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Polynomial and rational functions Quiz 2

mathematics

 If $2 - i$ is a root of the equation ${x^2} + ax + b = 0$ then the values of $a\,{\text{and }}b$ are

 Complex roots are always in pairs and ${i^2} =  - 1$ .

 As it complex number $ \\ x = 2 - i,\,2 + i \\   \Rightarrow \left( {2 - i} \right)\left( {2 + i} \right) = b \\    \Rightarrow  {2^2} + {\left( i \right)^2} = b \\  \mathrm{Apply\:complex\:arithmetic\:rule} \left(a+bi\right)\left(a-bi\right)=a^2+b^2 \\  \Rightarrow b=2^2+\left(-1\right)^2 \\  \Rightarrow b=5 \\ As,\,(2 - i) + (2 + i) =  - a \\  \Rightarrow  2-i+2+i=-a \\  \Rightarrow  2+2=-a \\  \Rightarrow a = -4 $ 

 Calculate the roots of a given equation $3{p^3} - 3{p^2} - 90p = 0$ ?

$  3{p^2} - 3{p^2} - 90p = 0 \\   \Rightarrow 3p({p^2} - p - 30) = 0 \\  \Rightarrow 3p\left[ {\left( {{p^2} - 6p} \right) + \left( {5p - 30} \right)} \right] \\ $ Factorization, $ {\text{3p}}\left[ {(p(p - 6) + 5(p - 6)} \right] \\   \Rightarrow 3p\left( {p - 6} \right)\left( {p + 5} \right) \\   \Rightarrow 3p\left( {p - 6} \right)\left( {p + 5} \right) = 0 \\   \Rightarrow p =  - 5,0,6 $ 

 Calculate the value of $x$ in the given equation $\sqrt {x + 3 - 4\sqrt {x - 1} }  + \sqrt {x + 8 - 6\sqrt {x - 1} }  = 1$ 

 Assume any function as a variable to simplify our equation.

 $\sqrt {x + 3 - 4\sqrt {x - 1} }  + \sqrt {x + 8 - 6\sqrt {x - 1} }  = 1 \\ $ Put the value In $x - 1 = {t^2}$ in the given equation, we get, $ \\ \sqrt {{t^2} + 1 + 3 - 4t}  + \sqrt {{t^2} + 1 + 8 - 6t}  = 1 \\   \Rightarrow \left| {t - 2} \right| + \left| {t - 3} \right| = 1 \\   \Rightarrow t \in \left[ {2,3} \right] \\   \Rightarrow x \in \left[ {5,10} \right] $ 

 Calculate the vertical asymptotes of this rational function is given equation  $f(x) = \dfrac{{{x^2} + 2}}{{4{x^2} - 8x - 5}}$  

 Vertical asymptotes of a rational function are always the zeroes of the polynomial in the denominator. So to solve the problem all we need to do is get the zeroes of $4{x^2} - 8x - 5$ .

 Use the quadratic formula, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$  $ \\ $ In this case, value put in the formula ${\text{a}} = 4,\,{\text{b}} =  - 8,\,{\text{c}} =  - 5$ $ \\ \dfrac{{ - ( - 8) \pm \sqrt {{{( - 8)}^2} - 4 \times 4( - 5)} }}{{2 \times 4}} \\ = \dfrac{{8 \pm \sqrt {64 + 80} }}{8} \\ = \dfrac{{8 \pm 12}}{8} \\   \Rightarrow x =  - \dfrac{1}{2},x = \dfrac{5}{2}  $ 

 Calculate the value of $x$ in the given equation ${\left| {x - 4} \right|^2} + \left| {x - 4} \right| - 6 = 0$ 

 Modules are always open with positive value.

 Let $ \left| {x - 4} \right| = t \\   \Rightarrow {t^2} + t - 6 = 0 \\   \Rightarrow t = \dfrac{{ - 1 \pm \sqrt {1 + 24} }}{2} = 2, - 3 \\  t = 2\,{\text{and }}t =  - 3 \\  {\text{Hence, }}\left| {x - 4} \right| = 2\,{\text{and }}\left| {x - 4} \right| =  - 3 \\  {\text{So,  }}\left| {x - 4} \right| =  - 3\,{\text{is not possible}} \\   \Rightarrow \left| {x - 4} \right| = 2 \\   \Rightarrow x - 4 = 2 \\  \, \Rightarrow x = 6 \\ \text{And} \\   \Rightarrow x - 4 =  - 2\, \\   \Rightarrow x = 2 \\   \\  $ 