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Anon25 [30]
3 years ago
11

What is the sum of the roots of the polynomial shown below? f(x) = x^3 + 2x^2 - 11x-12

Mathematics
1 answer:
svp [43]3 years ago
4 0

<u>Answer:  </u>

Sum of the roots of the polynomial x^{3}+2 x^{2}-11 x-12 \text { is }-2

<u>Solution:</u>

The general form of cubic polynomial is a x^{3}+b x^{2}+c x+d=0 ---- (1)

If we have any cubic polynomial a x^{3}+b x^{2}+c x+d=0 having roots \alpha , \beta , \theta

Sum of roots \alpha + \beta + \theta = \frac{-b}{a} ---(2)

From question given that,

x^{3}+2 x^{2}-11 x-12 --- (3)

On comparing equation (1) and (3), we get a = 1, b = 2, c = -11 and d = -12

Hence the sum of roots using eqn 2 is given as,

= \frac{-2}{1}

= -2

Hence the sum of the roots of the polynomial x^{3}+2 x^{2}-11 x-12 \text { is }-2

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Answer:

\tan(\theta)=\frac{-\sqrt{5}}{2}

Step-by-step explanation:

Since we are in quadrant 2, sine is positive.  Since sine is positive and cosine is negative, then tangent is negative.

Now I'm going to find the sine value of this angle given using one of the Pythagorean Identities, namely \sin^2(\theta)+\cos^2(\theta)=1.

If given \cos(\theta)=\frac{-2}{3}, then we have \sin^2(\theta)+(\frac{-2}{3})^2=1 by substitution of \cos(\theta)=\frac{-2}{3}.

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Subtract 4/9 on both sides:

\sin^2(\theta)=1-\frac{4}{9}

Simplify:

\sin^2(\theta)=\frac{5}{9}

Square root both sides:

\sin(\theta)=\sqrt{\frac{5}{9}}

\sin(\theta)=\frac{\sqrt{5}}{\sqrt{9}}

\sin(\theta)=\frac{\sqrt{5}}{3}

===========

\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=\frac{\frac{\sqrt{5}}{3}}{\frac{-2}{3}}

Multiplying top and bottom by 3 gives:

\tan(\theta)=\frac{\sqrt{5}}{-2}

I'm going to move the factor of -1 to the top:

\tan(\theta)=\frac{-\sqrt{5}}{2}

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3 years ago
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