Common Roots of Quadratic Equations

Case 1: Both Roots are Common

This is the most straightforward scenario. Consider two quadratic equations:
$a_1{x}^2+b_{1}x+c_1=0$
$a_2{x}^2+b_{2}x+c_2=0$

If both equations have the same two roots, say $\alpha$ and $\beta$, then from Vieta's formulas, we know that for the first equation, $\alpha +\beta =-b_1/a_1$ and $\alpha \beta =c_1/a_1$. For the second, $\alpha +\beta =-b_2/a_2$ and $\alpha \beta =c_2/a_2$.

This implies that $-b_1/a_1=-b_2/a_2$ and $c_1/a_1=c_2/a_2$. Rearranging these gives a very simple condition: the coefficients of the two equations must be proportional.

Condition for both roots common:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

This means the two equations are essentially identical, just possibly multiplied by a constant factor.

Let's look at a problem that uses this principle.

The logic shown in the resource is that if the coefficients are in cyclic order (a, b, c for the first equation and b, c, a for the second), and a common root is specified, it's a strong hint to first check the condition for both roots being common, as it often leads to a quick solution. In this case, it leads to $a=b=c$, which simplifies the expression $\frac{a^3+b^3+c^3}{abc}$ to $\frac{3a^3}{a^3}=3$.

Case 2: One Common Root

This scenario requires more algebraic manipulation. Let the common root be $\alpha$. Since $\alpha$ satisfies both equations, we have:
$a_1{\alpha }^2+b_1\alpha +c_1=0\dots (1)$
$a_2{\alpha }^2+b_2\alpha +c_2=0\dots (2)$

We now have a system of two equations with $\alpha$ as the variable. There are two main ways to solve this system to find the condition on the coefficients.

Method 1: Elimination by Subtraction

This is often the quickest method, especially when $a_1=a_2=1$, which is common in JEE problems. By subtracting one equation from the other, we can eliminate the ${\alpha }^2$ term and solve for $\alpha$.

Let's apply this to a typical problem.

As you can see, the process is:

1. Let the common root be $\alpha$. Write the two equations with $\alpha$.
2. Subtract one equation from the other to get a linear equation in $\alpha$.
3. Solve for $\alpha$. (Important: ensure you don't divide by zero. In the example, we can divide by (a-b) assuming a ≠ b, which is given in a similar problem, Byjus Question 10).
4. Substitute the value of $\alpha$ back into either of the original quadratic equations to get the final condition on the coefficients.

Method 2: The General Condition (Cross-Multiplication)

This method is more general and works for any set of coefficients. It treats the system of equations
$a_1{\alpha }^2+b_1\alpha +c_1=0$
$a_2{\alpha }^2+b_2\alpha +c_2=0$
as a system of linear equations in the "variables" ${\alpha }^2$ and $\alpha$.

Using the rule of cross-multiplication, we get:
$\frac{{\alpha }^2}{b_1{c}_2-b_2{c}_1}=\frac{\alpha }{c_1{a}_2-c_2{a}_1}=\frac{1}{a_1{b}_2-a_2{b}_1}$

From this, we can express $\alpha$ and ${\alpha }^2$ in terms of the coefficients:
$\alpha =\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1}\text{and}{\alpha }^2=\frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1}$

Since $(\alpha {)}^2={\alpha }^2$, we can equate the expressions:
${\left(\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1}\right)}^2=\frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1}$

This simplifies to the general condition for one common root:
$(c_1{a}_2-c_2{a}_1{)}^2=(b_1{c}_2-b_2{c}_1)(a_1{b}_2-a_2{b}_1)$

While this formula always works, it can be cumbersome to memorize and apply. The subtraction method is often faster for simpler problems. However, you should understand how this general condition is derived.

The following resources show this method in action.

Now, let's see it applied to solve a problem from the JEE Advanced question bank.

Practice Problems

Your goal is practice, so let's put these methods to work. The following problems are good examples of what you can expect in the JEE Advanced. For each problem, decide whether the subtraction method or the general cross-multiplication method is more efficient.

1. From Quadratic Equation JEE Advanced Previous Year Questions (ID: LINK):

   • Solve Question 10. It is very similar to the one we analyzed using the subtraction method. Try solving it yourself first before looking at the solution.

2. From Common Roots MCQ (ID: LINK):

   • Solve Question 15. Here, the coefficients of $x^2$ are the same, so the subtraction method is very effective.
   • Solve Question 1. This is a slightly more complex problem where the coefficients are in A.P. Use the A.P. condition to simplify the equations before looking for the common root.
   • For a tougher challenge, tackle Question 3. This problem involves finding the common root first, using it to solve for a parameter $\lambda$, and then using Vieta's formulas to find the other roots.

Conclusion

You've now learned the standard techniques for one of the classic problem types in quadratic equations. The methods are systematic and can be applied reliably.

Key Takeaways:

• Both Roots Common: If two quadratic equations share both roots, their corresponding coefficients must be proportional: $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$.
• One Common Root:
  • Method 1 (Elimination): If possible, subtract one equation from the other to eliminate the $x^2$ term, solve for the common root $x=\alpha$, and substitute it back into an original equation. This is usually fastest when the leading coefficients are equal.
  • Method 2 (General Condition): For any two quadratics, the condition $(c_1{a}_2-c_2{a}_1{)}^2=(b_1{c}_2-b_2{c}_1)(a_1{b}_2-a_2{b}_1)$ holds. This is derived using cross-multiplication on the system of equations for the common root $\alpha$.

Our next lesson will shift gears to a different type of algebraic manipulation: solving equations that can be reduced to a quadratic form using clever substitutions. This will further expand your problem-solving arsenal.