What are the solutions to the equation 5(3k+2)=k^2(k-7) ?

The solutions to the equation 5(3k+2)=k^2(k-7) are k\approx 8.82755…

Find the zeros of 5(3k+2)=k^2(k-7)

The zeros of 5(3k+2)=k^2(k-7) are k\approx 8.82755…

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5 (3 k + 2) = k^{2} (k - 7)

k \approx 8.82755 \dots

Solution steps

5 (3 k + 2) = k^{2} (k - 7)

Expand

5 (3 k + 2) : 15 k + 10

Expand

k^{2} (k - 7) : k^{3} - 7 k^{2}

15 k + 10 = k^{3} - 7 k^{2}

Switch sides

k^{3} - 7 k^{2} = 15 k + 10

Move

10

to the left side

k^{3} - 7 k^{2} - 10 = 15 k

Move

15 k

to the left side

k^{3} - 7 k^{2} - 10 - 15 k = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

k^{3} - 7 k^{2} - 15 k - 10 = 0

Find one solution for

k^{3} - 7 k^{2} - 15 k - 10 = 0

using Newton-Raphson:

No Solution for

k \in R

k \approx 8.82755 \dots

The solution is

k \approx 8.82755 \dots

27 x^{6} + 26 x^{3} - 1 = 0

- x^{3} - 3 x^{2} + 4 = 0

x^{3} - 4 x^{2} + 2 x + 4 = 0

t^{3} - 64 = 0

z^{4} + 15 z^{2} - 16 = 0

What are the solutions to the equation 5(3k+2)=k^2(k-7) ?
The solutions to the equation 5(3k+2)=k^2(k-7) are k\approx 8.82755…
Find the zeros of 5(3k+2)=k^2(k-7)
The zeros of 5(3k+2)=k^2(k-7) are k\approx 8.82755…