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0 votes
Does the quadratic function

f(x) = 5x² + 5x + 21 have one,
two, or no real zeros? Utilize the
quadratic formula to determine
the answer.

2 Answers

3 votes
5x^2 + 5x + 21 = 0

No real solution

no x - intercept
User RusHughes
by
7.2k points
1 vote

Answer:

No real zeros.

Explanation:


\boxed{\begin{minipage}{7.4 cm}\underline{Discriminant of the Quadratic Formula}\\\\$b^2-4ac$ \quad when $ax^2+bx+c=0$\\\\When $b^2-4ac > 0 \implies$ two real zeros.\\When $b^2-4ac=0 \implies$ one real zero.\\When $b^2-4ac < 0 \implies$ no real zeros.\\\end{minipage}}

The value of the discriminant shows how many zeros the function has.

Given quadratic function:


f(x)=5x^2+5x+21

Therefore:

  • a = 5
  • b = 5
  • c = 21

Substitute the values into the discriminant and solve:


\begin{aligned}\implies b^2-4ac&amp;=(5)^2-4(5)(21)\\&amp; = 25 - 20(21)\\&amp;=25-420\\&amp;=-395\end{aligned}

Therefore, as -395 < 0, there are no real zeros.

User David Vicente
by
7.3k points
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