Question

Sachin and Rahul attempted to solve a quadratic equation. Sachin made a mistake in writing down the constant term and ended up in roots \[\left( 4,3 \right)\]. Raul made a mistake in writing down the coefficient of \[x\] to get to the roots \[\left( 3,2 \right)\]. The correct roots of the equation are?

Answer 1

Hint: Firstly we have to find the sum and product of the roots respectively of the equations. As the sum of the roots gives us the coefficient of \[x\] and the product of roots gives us the constant term. We will be finding the sum of the roots obtained by Sachin and in the same way the product of roots of the roots obtained by Rahul. And then we will be substituting these values in the general equation and obtaining the equation.

Complete step by step answer:
Now let us have a brief regarding the roots of quadratic equations. There exists three types of roots to a quadratic equation. They are : real, rational and imaginary roots. We can determine if the roots are real or imaginary or rational by calculating the value of the discriminant. Discriminant is denoted by \[\Delta \]. If \[\Delta <0\], then the roots are imaginary. If \[\Delta \ge 0\], then the roots are real. If \[\Delta =0\], then the roots are equal.
Now let us find the correct roots of the equation.
We can find the equation when roots are given as\[\left( \alpha ,\beta \right)\] by \[\left( x-\alpha \right)\left( x-\beta \right)=0\]
Now we will be applying the same method with the roots we have.
The equation Sachin gets is,
The roots are \[\left( 4,3 \right)\]. So the equation would be
\[\begin{align}
  & \left( x-4 \right)\left( x-3 \right)=0 \\
 & \Rightarrow {{x}^{2}}-7x+12=0 \\
\end{align}\]
The equation Rahul gets is,
The roots are \[\left( 3,2 \right)\]. So the equation would be
\[\begin{align}
  & \left( x-3 \right)\left( x-2 \right)=0 \\
 & \Rightarrow {{x}^{2}}-5x+6=0 \\
\end{align}\]
We know that Sachin has written his constant term wrong, so his constant term should be \[6\]. In the same way, Rahul has written the coefficient of \[x\] wrongly, so his coefficient of \[x\] should be \[-7\].
By considering these factors, the correct equation would be
\[\Rightarrow {{x}^{2}}-7x+6=0\]
Now let us find the roots of this equation. We get the roots as
\[\begin{align}
  & \Rightarrow {{x}^{2}}-7x+6 \\
 & \Rightarrow {{x}^{2}}-x-6x+6 \\
 & \Rightarrow x\left( x-1 \right)-6\left( x-1 \right) \\
 & \Rightarrow \left( x-6 \right)\left( x-1 \right) \\
 & \Rightarrow x=1,6 \\
\end{align}\]
\[\therefore \] The correct roots are \[1,6\].

Note: If the roots are unknown, we can find them by either the prime factorization method or by the formula of finding out the roots i.e. \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. The common error could be trying to find out the roots before finding whether the roots exist or not. So for any quadratic equation, we are supposed to find if the roots exist and then find them.