Theory of EquationshardFree

Theory of Equations: Let Root Root Real Equation Root

JEE Maths question with a full step-by-step solution.

Question

Let

\alpha

be a root of

ax^2 + bx + c = 0

and

\beta

be a root of

-ax^2 + bx + c = 0

, where

a, b, c

are real and

a \neq 0

. Then the equation

\dfrac{a}{2}x^2 + bx + c = 0

has a root

\gamma

such that

\gamma < \min\{\alpha, \beta\}

\gamma > \max\{\alpha, \beta\}

\gamma

lies between

\alpha

and

\beta

correct

\gamma

does not lie between

\alpha

and

\beta

Solution

Let

f(x) = \dfrac{a}{2}x^2 + bx + c

. Step 1: From

a\alpha^2 + b\alpha + c = 0

b\alpha + c = -a\alpha^2

. So:

f(\alpha) = \frac{a}{2}\alpha^2 + b\alpha + c = \frac{a}{2}\alpha^2 - a\alpha^2 = -\frac{a}{2}\alpha^2

Step 2: From

-a\beta^2 + b\beta + c = 0

b\beta + c = a\beta^2

. So:

f(\beta) = \frac{a}{2}\beta^2 + b\beta + c = \frac{a}{2}\beta^2 + a\beta^2 = \frac{3a}{2}\beta^2

Step 3:

f(\alpha) \cdot f(\beta) = \left(-\frac{a}{2}\alpha^2\right)\left(\frac{3a}{2}\beta^2\right) = -\frac{3a^2\alpha^2\beta^2}{4} < 0

(assuming

\alpha \neq 0

\beta \neq 0

a \neq 0

). Step 4: Since

f

is continuous and

f(\alpha) \cdot f(\beta) < 0

, by the Intermediate Value Theorem

f

has at least one root

\gamma

strictly between

\alpha

and

\beta

. Correct answer: (3)

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