If the quadratic equations x² + ax + b = 0 and x² + bx + a = 0 (a ≠ b) have a common root, then the

Q: Wait, why can't we just use the discriminant to find the common root?

That is a common trap! The discriminant only tells you about the nature of the roots (real, imaginary, equal). It doesn't directly give you the value of the root itself. In common root problems, the most robust method is to treat the root as a variable and use algebraic manipulation to isolate it.

Q: Why is the condition $a \neq b$ so important?

Great catch! When we subtract the equations, we arrive at $(a - b)\alpha = (a - b)$. If $a$ were equal to $b$, the equation would become $0 = 0$, which is true for any $x$, but it wouldn't help us find a specific value for $\alpha$. The condition $a \neq b$ ensures that $(a - b)$ is non-zero, allowing us to divide by it safely to find $\alpha = 1$.

Q: Can I add the equations instead of subtracting them?

You can, but it won't help you eliminate the $\alpha^2$ term. Adding them would give you $2\alpha^2 + (a+b)\alpha + (a+b) = 0$, which is much harder to solve. Always look for ways to eliminate the highest-degree term first—subtraction is your best friend here!

Question

If the quadratic equations $x^2 + ax + b = 0$ and $x^2 + bx + a = 0$ ($a 
eq b$) have a common root, then the numerical value of $a + b$ is .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are diving into a beautiful piece of algebraic symmetry. We are looking at two quadratic equations: $x^2 + ax + b = 0$ and $x^2 + bx + a = 0$.

At first glance, they look almost identical, just with the coefficients $a$ and $b$ swapped. We are told they share a common root, and we are given the crucial constraint that $a 
eq b$. Let's unravel this mystery together.

### Defining the Common Root

Let's call this mysterious common root $\alpha$. Geometrically, if you were to plot these two quadratic equations as parabolas on a Cartesian plane, the roots are the points where the curves intersect the x-axis.

A common root means that both parabolas share the exact same x-intercept at $x = \alpha$. Since $\alpha$ is a root, it must satisfy both equations:

1) $\alpha^2 + a\alpha + b = 0$

2) $\alpha^2 + b\alpha + a = 0$

### The Strategy of Elimination

Now, we have a system of two equations. When you see two equations with the same variable, your first instinct should be to simplify. We have an $\alpha^2$ term in both, so let's perform a strategic subtraction:

$$(\alpha^2 + a\alpha + b) - (\alpha^2 + b\alpha + a) = 0$$

Watch what happens. The $\alpha^2$ terms vanish into thin air, leaving us with:

$$\alpha^2 - \alpha^2 + a\alpha - b\alpha + b - a = 0$$

This simplifies beautifully to:

$$(a - b)\alpha + (b - a) = 0$$

### Solving for the Root

Now, let's group the terms. We can rewrite $(b - a)$ as $-(a - b)$, so our equation becomes:

$$(a - b)\alpha - (a - b) = 0$$

Moving the constant to the other side, we get:

$$(a - b)\alpha = (a - b)$$

Here is where that condition $a 
eq b$ saves the day. Because $a 
eq b$, we know that $(a - b)$ is definitely not zero, allowing us to divide both sides by $(a - b)$ to find:

$$\alpha = 1$$

### The Final Reveal

We have found our common root! Now, we just need to find the value of $a + b$. Since $\alpha = 1$ is a root of the first equation, we plug it back in:

$$1^2 + a(1) + b = 0$$

Simplifying this, we get:

$$1 + a + b = 0$$

Which leads us directly to our final answer:

**$$a + b = -1$$**

Isn't that satisfying? By simply subtracting the equations, we cut through the complexity to find the answer. Remember, in JEE Advanced, the most complex-looking problems often yield to the most elegant, fundamental techniques.

Animated Solution for Mathematics - Quadratic Equations: If the quadratic equations $x^{2} + a x + b = 0$ and $x^{2} + b x + a = 0$ ( $a \neq = b$ ) have a common root, then the numerical value of $a + b$ is .........

Enter Numerical Value:

Visualized Solution

The Two Equations

Defining the Common Root $α$

Substitution into Equations

Eliminating $α^{2}$

Grouping the Terms

Solving for $α$

Finding $a + b$

Final Conclusion

The Sigma Insight: Common Roots

Analyzing the Setup

Defining the Common Root

The Strategy of Elimination

Solving for the Root

The Final Reveal

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Let $λ \neq = 0$ be in $R$ . If $α$ and $β$ are the roots of the equation $x^{2} - x + 2 λ = 0$ , and $α$ and $γ$ are the roots of equation $3 x^{2} - 10 x + 27 λ = 0$ , then $\frac{β γ}{λ}$ is equal to .

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STATEMENT-1: $(p^{2} - q) (b^{2} - a c) \geq 0$ and STATEMENT-2: $b \neq = p a$ or $c \neq = q a$

Enter Numerical Value:

Visualized Solution

The Two Equations

Defining the Common Root .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Substitution into Equations

Eliminating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2

Grouping the Terms

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a+b