Question

Find the value of $2{{x}^{4}}+5{{x}^{3}}+7{{x}^{2}}-x+41$ where $x=-3+\sqrt{2+2x}$.

Answer 1

Hint: In this problem, first we have to find the value of $x$ by solving the expression $x=-3+\sqrt{2+2x}$ as given in the question. This is done by rearranging this expression and squaring the both sides to remove the root. A quadratic equation is formed first and then it is solved using the quadratic formula,
$\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
The values of $x$ obtained from this formula is then substituted in the function $2{{x}^{4}}+5{{x}^{3}}+7{{x}^{2}}-x+41$ and the corresponding values are obtained.

Complete step-by-step answer:
Here in this problem we have,
$\Rightarrow x=-3+\sqrt{2+2x}.........(i)$
And we have to find the value of,
$2{{x}^{4}}+5{{x}^{3}}+7{{x}^{2}}-x+41..........(ii)$
Now, the first step is to find the values of $x$ .
For this, we have to first rearrange equation (i) such that the term having the root is taken to the RHS and all other terms are taken to the LHS of the equation,
$\Rightarrow x+3=\sqrt{2+2x}$
Now, in order to remove the root, the above equation is squared on both sides of the equation. On squaring the equation becomes,
$\begin{aligned} & \Rightarrow {{(x+3)}^{2}}={{\left( \sqrt{2+2x} \right)}^{2}} \\\ & \Rightarrow {{x}^{2}}+6x+9=2+2x........(iii) \\\ \end{aligned}$
Now, on rearranging equation (iii) we have,
$\Rightarrow {{x}^{2}}+(6x-2x)+(9-2)=0........(iv)$
Simplifying equation (iv) we get,
$\Rightarrow {{x}^{2}}+4x+7=0........(v)$
Now, we have obtained equation (v) in the form of a quadratic equation, $a{{x}^{2}}+bx+c=0$ .
Now, the next step is to find the values of $x$ , using the quadratic formula.
For this we have the quadratic formula,
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}........(vi)$
On comparing the quadratic formula with equation (v) we know,
$\begin{aligned} & \Rightarrow a=1......(vii) \\\ & \Rightarrow b=4.......(viii) \\\ & \Rightarrow c=7.......(ix) \\\ \end{aligned}$
Substituting the values in equations (vi), (vii) and (viii), in equation (v) we get,
$\Rightarrow x=\dfrac{-4\pm \sqrt{{{4}^{2}}-(4\times 1\times 7)}}{2\times 1}$
Simplifying the above equation, we get,
$\begin{aligned} & \Rightarrow x=\dfrac{-4\pm \sqrt{16-28}}{2} \\\ & \Rightarrow x=\dfrac{-4\pm \sqrt{-12}}{2}.........(x) \\\ \end{aligned}$
In equation (ix) we have $\sqrt{-12}$ . According to the property of complex numbers we know,
$\sqrt{-1}=i$ .
$\therefore$ $\sqrt{-12}=\sqrt{12\times -1}=\sqrt{12}\times \sqrt{-1}=\sqrt{12}i$ .
Substituting this in equation (ix) we get,
$\begin{aligned} & \Rightarrow x=\dfrac{-4+\sqrt{12}i}{2} \\\ & \Rightarrow x=\dfrac{-4}{2}\pm \dfrac{\sqrt{12}}{2}i \\\ & \Rightarrow x=-2\pm \dfrac{\sqrt{4}\times \sqrt{3}}{2}i \\\ & \therefore x=-2\pm \sqrt{3}i............(xi) \\\ \end{aligned}$
