What is the possible root combinations of a degree of 6?

Mathematics

2 answers:

klio [65]3 years ago

8 0

Answer:

2sqrt9 would be the answer

Step-by-step explanation:

1. The square root of 9 is 3.

2. When you multiply 3 x 2 you get 6.

Please mark brainliest and have a great day!

grin007 [14]3 years ago

5 0

Answer:

A polynomial can't have more roots than the degree. So, a sixth degree polynomial, has at most 6 distinct real roots. I may be wrong

Step-by-step explanation:

no step

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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 $k$ :