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White raven [17]
3 years ago
9

Solve the equation by completing the square 5x(x+6)=-50

Mathematics
1 answer:
motikmotik3 years ago
3 0
First lets distribute the 5x 
5x^2 + 30x = -50

Lets divide every term by 5 
x^2 + 6x = -10 

To complete the square we have to half the b value, which in this case is 6. Then square it. 
Half of 6 is 3, 3 squared is 9 

Add that to both sides of the equation 
x^2 + 6x + 9 = -1 

Find the binomial squared
(x+3)^2 (If you're wondering how i got that please comment) 

(x+3)^2 = -1 

Take the square root of the equation of both sides
(x+3) = +/- i
x = -3 +/- i
x = -3 - i
and 
x = -3 + i


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Find the slope of the tangent line to the given polar curve at the point specified by the value of θ
MariettaO [177]

The given curve has equation

<em>r(θ)</em> = 9 + 8 cos(<em>θ</em>)

and its derivative is

d<em>r</em>/d<em>θ</em> = -8 sin(<em>θ</em>)

When <em>θ</em> = <em>π</em>/3, we have <em>r</em> (<em>π</em>/3) = 13, and d<em>r</em>/d<em>θ</em> (<em>π</em>/3) = -4√3.

Differentiate these with respect to <em>θ</em> :

d<em>y</em>/d<em>θ</em> = d<em>r</em>/d<em>θ</em> sin(<em>θ</em>) + <em>r(θ)</em> cos(<em>θ</em>)

d<em>x</em>/d<em>θ</em> = d<em>r</em>/d<em>θ</em> cos(<em>θ</em>) - <em>r(θ</em>) sin(<em>θ</em>)

In polar coordinates, we have

<em>y(θ)</em> = <em>r(θ)</em> sin(<em>θ</em>)

<em>x(θ)</em> = <em>r(θ)</em> cos(<em>θ</em>)

and when <em>θ</em> = <em>π</em>/3, we have <em>y</em> (<em>π</em>/3) = 13√3/2 and <em>x</em> (<em>π</em>/3) = 13/2.

The slope of the tangent line to the curve is d<em>y</em>/d<em>x</em>. By the chain rule,

d<em>y</em>/d<em>x</em> = d<em>y</em>/d<em>θ</em> • d<em>θ</em>/d<em>x</em> = (d<em>y</em>/d<em>θ</em>) / (d<em>x</em>/d<em>θ</em>)

d<em>y</em>/d<em>x</em> = (d<em>r</em>/d<em>θ</em> sin(<em>θ</em>) + <em>r(θ)</em> cos(<em>θ</em>)) / (d<em>r</em>/d<em>θ</em> cos(<em>θ</em>) - <em>r(θ</em>) sin(<em>θ</em>))

When <em>θ</em> = <em>π</em>/3, the slope is

d<em>y</em>/d<em>x</em> = (-4√3 sin(<em>π</em>/3) + 13 cos(<em>π</em>/3)) / (-4√3 cos(<em>π</em>/3) - 13 sin(<em>π</em>/3))

d<em>y</em>/d<em>x</em> = (-4√3 (√3/2) + 13 (1/2)) / (-4√3 (1/2) - 13 (√3/2))

d<em>y</em>/d<em>x</em> = - 1/(17√3)

So, the tangent line has slope -1/(17√3) and passes through (13/2, 13√3/2). Using the point-slope formula, its equation is

<em>y</em> - 13√3/2 = -1/(17√3) (<em>x</em> - 13/2)

<em>y</em> = -(<em>x</em> - 338)/(17√3)

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