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Licemer1 [7]
3 years ago
5

The polynomial equation x^3-4x^2+2x+10=x^2-5x-3 has complex roots 3+2j What is the other root? Use a graphing calculator and a s

ystem of equations.
–3
–1
3
10

Mathematics
2 answers:
Flura [38]3 years ago
8 0

Answer:

  -1

Step-by-step explanation:

Using a graphing calculator, it is convenient to find where the difference between the two sides of the equation is zero. The associated real root is ...

  x = -1

Rainbow [258]3 years ago
6 0

Answer:

-1

Step-by-step explanation:

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Answer:

8. Identify the common denominator; express each fraction using that denominator; combine the numerators of those rewritten fractions and express the result over the common denominator. Factor out any common factors from numerator and denominator in your result. (It's exactly the same set of instructions that apply for completely numerical fractions.)

9. As with numerical fractions, multiply the numerator by the inverse of the denominator; cancel common factors from numerator and denominator.

10. The method often recommended is to multiply the equation by a common denominator to eliminate the fractions. Then solve in the usual way. Check all answers. If one of the answers makes your multiplier (common denominator) be zero, it is extraneous. (10a cannot have extraneous solutions; 10b might)

Step-by-step explanation:

For a couple of these, it is helpful to remember that (a-b) = -(b-a).

<h3>8d.</h3>

\dfrac{5}{x+2}+\dfrac{25-x}{x^2-3x-10}=\dfrac{5(x-5)}{(x+2)(x-5)}+\dfrac{25-x}{(x+2)(x-5)}\\\\=\dfrac{5x-25+25-x}{(x+2)(x-5)}=\dfrac{4x}{x^2-3x-10}

___

<h3>9b.</h3>

\displaystyle\frac{\left(\frac{x}{x-2}\right)}{\left(\frac{2x}{2-x}\right)}=\frac{x}{x-2}\cdot\frac{-(x-2)}{2x}=\frac{-x(x-2)}{2x(x-2)}=-\frac{1}{2}

___

<h3>10b.</h3>

\dfrac{3}{x-1}+\dfrac{6}{x^2-3x+2}=2\\\\\dfrac{3(x-2)}{(x-1)(x-2)}+\dfrac{6}{(x-1)(x-2)}=\dfrac{2(x-1)(x-2)}{(x-1)(x-2)}\\\\3x-6+6=2(x^2-3x+2) \qquad\text{multiply by the denominator}\\\\2x^2-9x+4=0 \qquad\text{subtract 3x}\\\\(2x-1)(x-4)=0 \qquad\text{factor; x=1/2, x=4}

Neither solution makes any denominator be zero, so both are good solutions.

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