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3 years ago

15

What are the values of x in this equation?

1 answer:

NNADVOKAT [17]3 years ago

4 0

Answer:

A. x = 0, x = -3, x = 2, and x = -2

Step-by-step explanation:

$x^{4} +12x=4x^{2} +3x^{3}$

by moving the terms on the left side to the right side

$x^{4} -3x^{3} -4x^{2} +12x = 0$

factor an $x^{3}$ from $x^{4} -3x^{3}$

and factor - 4 $x$ from $-4x^{2} +12x$

the equation becomes :

$x^{3} (x - 3 )-4x (x - 3)$ = 0

take $(x - 3)$ as a common factor the equation becomes :

$(x-3)(x^{3}-4x )$ = 0

thus

$x - 3 = 0$ ⇒ $x = 3$

and

$x^{3} - 4x = 0$ by factoring an x we get :

$x(x^{2}-4 ) =0$ ⇒ $x$ = 0 and $x^{2} - 4 = 0$

$x^{2} - 4 =0$ ⇒ $x = 2$ and $x = -2$

Thus

the values are :

x = 0, x = -3, x = 2, and x = -2

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b.

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Step-by-step explanation:

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And the normal vector is given by

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a.

For this case, using the chain rule

$r'(t) = ( -10*2tsin(t^2) , 102t cos(t^2) )\\$

And also remember that

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$T'(t) = ( -2t cos(t^2) , -2t sin(t^2))$

And also

$|T'(t)| = \sqrt{ ( -2t cos(t^2))^2 + (-2t sin(t^2))^2} = \sqrt{ 4t^2 ( ( cos(t^2))^2 + ( sin(t^2))^2)} = \sqrt{4t^2} \\ = 2t$

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Simlarly

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