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Anni [7]
3 years ago
6

Write the quadratic equation whose roots are

Mathematics
1 answer:
maks197457 [2]3 years ago
5 0
The quadratic equation y = ax^2 +bx+c can be written in the form:
y = a(x-p)(x-q)
Where p and q are the roots of the equation.
Given p = 4, q = -2, a = 3, the quadratic will look like this:
y = 3(x-4)(x+2)
Finally, distribute and combine like terms to put it in standard form.
y = 3(x^2 - 4x+2x -8) \\  \\ y = 3(x^2 -2x-8) \\  \\ y = 3x^2 -6x -24
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4) (a) For these problems, you should take time to familiarize yourself with common fractions that appear on the unit circle.  \frac{\sqrt{3}} {3} does not appear in the unit circle unless you take the quotient 1/2 divided by sqrt(3)/2 which gives you 1/sqrt(3) which is the same as sqrt(3)/3.  So our numerator is 1/2 and our denominator is sqrt(3)/2.

And remember tangent is just sin/cos.  So what degree has sinx as 1/2 and and cosx as sqrt(3)/2?  Well, 30 degrees does, but 30 degrees is not within the range we are given.  That means they are looking for a sinx that gives us -1/2 and a cosx that gives us -sqrt(3)/2 and that is 210 degrees.

And 210 degrees in radians is 7pi/6.

I hoped that made sense.

(b) This is a lot easier. What angle gives us a cos x of -sqrt(3)/2?  According to the unit circle, 150 degrees and 210 degrees does.  They usually want these in radians, so the answer is 5pi/6 and 7pi/6, respectively.

5) What quadrant is radian measure 5 in?

Well 2pi or roughly 6.28 is a full circle. And 5 is slightly less than 6.28, so it is probably in quadrant IV.

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7 0
4 years ago
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