What multiplies to -26 and adds to 5

1 answer:

6 0

The two numbers are (5+ √(129))/2 and (5-√(129))/2.

(5+ √(129))/2+ (5-√(129))/2= 5
(5+ √(129))/2* (5-√(129))/2= -26

Hope this helps~

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Find r(t) if r'(t) = 5t4i + 6t5j + t k and r(1) = i + j.

Vladimir [108]

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$\begin{cases}1^5+C_1=1\\\\1^6+C_2=1\\\\\dfrac121^2+C_3=0\end{cases}\implies C_1=0,C_2=0,C_3=-\dfrac12$

so that the particular solution is

$\mathbf r(t)=t^5\,\mathbf i+t^6\,\mathbf j+\dfrac12(t^2-1)\,\mathbf k$