
Evaluate the following: $2{{x}^{4}}+5{{x}^{3}}+7{{x}^{2}}-x+41$ when x = $-\left( 2+\sqrt{3}\text{i} \right)$.
Answer
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Hint: In this question, put the given value of x in the bi quadratic equation and solve the powers. We also have to keep in mind the rules for the imaginary part. Also, we have to keep in mind that only the real part of the number interacts with the real part of the other number and the same for imaginary.
Complete step-by-step answer:
We are given the equation: $2{{x}^{4}}+5{{x}^{3}}+7{{x}^{2}}-x+41$ and we have to find the value of the following equation at x = $-\left( 2+\sqrt{3}\text{i} \right)$.So, we solve for each of the individual terms in the equation
$2{{x}^{4}}$ = $2{{\left( -1(2+\sqrt{3}\text{i)} \right)}^{2}}{{\left( -1(2+\sqrt{3}\text{i)} \right)}^{2}}$
= $2{{\left( 4-3+4\sqrt{3}\text{i} \right)}^{2}}$
= $2{{\left( 1+4\sqrt{3}\text{i} \right)}^{2}}$
= $2\left( 1-48+8\sqrt{3}\text{i} \right)$
= $-94+16\sqrt{3}\text{i}$
$5{{x}^{3}}$ = $-5\left( 2+\sqrt{3}\text{i} \right){{\left( -1(2+\sqrt{3}\text{i)} \right)}^{2}}$
= $-5\left( 1+4\sqrt{3}\text{i} \right)\left( 2+\sqrt{3}\text{i} \right)$
= $-5\left( 2+\sqrt{3}\text{i}+8\sqrt{3}\text{i}-12 \right)$
= $-5\left( -10+9\sqrt{3}\text{i} \right)$
= $50-45\sqrt{3}\text{i}$
$7{{x}^{2}}$ = $7{{\left( -1(2+\sqrt{3}\text{i)} \right)}^{2}}$
= $7\left( 1+4\sqrt{3}\text{i} \right)$
= $7+28\sqrt{3}\text{i}$
Now that we have all the complex terms simplified to the basic terms, we perform the addition and subtraction operations.
$-94+16\sqrt{3}\text{i}$+ $50-45\sqrt{3}\text{i}$+ $7+28\sqrt{3}\text{i}$- ($-\left( 2+\sqrt{3}\text{i} \right)$) + 41
= 6 - $\sqrt{3}i$.
So, the value of $2{{x}^{4}}+5{{x}^{3}}+7{{x}^{2}}-x+41$ at x = $-\left( 2+\sqrt{3}\text{i} \right)$ is 6 - $\sqrt{3}i$.
Note: We can also solve the following question by factorising our equation and when it is simplified, we put the value of the x in the simplified equation and obtain the results. We avoided that approach because that generally takes a lot of effort, even though our solution is short but it would take a lot of time to follow through that approach.
Complete step-by-step answer:
We are given the equation: $2{{x}^{4}}+5{{x}^{3}}+7{{x}^{2}}-x+41$ and we have to find the value of the following equation at x = $-\left( 2+\sqrt{3}\text{i} \right)$.So, we solve for each of the individual terms in the equation
$2{{x}^{4}}$ = $2{{\left( -1(2+\sqrt{3}\text{i)} \right)}^{2}}{{\left( -1(2+\sqrt{3}\text{i)} \right)}^{2}}$
= $2{{\left( 4-3+4\sqrt{3}\text{i} \right)}^{2}}$
= $2{{\left( 1+4\sqrt{3}\text{i} \right)}^{2}}$
= $2\left( 1-48+8\sqrt{3}\text{i} \right)$
= $-94+16\sqrt{3}\text{i}$
$5{{x}^{3}}$ = $-5\left( 2+\sqrt{3}\text{i} \right){{\left( -1(2+\sqrt{3}\text{i)} \right)}^{2}}$
= $-5\left( 1+4\sqrt{3}\text{i} \right)\left( 2+\sqrt{3}\text{i} \right)$
= $-5\left( 2+\sqrt{3}\text{i}+8\sqrt{3}\text{i}-12 \right)$
= $-5\left( -10+9\sqrt{3}\text{i} \right)$
= $50-45\sqrt{3}\text{i}$
$7{{x}^{2}}$ = $7{{\left( -1(2+\sqrt{3}\text{i)} \right)}^{2}}$
= $7\left( 1+4\sqrt{3}\text{i} \right)$
= $7+28\sqrt{3}\text{i}$
Now that we have all the complex terms simplified to the basic terms, we perform the addition and subtraction operations.
$-94+16\sqrt{3}\text{i}$+ $50-45\sqrt{3}\text{i}$+ $7+28\sqrt{3}\text{i}$- ($-\left( 2+\sqrt{3}\text{i} \right)$) + 41
= 6 - $\sqrt{3}i$.
So, the value of $2{{x}^{4}}+5{{x}^{3}}+7{{x}^{2}}-x+41$ at x = $-\left( 2+\sqrt{3}\text{i} \right)$ is 6 - $\sqrt{3}i$.
Note: We can also solve the following question by factorising our equation and when it is simplified, we put the value of the x in the simplified equation and obtain the results. We avoided that approach because that generally takes a lot of effort, even though our solution is short but it would take a lot of time to follow through that approach.
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