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jek_recluse [69]
3 years ago
12

According to the Fundamental Theorem of Algebra, which polynomial function has exactly 8 roots?

Mathematics
1 answer:
kobusy [5.1K]3 years ago
7 0

Answer:

Therefore, the polynomial function with exactly 8 roots is f(x) = (3x^2 - 4x - 5)(2x^6 - 5).

Step-by-step explanation:

brainliest plzz

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Find the roots of h(t) = (139kt)^2 − 69t + 80
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Answer:

The positive value of k will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80, roots are those values of t so that h(t) = 0. That is:

19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0 (1)

Roots are determined analytically by the Quadratic Formula:

t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}

t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }

The smaller root is t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }, and the larger root is t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }.

h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80 has one real root when \frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0. Then, we solve the discriminant for k:

\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}

k \approx \pm 0.028

The positive value of k will result in exactly one real root is approximately 0.028.

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