Let a, b, c be real numbers, a ≠ 0. If α is a root of a²x² + bx + c = 0. β is the root of a²x² - bx

Q: Wait, why can't we just solve for $\alpha$ and $\beta$ using the quadratic formula and compare them directly?

That is a classic trap! While you could technically solve for the roots, the resulting expressions would be filled with messy radicals and nested square roots. In JEE Advanced, we value elegance over brute force. By using the properties of the roots and the Intermediate Value Theorem, we bypass the algebra entirely and arrive at the solution with much higher precision and less chance of a calculation error.

Q: How do I know for sure that $f(\alpha)$ is negative?

Great question! We rely on the algebraic substitution derived from the given condition. Since $\alpha$ is a root of $a^2x^2 + bx + c = 0$, we know $b\alpha + c = -a^2\alpha^2$. When we plug this into our target function $f(x)$, the quadratic terms combine to $-a^2\alpha^2$. Since $a \neq 0$ and $\alpha > 0$, the square $a^2\alpha^2$ is strictly positive, making $-a^2\alpha^2$ strictly negative. It is a rigorous proof, not just an assumption!

Q: Is the Intermediate Value Theorem really applicable here?

Absolutely! The Intermediate Value Theorem is one of the most powerful tools in your calculus toolkit. Because $f(x) = a^2x^2 + 2bx + 2c$ is a polynomial, it is continuous everywhere on the real number line. If a continuous function changes sign from negative to positive over an interval, it must cross the x-axis at least once. This is the geometric guarantee that a root $\gamma$ exists between $\alpha$ and $\beta$.

Question

Let $a, b, c$ be real numbers, $a 
eq 0$. If $\alpha$ is a root of $a^2x^2 + bx + c = 0$. $\beta$ is the root of $a^2x^2 - bx - c = 0$ and $0 < \alpha < \beta$, then the equation $a^2x^2 + 2bx + 2c = 0$ has a root $\gamma$ that always satisfies

Options:

(A) $\gamma = \frac{\alpha+\beta}{2}$

(B) $\gamma = \alpha + \frac{\beta}{2}$

(C) $\gamma = \alpha$

(D) $\alpha < \gamma < \beta$

Sigma Percentile · Accepted Answer

### Analyzing the Setup When you look at a quadratic equation like $a^2x^2 + 2bx + 2c = 0$, stop seeing it as a collection of symbols and start seeing it as a physical landscape. Since the coefficient of $x^2$ is $a^2$ and $a eq 0$, this parabola opens upwards, forming a valley. Our goal is to find where this valley crosses the ground—the $x$-axis. We are given two markers, $\alpha$ and $\beta$, with the constraint $0 < \alpha < \beta$. We need to locate the root $\gamma$ of our target function $f(x) = a^2x^2 + 2bx + 2c$. ### The Detective Work at $\alpha$ We are given that $\alpha$ is a root of $a^2x^2 + bx + c = 0$. This implies that $a^2\alpha^2 + b\alpha + c = 0$, which allows us to isolate the expression: $$b\alpha + c = -a^2\alpha^2$$ Now, let us evaluate our target function $f(x) = a^2x^2 + 2bx + 2c$ at $x = \alpha$: $$f(\alpha) = a^2\alpha^2 + 2(b\alpha + c)$$ Substituting our key expression into the equation, we obtain: $$f(\alpha) = a^2\alpha^2 + 2(-a^2\alpha^2) = -a^2\alpha^2$$ Since $a eq 0$ and $\alpha > 0$, it follows that $f(\alpha) < 0$. Our parabola is strictly below the $x$-axis at $x = \alpha$. ### The Contrast at $\beta$ Next, we consider $\beta$, which is a root of $a^2x^2 - bx - c = 0$. This implies $a^2\beta^2 - b\beta - c = 0$, or rearranged: $$b\beta + c = a^2\beta^2$$ Evaluating our target function $f(x)$ at $x = \beta$ yields: $$f(\beta) = a^2\beta^2 + 2(b\beta + c)$$ Substituting our second key expression, we find: $$f(\beta) = a^2\beta^2 + 2(a^2\beta^2) = 3a^2\beta^2$$ Since $a eq 0$ and $\beta > 0$, it follows that $f(\beta) > 0$. Our parabola is strictly above the $x$-axis at $x = \beta$. ### The Grand Finale We have established that at $x = \alpha$, the function is negative ($f(\alpha) < 0$), and at $x = \beta$, the function is positive ($f(\beta) > 0$). Because $f(x)$ is a polynomial, it is continuous across the entire real line. The **Intermediate Value Theorem** dictates that if a continuous function changes sign over an interval, it must cross the $x$-axis at some point within that interval. Therefore, there must exist a root $\gamma$ such that: $$\alpha < \gamma < \beta$$ We have successfully mapped the behavior of the function. This is the essence of JEE Advanced mathematics—using logic and theorems to determine the location of roots through the analysis of function signs.

Select Answer:

Visualized Solution

Defining the Target Function $f (x)$

Analyzing the First Given Root $α$

Isolating the Linear Terms $b α + c$

Finding the Sign of $f (α)$

Analyzing the Second Given Root $β$

Isolating the Linear Terms $b β + c$

Finding the Sign of $f (β)$

Locating the Root $γ$ using IVT

The Sigma Insight: Location of Roots

Analyzing the Setup

The Detective Work at $α$

The Contrast at $β$

The Grand Finale

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Visualized Solution

Defining the Target Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

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

Isolating the Linear Terms .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bα+c

Finding the Sign of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(α)

Analyzing the Second Given Root .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Isolating the Linear Terms .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bβ+c

Finding the Sign of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(β)

Locating the Root .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } γ using IVT

The Sigma Insight: Location of Roots

Analyzing the Setup

The Detective Work at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

The Contrast at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β