Question

Find the products of roots of the equation ${{x}^{2}}+18x+30=2\sqrt{{{x}^{2}}+18x+45}$ is given that all the roots of this equation are real:
(a). 1080
(b). -1080
(c). 1008
(d). -1008

Answer 1

Hint: First we will square both the sides of the equation and then we will rearrange the terms to take all the terms in the left hand side and then we will find the products of roots using the formula.

Complete step-by-step solution -
Let’s square both the sides of the equation and see what we get,
After that we will rearrange the terms.
\[{{\left( {{x}^{2}}+18x+30 \right)}^{2}}={{\left( 2\sqrt{{{x}^{2}}+18x+45} \right)}^{2}}\]
Using the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ to solve our above equation we get,
$\begin{align}
  & {{x}^{4}}+{{\left( 18x+30 \right)}^{2}}+2{{x}^{2}}\left( 18x+30 \right)=4\left( {{x}^{2}}+18x+45 \right) \\
 & {{x}^{4}}+324{{x}^{2}}+900+1080x+36{{x}^{3}}+60{{x}^{2}}-4{{x}^{2}}-72x+180=0 \\
 & {{x}^{4}}+36{{x}^{3}}+380{{x}^{2}}+1008x+1080=0 \\
\end{align}$
Now we have arranged all our terms in the order.
Now we are going to use the formula that will work for any polynomial equation for finding the product of roots of the given equation,
The formula is:
Product of root = ${{\left( -1 \right)}^{n}}\dfrac{\text{constant term}}{\text{coefficient of highest power of x}}$
Here n = degree of the given equation.
As per the given equation the value of n = 4,
As per the given equation the value of constant term is 1080.
As per the given equation the value of coefficient of highest power of x is 1.
Now putting all the values in the formula.

So, the product of root will be: $\dfrac{{{\left( -1 \right)}^{4}}1080}{1}=1080$
Hence, the correct option will be (a).

Note: The method that we have used is the shortest which doesn’t require us to find the all the root of this equation, we can solve this question by finding all the roots of the equation and then multiply it one by one to get the final answer.