← distribute parenthesis on left side
n² + n = 110 ( subtract 110 from both sides )
n² + n - 110 = 0 ← in standard form
Consider the factors of the constant term (- 110) which sum to give the coefficient of the n- term (+ 1)
the factors are + 11 and - 10 , since
11 × - 10 = - 110 and 11 - 10 = + 1 , then
(n + 11)(n - 10) = 0 ← in factored form
equate each factor to zero and solve for n
n + 11 = 0 ⇒ n = - 11
n - 10 = 0 ⇒ n = 10
However, n > 0 , then n = 10
number of terms which sum to 55 is 10