Question & Answer
QUESTION

If \[{{x}^{2}}+ax+bc=0~\] and \[{{x}^{2}}+bx+ca=0\] $\left( a\ne b \right)$ have a common root, then prove that their other roots satisfy the equation \[{{x}^{2}}+cx+ab=0\].

ANSWER Verified Verified
Hint: For solving this problem, we fast evaluate the roots of a given equation by using the determinant method. Then after meeting the common root we will satisfy the other routes in the given equation. By using this methodology, we can easily solve the question.

Complete step-by-step answer:
In algebra, a quadratic equation is any equation that can be rearranged in standard form as $a{{x}^{2}}+bx+c=0$, where x represents an unknown and a, b and c represents known numbers and a ≠ 0. By the quadratic formulas, the two roots of any equation of the form $a{{x}^{2}}+bx+c=0$ can be generalized as:
\[\begin{align}
  & {{r}_{1}}=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\text{ and }{{r}_{2}}=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a} \\
 & {{r}_{1}}+{{r}_{2}}=\dfrac{-b}{a},{{r}_{1}}\times {{r}_{2}}=\dfrac{c}{a} \\
\end{align}\]
Here, r1 and r2 are the zeroes of the equation. Also, \[{{r}_{1}}+{{r}_{2}}\] is the sum of zeroes and \[{{r}_{1}}\times {{r}_{2}}\] is the product of zeroes.
According to our problem, we have two equations \[{{x}^{2}}+ax+bc=0~\text{ and }{{x}^{2}}+bx+ca=0\].
Let, \[{{x}^{2}}+ax+bc=0~\] have roots α and β. Now, by using the above formula:
\[\begin{align}
  & \alpha +\beta =-a...(1) \\
 & \alpha \beta =bc...(2) \\
\end{align}\]
Let, \[{{x}^{2}}+bx+ac=0~\] have roots α and γ as both the equations have one root common. Now, again operating in the same manner, we get
\[\begin{array}{*{35}{l}}
   \alpha +\gamma =-b...(3) \\
   \alpha \gamma =ac...(4) \\
\end{array}\]
Subtracting equation (3) from equation (1), we get
\[\beta -\gamma =b-a...\left( 5 \right)\]
Dividing equation (2) and equation (3) we get:
\[\gamma \beta =ab...\left( 6 \right)\]
From the equation (5) and equation (6), by observation \[\beta =b,\gamma =a\].
From equation (2) we get:
\[\begin{align}
  & \alpha .\beta =bc \\
 & \Rightarrow \alpha =c \\
\end{align}\]
To form a quadratic equation having two roots a and b is: ${{x}^{2}}-(a+b)x+ab=0$.
Now, equation having other roots $\beta \text{ and }\gamma $ can be formed as:
\[\begin{align}
  & {{x}^{2}}-\left( \beta +\gamma \right)x+\beta \gamma =0 \\
 & {{x}^{2}}-\left( a+b \right)x+ab...\left( 7 \right)~ \\
\end{align}\]
Since, \[\beta =b\] is the root of \[{{x}^{2}}+ax+bc=0\], so satisfying the root to get,
\[\begin{align}
  & {{b}^{2}}+ab+bc=0 \\
 & \Rightarrow a+b=-c \\
\end{align}\]
Putting this in equation (7) we get:
\[~{{x}^{2}}+cx+ab=0\]
Hence, we proved that other roots of the given equation satisfy the equation ${{x}^{2}}+cx+ab=0$.

Note: The key concept for solving this problem is the knowledge of quadratic roots of an algebraic equation. Using some assumption in roots we evaluated the result. So, students must take care of this assumption.