Question

Find a quadratic equation whose one root is square root of $-47+8\sqrt{-3}$?

Answer 1

Hint: We find the square root of the given complex number $-47+8\sqrt{-3}$. The square root is considered as the value of the variable $x$. Then we use the conjugate theorem to find the other root for the equation. We use the quadratic equation ${{x}^{2}}-\left( p+q \right)x+pq=0$ to find the required equation.

Complete step by step answer:
We need to find the quadratic equation whose one root is the square root of $-47+8\sqrt{-3}$.
So, if the variable is $x$, then the value of $x$ is square root of $-47+8\sqrt{-3}$.
We need to find the square root of $-47+8\sqrt{-3}$.
Here we have a complex number and we denote that as \[\sqrt{-1}=i\]. We have the relations for imaginary $i$ where ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$.
So, the number becomes \[-47+8\sqrt{-3}=-47+8\sqrt{3}i\].
We express it in the form of identity form of ${{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}$.
For our given expression $-47+8\sqrt{3}i$, we convert $-47$ for the form ${{a}^{2}}+{{b}^{2}}$ of ${{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}$. Then we convert $8\sqrt{3}i$ for the form $2ab$ of ${{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}$.
We break $-47$ as $-47=-48+1=48{{i}^{2}}+1$. We have the sum of two squares. We know $48=4\sqrt{3}$.
So, \[-47=48{{i}^{2}}+1={{\left( 4\sqrt{3}i \right)}^{2}}+{{1}^{2}}\]. Also, we have $8\sqrt{3}i=2\times 4\sqrt{3}i\times 1$.
For our identity ${{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}$, we got \[a=4\sqrt{3}i,b=1\].
So, $-47+8\sqrt{-3}={{\left( 4\sqrt{3}i+1 \right)}^{2}}$.
We can express \[\sqrt{\left( -47+8\sqrt{-3} \right)}=1+4\sqrt{3}i\].
We know that in a quadratic equation the complex number always remains with its conjugate pair if the coefficients of the equation are real.
So, the required quadratic equation will have the roots as \[1+4\sqrt{3}i\] and its conjugate \[1-4\sqrt{3}i\].
We know that the quadratic equation with roots $p$ and $q$ will be ${{x}^{2}}-\left( p+q \right)x+pq=0$.
So, our required equation will be \[{{x}^{2}}-\left\{ \left( 1-4\sqrt{3}i \right)+\left( 1+4\sqrt{3}i \right) \right\}x+\left( 1-4\sqrt{3}i \right)\left( 1+4\sqrt{3}i \right)=0\].
Simplifying we get the equation as \[{{x}^{2}}-2x+49=0\].
Therefore, the quadratic equation whose one root is square root of $-47+8\sqrt{-3}$ is \[{{x}^{2}}-2x+49=0\].

Note:
We also use the identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ for the multiplication of the roots. The equation ${{x}^{2}}-\left( p+q \right)x+pq=0$ can be broken into two parts where $\left( x-p \right)\left( x-q \right)=0$ giving two roots as $p$ and $q$.