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Softa [21]
2 years ago
14

The equation x^2-13x+q=0 has one root that is equal to 12.5. Find the other root and the value of the coefficient q.

Mathematics
1 answer:
Eva8 [605]2 years ago
6 0
You can find all of this in one step using synthetic division.  If one root is x = 12.5, that means when we use 12.5 in our synthetic division we will have a remainder of 0.  Put 12.5 outside the "box" and the leading coefficients of our polynomial inside like this:  12.5   (1   -13   q).  Bring down the first 1.  Multiply it by the 12.5 to get 12.5.  Put that 12.5 up under the -13 and add to get -.5.  Multiply that -.5 by the 12.5 to get -6.25.  Put that -6.25 up under the q.  Since we need this addition to equal 0, q has to be 6.25.  That is the coefficient that will take the place of q.  What we have left are the bolded numbers that represent the depressed polynomial that is one degree less than the polynomial we started with.  It has an equation of x - .5 = 0.  We solve this for x to get the other root of the polynomial.  x = .5.  And there you go!
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Using trigonometric ratio formula:

$\sin\theta =\frac{\text{opposite }}{\text{hypotenuse}}

$\sin\theta =\frac{4\sqrt{14} }{18}

$\csc\theta =\frac{\text{hypotenuse}}{\text{opposite }}

$\csc\theta =\frac{18}{4\sqrt{14} }

$\cos \theta=\frac{\text { adjacent side }}{\text { hypotenuse }}

$\cos \theta=\frac{10}{18}

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$\tan \theta=\frac{\text { opposite side }}{\text { adjacent side }}

$\tan \theta=\frac{4\sqrt{14} }{10}

$\cot \theta=\frac{\text { adjacent side }}{\text { opposite side }}

$\cot \theta=\frac{10}{4\sqrt{14} }

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