Question

If the roots of \[2{{x}^{2}}-ax-{{a}^{2}}=0\] are \[{{x}_{1}},{{x}_{2}}\] then,
A. ${{x}_{1}}=-2{{x}_{2}}$
B. ${{x}_{1}}=\dfrac{-{{a}_{2}}}{2{{x}_{2}}}$
C. ${{x}_{1}}=-{{x}_{2}}$
D. ${{x}_{1}}=2x$

Answer 1

Hint: We will start by using the fact the roots of a quadratic $a{{x}^{2}}+bx+c$ are $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Then we will find the roots of the equation given to us and then try to find the relation between \[{{x}_{1}},{{x}_{2}}\].
Complete step-by-step answer:
Now, we have been given the equation that \[2{{x}^{2}}-ax-{{a}^{2}}=0\].
Now, we know that the roots of a quadratic equation $a{{x}^{2}}+bx+c=0$ can be found as $\dfrac{-b\pm \sqrt{D}}{2a}$ where D is ${{b}^{2}}-4ac$ and known as discriminant of the quadratic equation $a{{x}^{2}}+bx+c$.
Now, we have the discriminant D of the quadratic equation \[2{{x}^{2}}-ax-{{a}^{2}}=0\] as $\begin{align}
  & D={{\left( -a \right)}^{2}}-4\left( 2 \right){{ -a}^{2}} \\
 & D={{a}^{2}}+8{{a}^{2}} \\
 & D=9{{a}^{2}} \\
\end{align}$
Now, we have the roots of the quadratic equation as,
$\begin{align}
  & {{x}_{1}}=\dfrac{a+\sqrt{9{{a}^{2}}}}{2\times 2} \\
 & =\dfrac{a+3a}{4} \\
 & =\dfrac{4a}{4} \\
 & {{x}_{1}}=a.......\left( 1 \right) \\
 & {{x}_{2}}=\dfrac{a-\sqrt{9{{a}^{2}}}}{2\times 2} \\
 & =\dfrac{a-3a}{4} \\
 & =\dfrac{-2a}{4} \\
 & {{x}_{2}}=\dfrac{-a}{2}.......\left( 2 \right) \\
\end{align}$
Now, from (1) and (2) if we eliminate ‘a’ we have,
${{x}_{1}}=-2{{x}_{2}}$
Hence, the correct option is (A).

Note: It is important to note that we have used the fact that the roots of a standard quadratic equation is $\dfrac{-b\pm \sqrt{D}}{2a}$. Also, it is important to note that we have randomly taken ${{x}_{1}}=\dfrac{a+\sqrt{9{{a}^{2}}}}{2\times 2}$ and ${{x}_{2}}=\dfrac{a-\sqrt{9{{a}^{2}}}}{2\times 2}$ if we would have taken the reverse that would also be correct but according to the options we would have got a similar expression but the variable interchanged i.e. ${{x}_{2}}=-2{{x}_{1}}$. Hence, we have taken \[{{x}_{1}},{{x}_{2}}\] as such that we have exactly the same relation.