Question

Solve the following quadratic equation and find its roots:
$\begin{align}
  & {{x}^{2}}-2x+360=0 \\
 & \\
 & \left( a \right)x=1\pm \sqrt{-359} \\
 & \left( b \right)x=2\pm \sqrt{-359} \\
 & \left( c \right)x=1\pm \sqrt{359} \\
 & \left( d \right)\text{none of these} \\
\end{align}$

Answer 1

Hint: To find the value of x in the above question, we will solve the quadratic equation with the help of quadratic formula. The quadratic formula for calculating the roots of the equation is
given by: $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Where, x is the root of the quadratic equation $a{{x}^{2}}+bx+c=0$. Before finding the roots, we will find out about the nature of roots.
Complete step-by-step answer:
Before solving the question, we must know what a quadratic equation is. A quadratic equation in x is an equation which has the highest power of x as 2. The general form of any quadratic equation $a{{x}^{2}}+bx+c=0$. Now, we have to solve the quadratic given in question. By solving, we mean we have to find the roots of the given quadratic equation. The quadratic equation given in question is: ${{x}^{2}}-2x+360=0$
We can see that, here we cannot do the factors of the quadratic equation. So, we will use a quadratic formula to calculate the roots. Before calculating the roots, we will find the nature of the roots of the equation. It is calculated by the given formula:
$\begin{align}
  & D={{b}^{2}}-4ac \\
 & \Rightarrow D={{\left( -2 \right)}^{2}}-4\left( 360 \right)\left( 1 \right) \\
 & \Rightarrow D=4-4\left( 360 \right) \\
 & \Rightarrow D=4-1440 \\
 & \Rightarrow D=-1436 \\
\end{align}$
Here, D is negative, so both the roots of the equation will be imaginary. Now, we will calculate the roots of the given quadratic equation with the help of quadratic formula. The quadratic formula for calculating the roots of $a{{x}^{2}}+bx+c=0$ is
given by: $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
In our case, b=2, a=1, c=360. Thus, we have:
$\begin{align}
  & x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 360 \right)\left( 1 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{2\pm \sqrt{4-4\left( 360 \right)}}{2} \\
 & \Rightarrow x=\dfrac{2\pm 2\sqrt{1-360}}{2} \\
 & \Rightarrow x=1\pm \sqrt{1-360} \\
 & \Rightarrow x=1\pm \sqrt{-35}9 \\
 & \\
\end{align}$
Hence, option (a) is correct.

Note: We can solve the above quadratic equation by completing the square method. This method is as shown below: ${{x}^{2}}-2x+360=0\Rightarrow {{x}^{2}}-2x+1+359=0.$
Now, we will use the identity ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}.$ Therefore, we will get:
$\begin{align}
  & \Rightarrow {{\left( x-1 \right)}^{2}}+359=0 \\
 & \Rightarrow {{\left( x-1 \right)}^{2}}+{{\left( \sqrt{359} \right)}^{2}}=0 \\
\end{align}$
Here, we will use the formula: \[{{a}^{2}}+{{b}^{2}}=\left( a+b \right)\left( a-b \right).\] Thus, we have:
$\begin{align}
  & =\left( x-1+i\sqrt{359} \right)\left( x-1-i\sqrt{359} \right)=0 \\
 & \Rightarrow x=1\pm \sqrt{359i} \\
 & \Rightarrow x=1\pm \sqrt{-359.} \\
\end{align}$