Let α, β be the roots of the equation (x-a)(x-b) = c, c ≠ 0. Then the roots of the equation (x-α)(x-

Q: Why can't I just expand $(x-\alpha)(x-\beta)$ and use the sum and product of roots? It seems like a lot of work.

You *could* do that, but it is a trap! It leads to a mess of coefficients and a high probability of calculation errors. In JEE Advanced, we value elegance and structural insight. Using the identity $(x-a)(x-b) - c = (x-\alpha)(x-\beta)$ is faster, cleaner, and demonstrates a deeper understanding of polynomial structure.

Q: Is it always true that $(x-a)(x-b) - c = (x-\alpha)(x-\beta)$?

Yes, provided the leading coefficient is 1. Since both expressions are monic quadratics (the coefficient of $x^2$ is 1) and they share the same roots $\alpha$ and $\beta$, they must be identical polynomials. This is a fundamental property of polynomials: a polynomial of degree $n$ is uniquely determined by its roots and its leading coefficient.

Q: Why does the answer turn out to be $a$ and $b$ again? Does this mean the transformation is reversible?

Exactly! You have spotted the symmetry. The transformation $P(x) = c$ and $Q(x) = -c$ are essentially inverse operations in this context. It is a beautiful cycle where the parabola shifts, and then shifts back. Recognizing this symmetry is what separates a good student from a great one.

Question

Let $\alpha, \beta$ be the roots of the equation $(x-a)(x-b) = c, c 
eq 0$. Then the roots of the equation $(x-\alpha)(x-\beta) + c = 0$ are

Options:

(A) $a, c$

(B) $b, c$

(C) $a, b$

(D) $a+c, b+c$

Sigma Percentile · Accepted Answer

### The Dance of the Parabolas: A Journey into Symmetry

Welcome, future engineers. Today, we are not just solving a quadratic equation; we are stepping inside the geometry of algebra.

Many students approach a problem like $(x-a)(x-b) = c$ and immediately reach for their pens to expand the brackets. They see $x^2 - (a+b)x + ab - c = 0$ and start calculating discriminants.

While that is a valid path, it is the path of brute force. We are here to find the path of the architect.

### Phase 1: The Original Landscape

Let us define our starting point. We have a polynomial $P(x) = (x-a)(x-b)$. Imagine this as a blue parabola on your coordinate plane.

Its roots, where it kisses the x-axis, are $x = a$ and $x = b$. This is our baseline—a simple, elegant curve.

The problem introduces a constant $c$, where $c 
eq 0$. The equation $(x-a)(x-b) = c$ is not just an algebraic statement; it is a geometric event representing the intersection of our blue parabola with the horizontal line $y = c$.

When this line slices through the parabola, it creates two intersection points. The x-coordinates of these points are given to us as $\alpha$ and $\beta$, which are the roots of the equation $(x-a)(x-b) - c = 0$.

### Phase 2: The Power of the Identity

Here is where the magic happens. We know that $\alpha$ and $\beta$ are the roots of the quadratic $(x-a)(x-b) - c = 0$.

Because this is a monic quadratic (the coefficient of $x^2$ is 1), we can write it in its factored form. This is the most powerful tool in your algebraic toolkit: if you know the roots, you know the polynomial.

Therefore, we can establish the identity:

$$ (x-a)(x-b) - c = (x-\alpha)(x-\beta) $$

Pause for a moment and appreciate this. We have just bridged the gap between the original parabola and a new, shifted parabola.

Geometrically, $(x-\alpha)(x-\beta)$ represents a new curve—let us call it $Q(x)$—which is simply our original blue parabola shifted downwards by $c$ units. This is the "Red Parabola" of our transformation.

### Phase 3: The Target Equation

Now, the problem asks us to find the roots of the equation $(x-\alpha)(x-\beta) + c = 0$.

If you look at this through the lens of our identity, the path becomes blindingly clear. We are not looking for a complex solution involving the quadratic formula; we are looking for a substitution.

We have an expression for $(x-\alpha)(x-\beta)$ sitting right there in our identity. Let us rearrange our identity:

$$ (x-\alpha)(x-\beta) = (x-a)(x-b) - c $$

Now, substitute this into our target equation:

$$ [(x-a)(x-b) - c] + c = 0 $$

### Phase 4: The Elegant Cancellation

Look at that! The $-c$ and the $+c$ vanish into thin air. They cancel out perfectly, leaving us with:

$$ (x-a)(x-b) = 0 $$

We have come full circle. We started with the original parabola, shifted it, and then shifted it back.

The equation simplifies back to the original polynomial we started with. Therefore, the roots of this new equation are exactly the same as the roots of the original: **$x = a$ and $x = b$**.

### The JEE Mindset

Why did we do this? Why not just expand everything? Because in the heat of a JEE Advanced exam, time is your most precious currency.

If you had expanded the terms, you would have been dealing with complex expressions involving $\alpha$ and $\beta$, likely leading to a calculation error. By visualizing the geometry—the shifting of the parabola—and using the identity, we bypassed the calculation entirely.

This is the essence of advanced mathematics. It is not about how hard you can calculate; it is about how clearly you can see the structure of the problem.

You have just mastered a technique that turns a complex-looking algebraic mess into a simple, beautiful symmetry. Keep this mindset. When you see a problem, don't just calculate—look for the structure, look for the symmetry, and enjoy the beauty of the math.

Animated Solution for Mathematics - Quadratic Equations: Let $α, β$ be the roots of the equation $(x - a) (x - b) = c, c \neq = 0$ . Then the roots of the equation $(x - α) (x - β) + c = 0$ are

Select Answer:

Visualized Solution

Defining the Original Polynomial

Roots of the Original Polynomial

The First Equation

Roots $α$ and $β$

The Polynomial Identity

Geometric Interpretation

The Target Equation

Algebraic Substitution

Simplifying the Equation

The Final Roots

The Sigma Insight: Relation Between Roots and Coefficients

The Dance of the Parabolas

Phase 1

Phase 2

Phase 3

Phase 4

The JEE Mindset

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