Now, we have obtained the values of $x$ .
The question is to find the value of $2{{x}^{4}}+5{{x}^{3}}+7{{x}^{2}}-x+41$ .
Here we have two values of $x$ .
First, we will find the value of equation (ii), when $x=-2+\sqrt{3}i$ .
We have to find the value of $2{{x}^{4}}+5{{x}^{3}}+7{{x}^{2}}-x+41$ .
First, let us find ${{x}^{2}}$ .
$\begin{aligned} & \Rightarrow {{x}^{2}}={{\left( -2+\sqrt{3}i \right)}^{2}} \\\ & \Rightarrow {{x}^{2}}=\left( 4-2\times 2\times \sqrt{3}i+{{\left( \sqrt{3}i \right)}^{2}} \right) \\\ & \Rightarrow {{x}^{2}}=\left( 4-4\sqrt{3}i+3\times {{i}^{2}} \right) \\\ \end{aligned}$
We know, ${{i}^{2}}=-1$. Therefore, the above equation becomes,
$\begin{aligned} & \Rightarrow {{x}^{2}}=4-4\sqrt{3}i-3 \\\ & \Rightarrow {{x}^{2}}=1-4\sqrt{3}i..........(xii) \\\ \end{aligned}$
Now, let us find ${{x}^{3}}$ .
$\begin{aligned} & \Rightarrow {{x}^{3}}={{x}^{2}}\cdot x \\\ & \Rightarrow {{x}^{3}}=\left( 1-4\sqrt{3}i \right)\cdot \left( -2+\sqrt{3}i \right) \\\ \end{aligned}$
Now, opening the brackets and simplifying we get,
$\begin{aligned} & \Rightarrow {{x}^{3}}=\left( -2+1\sqrt{3}i+8\sqrt{3}i-\left( 4\sqrt{3}i\times \sqrt{3}i \right) \right) \\\ & \Rightarrow {{x}^{3}}=\left( -2+9\sqrt{3}i-\left( 4\times 3\times {{i}^{2}} \right) \right) \\\ & \Rightarrow {{x}^{3}}=\left( -2+9\sqrt{3}i+12 \right) \\\ & \Rightarrow {{x}^{3}}=\left( 10+9\sqrt{3}i \right)..........(xiii) \\\ \end{aligned}$
Now, let us find ${{x}^{4}}$.
$\begin{aligned} & \Rightarrow {{x}^{4}}={{({{x}^{2}})}^{2}} \\\ & \Rightarrow {{x}^{4}}={{\left( 1-4\sqrt{3}i \right)}^{2}} \\\ & \Rightarrow {{x}^{4}}=\left( {{1}^{2}}-2\times 1\times 4\sqrt{3}i+{{\left( 4\sqrt{3}i \right)}^{2}} \right) \\\ & \Rightarrow {{x}^{4}}=\left( 1-8\sqrt{3}i+\left( 16\times 3\times {{i}^{2}} \right) \right) \\\ & \Rightarrow {{x}^{4}}=\left( 1-8\sqrt{3}i-48 \right) \\\ & \Rightarrow {{x}^{4}}=\left( -47-8\sqrt{3}i \right).........(xiv) \\\ \end{aligned}$
Substituting the values of $x$, ${{x}^{2}}$ , ${{x}^{3}}$ and ${{x}^{4}}$ as obtained in the above equations in equation (ii) we get,
$\Rightarrow 2{{x}^{4}}+5{{x}^{3}}+7{{x}^{2}}-x+41=2\left( -47-8\sqrt{3}i \right)+5\left( 10+9\sqrt{3}i \right)+7\left( 1-4\sqrt{3}i \right)-(-2+\sqrt{3}i)+41$
Simplifying the above equation, we get,
$\begin{aligned} & \Rightarrow 2\left( -47-8\sqrt{3}i \right)+5\left( 10+9\sqrt{3}i \right)+7\left( 1-4\sqrt{3}i \right)-(-2+\sqrt{3}i)+41 \\\ & \Rightarrow \left( -94-16\sqrt{3}i \right)+\left( 50+45\sqrt{3}i \right)+\left( 7-28\sqrt{3}i \right)+\left( 2-\sqrt{3}i \right)+41 \\\ & \Rightarrow \left( -94+50+7+2+41 \right)+\left( -16\sqrt{3}i+45\sqrt{3}i-28\sqrt{3}i-\sqrt{3}i \right) \\\ & \Rightarrow 6+0 \\\ & \Rightarrow 6...........(xv) \\\ \end{aligned}$
Hence, the value of $2{{x}^{4}}+5{{x}^{3}}+7{{x}^{2}}-x+41$ when $x=-2+\sqrt{3}i$ is 6.

