Question

Find the zeros of the polynomial $3\sqrt{2}{{x}^{2}}+13x+6\sqrt{2}$ and verify the relationship between the zeros.
A. $x=\dfrac{\sqrt{2}}{3}\text{ and }\dfrac{-3}{\sqrt{2}}$
B. $x=2\sqrt{2}\text{ and }\dfrac{-3}{\sqrt{2}}$
C. $x=\dfrac{-2\sqrt{2}}{3}\text{ and }-\text{3}\sqrt{2}$
D. $x=\dfrac{2\sqrt{2}}{3}\text{ and }\dfrac{-3}{\sqrt{2}}$

Answer 1

Hint: We will compare the given polynomial with $a{{x}^{2}}+bx+c=0$ and list out the values of $a,b,c$. Now we will use the formula of the roots/zeros of the polynomial $a{{x}^{2}}+bx+c=0$ i.e. $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. In this formula we will substitute the values of $a,b,c$ to get roots/zeros. We have the relation between the roots as ${{x}_{1}}+{{x}_{2}}=\dfrac{-b}{a}$ and ${{x}_{1}}.{{x}_{2}}=\dfrac{c}{a}$ where ${{x}_{1}},{{x}_{2}}$ are the roots of the polynomial $a{{x}^{2}}+bx+c=0$. So, after calculating the roots of the given equation we will verify both the above conditions by calculating the all values involved in the relations.

Complete step-by-step answer:
Given that, $3\sqrt{2}{{x}^{2}}+13x+6\sqrt{2}$.
Comparing the above equation with $a{{x}^{2}}+bx+c=0$, then the values of $a,b,c$ are
$a=3\sqrt{2}$, $b=13$, $c=6\sqrt{2}$.
We know that the roots of the equation $a{{x}^{2}}+bx+c=0$ are $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now the roots of the equation $3\sqrt{2}{{x}^{2}}+13x+6\sqrt{2}=0$ are calculated by substituting the values of $a,b,c$ in $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
$\begin{align}
  & \Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow x=\dfrac{-13\pm \sqrt{{{13}^{2}}-4\left( 3\sqrt{2} \right)\left( 6\sqrt{2} \right)}}{2\left( 3\sqrt{2} \right)} \\
 & \Rightarrow x=\dfrac{-13\pm \sqrt{169-144}}{6\sqrt{2}} \\
 & \Rightarrow x=\dfrac{-13\pm \sqrt{25}}{6\sqrt{2}} \\
\end{align}$
Substituting $\sqrt{25}=5$, then we will get
$\begin{align}
  & \Rightarrow x=\dfrac{-13\pm 5}{6\sqrt{2}} \\
 & \Rightarrow x=\dfrac{-13+5}{6\sqrt{2}}\text{ or }\dfrac{-13-5}{6\sqrt{2}} \\
 & \Rightarrow x=\dfrac{-8}{6\sqrt{2}}\text{ or }\dfrac{-18}{6\sqrt{2}} \\
 & \Rightarrow x=\dfrac{-4}{3\sqrt{2}}\text{ or }\dfrac{-3}{\sqrt{2}} \\
\end{align}$
Multiplying and dividing $\sqrt{2}$ from the values of roots then
${{x}_{1}}=-\dfrac{4}{3\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}$ , ${{x}_{2}}=-\dfrac{3}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}$
We know that $\sqrt{a}\times \sqrt{a}=a$
$\begin{align}
  & \Rightarrow {{x}_{1}}=-\dfrac{4\sqrt{2}}{3\times 2} \\
 & \Rightarrow {{x}_{1}}=-\dfrac{2\sqrt{2}}{3} \\
\end{align}$, $\Rightarrow {{x}_{2}}=-\dfrac{3\sqrt{2}}{2}$
$\therefore $ The zeros or roots of the polynomial $3\sqrt{2}{{x}^{2}}+13x+6\sqrt{2}$ are $-\dfrac{2\sqrt{2}}{3}$ and $-\dfrac{3\sqrt{2}}{2}$.
Now the sum of the roots is
$\begin{align}
  & {{x}_{1}}+{{x}_{2}}=-\dfrac{2\sqrt{2}}{3}-\dfrac{3\sqrt{2}}{2} \\
 & \Rightarrow {{x}_{1}}+{{x}_{2}}=\dfrac{-2\times 2\sqrt{2}-3\times 3\sqrt{2}}{2\times 3} \\
 & \Rightarrow {{x}_{1}}+{{x}_{2}}=\dfrac{-4\sqrt{2}-9\sqrt{2}}{6} \\
 & \Rightarrow {{x}_{1}}+{{x}_{2}}=-\dfrac{13\sqrt{2}}{6} \\
\end{align}$
Multiplying and dividing with $\sqrt{2}$, then
$\begin{align}
  & {{x}_{1}}+{{x}_{2}}=-\dfrac{13\sqrt{2}}{6}\times \dfrac{\sqrt{2}}{\sqrt{2}} \\
 & \Rightarrow {{x}_{1}}+{{x}_{2}}=-\dfrac{13\times 2}{6\sqrt{2}} \\
 & \Rightarrow {{x}_{1}}+{{x}_{2}}=-\dfrac{13}{3\sqrt{2}} \\
 & \therefore {{x}_{1}}+{{x}_{2}}=-\dfrac{b}{a} \\
\end{align}$
Hence the sum of the roots is equal to the negative ratio of the coefficient $x$ to the coefficient of ${{x}^{2}}$.
Currently the product of the roots is given by
$\begin{align}
  & {{x}_{1}}\times {{x}_{2}}=\left( -\dfrac{2\sqrt{2}}{3} \right)\left( -\dfrac{3\sqrt{2}}{2} \right) \\
 & \Rightarrow {{x}_{1}}\times {{x}_{2}}=\dfrac{6\sqrt{2}\times \sqrt{2}}{6} \\
\end{align}$
Multiplying and dividing $\sqrt{2}$, then
$\begin{align}
  & {{x}_{1}}\times {{x}_{2}}=\dfrac{6\sqrt{2}\times \sqrt{2}}{6}\times \dfrac{\sqrt{2}}{\sqrt{2}} \\
 & \Rightarrow {{x}_{1}}\times {{x}_{2}}=\dfrac{6\sqrt{2}\times 2}{6\sqrt{2}} \\
 & \Rightarrow {{x}_{1}}\times {{x}_{2}}=\dfrac{6\sqrt{2}}{3\sqrt{2}} \\
 & \therefore {{x}_{1}}\times {{x}_{2}}=\dfrac{c}{a} \\
\end{align}$
Hence the product of the two roots is equal to the ratio of the constant to the coefficient of ${{x}^{2}}$.

Note: It is better to use the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ when there are some functions like square root and cube root for the coefficients. Sometimes they may also ask about the nature of the roots. The nature of the roots depends on the value of $\sqrt{{{b}^{2}}-4ac}$
If $\sqrt{{{b}^{2}}-4ac}>0$ roots are real and distinct.
If $\sqrt{{{b}^{2}}-4ac}=0$ roots are real and the same.
If $\sqrt{{{b}^{2}}-4ac}<0$ roots are imaginary.