Theory of EquationshardFree

Theory of Equations: Let Root Root Real Equation Root

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

Question
Let α\alpha be a root of ax2+bx+c=0ax^2 + bx + c = 0 and β\beta be a root of −ax2+bx+c=0-ax^2 + bx + c = 0, where a,b,ca, b, c are real and a≠0a \neq 0. Then the equation a2x2+bx+c=0\dfrac{a}{2}x^2 + bx + c = 0 has a root γ\gamma such that
Aγ<min⁡{α,β}\gamma < \min\{\alpha, \beta\}
Bγ>max⁡{α,β}\gamma > \max\{\alpha, \beta\}
Cγ\gamma lies between α\alpha and β\betacorrect
Dγ\gamma does not lie between α\alpha and β\beta
Solution
Let f(x)=a2x2+bx+cf(x) = \dfrac{a}{2}x^2 + bx + c. Step 1: From aα2+bα+c=0a\alpha^2 + b\alpha + c = 0: bα+c=−aα2b\alpha + c = -a\alpha^2. So:
f(α)=a2α2+bα+c=a2α2−aα2=−a2α2f(\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 −aβ2+bβ+c=0-a\beta^2 + b\beta + c = 0: bβ+c=aβ2b\beta + c = a\beta^2. So:
f(β)=a2β2+bβ+c=a2β2+aβ2=3a2β2f(\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(α)⋅f(β)=(−a2α2)(3a2β2)=−3a2α2β24<0f(\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 α≠0\alpha \neq 0, β≠0\beta \neq 0, a≠0a \neq 0). Step 4: Since ff is continuous and f(α)⋅f(β)<0f(\alpha) \cdot f(\beta) < 0, by the Intermediate Value Theorem ff has at least one root γ\gamma strictly between α\alpha and β\beta. Correct answer: (3)
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