Similarly, we will find the value of equation (ii), when $x=-2-\sqrt{3}i$ .
We have to find the value of $2{{x}^{4}}+5{{x}^{3}}+7{{x}^{2}}-x+41$ .
First, let us find ${{x}^{2}}$ .
$\begin{aligned} & \Rightarrow {{x}^{2}}={{\left( -2-\sqrt{3}i \right)}^{2}} \\\ & \Rightarrow {{x}^{2}}={{\left( -\left( 2+\sqrt{3}i \right) \right)}^{2}}={{\left( 2+\sqrt{3}i \right)}^{2}} \\\ & \Rightarrow {{x}^{2}}=\left( 4+2\times 2\times \sqrt{3}i+{{\left( \sqrt{3}i \right)}^{2}} \right) \\\ & \Rightarrow {{x}^{2}}=\left( 4+4\sqrt{3}i+3\times {{i}^{2}} \right) \\\ \end{aligned}$
We know, ${{i}^{2}}=-1$ . Therefore, the above equation becomes,
$\begin{aligned} & \Rightarrow {{x}^{2}}=4+4\sqrt{3}i-3 \\\ & \Rightarrow {{x}^{2}}=1+4\sqrt{3}i..........(xvi) \\\ \end{aligned}$
Now, let us find ${{x}^{3}}$ .
$\begin{aligned} & \Rightarrow {{x}^{3}}={{x}^{2}}\cdot x \\\ & \Rightarrow {{x}^{3}}=\left( 1+4\sqrt{3}i \right)\cdot \left( -2-\sqrt{3}i \right) \\\ \end{aligned}$
Now, opening the brackets and simplifying we get,
$\begin{aligned} & \Rightarrow {{x}^{3}}=\left( -2-1\sqrt{3}i-8\sqrt{3}i+\left( 4\sqrt{3}i\times -\sqrt{3}i \right) \right) \\\ & \Rightarrow {{x}^{3}}=\left( -2-9\sqrt{3}i-\left( 4\times 3\times {{i}^{2}} \right) \right) \\\ & \Rightarrow {{x}^{3}}=\left( -2-9\sqrt{3}i+12 \right) \\\ & \Rightarrow {{x}^{3}}=\left( 10-9\sqrt{3}i \right)..........(xvii) \\\ \end{aligned}$
Now, let us find ${{x}^{4}}$ .
$\begin{aligned} & \Rightarrow {{x}^{4}}={{({{x}^{2}})}^{2}} \\\ & \Rightarrow {{x}^{4}}={{\left( 1+4\sqrt{3}i \right)}^{2}} \\\ & \Rightarrow {{x}^{4}}=\left( {{1}^{2}}+2\times 1\times 4\sqrt{3}i+{{\left( 4\sqrt{3}i \right)}^{2}} \right) \\\ & \Rightarrow {{x}^{4}}=\left( 1+8\sqrt{3}i+\left( 16\times 3\times {{i}^{2}} \right) \right) \\\ & \Rightarrow {{x}^{4}}=\left( 1+8\sqrt{3}i-48 \right) \\\ & \Rightarrow {{x}^{4}}=\left( -47+8\sqrt{3}i \right).........(xviii) \\\ \end{aligned}$
Substituting the values of $x$ , ${{x}^{2}}$ , ${{x}^{3}}$ and ${{x}^{4}}$ as obtained in the above equations in equation (ii) we get,
$\Rightarrow 2{{x}^{4}}+5{{x}^{3}}+7{{x}^{2}}-x+41=2\left( -47+8\sqrt{3}i \right)+5\left( 10-9\sqrt{3}i \right)+7\left( 1+4\sqrt{3}i \right)-(-2-\sqrt{3}i)+41$
Simplifying the above equation, we get,
$\begin{aligned} & \Rightarrow 2\left( -47+8\sqrt{3}i \right)+5\left( 10-9\sqrt{3}i \right)+7\left( 1+4\sqrt{3}i \right)-\left( -2-\sqrt{3}i \right)+41 \\\ & \Rightarrow \left( -94+16\sqrt{3}i \right)+\left( 50-45\sqrt{3}i \right)+\left( 7+28\sqrt{3}i \right)+\left( 2+\sqrt{3}i \right)+41 \\\ & \Rightarrow \left( -94+50+7+2+41 \right)+\left( 16\sqrt{3}i-45\sqrt{3}i+28\sqrt{3}i+\sqrt{3}i \right) \\\ & \Rightarrow 6+0 \\\ & \Rightarrow 6............(xix) \\\ \end{aligned}$
Hence, the value of $2{{x}^{4}}+5{{x}^{3}}+7{{x}^{2}}-x+41$ when $x=-2-\sqrt{3}i$ is 6.
$\therefore$ from equations (xv) and (xix), we find that the value of $2{{x}^{4}}+5{{x}^{3}}+7{{x}^{2}}-x+41$ for both the values of $x$ is 6.
Hence, the value of $2{{x}^{4}}+5{{x}^{3}}+7{{x}^{2}}-x+41$ when $x=-3+\sqrt{2+2x}$ is $6$ .
$\therefore$ The answer is 6.

Note: In this problem the main idea is finding the values of $x$ and then substituting these values to get the expression of $2{{x}^{4}}+5{{x}^{3}}+7{{x}^{2}}-x+41$ . On solving $x=-3+\sqrt{2+2x}$ after rearranging we get two values for $x$ , $x=-2+\sqrt{3}i$ and $x=-2-\sqrt{3}i$ . Don’t forget to substitute both these values for solving the expression $2{{x}^{4}}+5{{x}^{3}}+7{{x}^{2}}-x+41$ .
The student should also know the properties of complex numbers in order to solve this problem.
The properties used in this problem are:
$\begin{aligned} & \Rightarrow i=\sqrt{-1} \\\ & \Rightarrow {{i}^{2}}=-1 \\\ & \Rightarrow {{(a+ib)}^{3}}={{(a+ib)}^{2}}\cdot (a+ib) \\\ & \Rightarrow {{(a+ib)}^{4}}={{\left( {{(a+ib)}^{2}} \right)}^{2}} \\\ \end{aligned}$