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Mathematics Question on Quadratic Equations

Let α\alpha and β\beta be the roots of the equation px2+qx−r=0px^2 + qx - r = 0, where p≠0p \neq 0. If p,q,p, q, and rr be the consecutive terms of a non-constant G.P. and 1α+1β=34,\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{4}, then the value of (α−β)2(\alpha - \beta)^2 is:

A

809\frac{80}{9}

B

9

C

203\frac{20}{3}

D

8

Answer

809\frac{80}{9}

Explanation

Solution

The given quadratic equation is: px2+qx−r=0with roots α and β.px^2 + qx - r = 0 \quad \text{with roots } \alpha \text{ and } \beta.
The coefficients are given as: p=A,q=AR,r=AR2.p = A, \quad q = AR, \quad r = AR^2.
Substituting these values into the equation: Ax2+ARx−AR2=0.Ax^2 + ARx - AR^2 = 0.
Dividing throughout by AA, we get: x2+Rx−R2=0.x^2 + Rx - R^2 = 0.
From the roots of the quadratic equation, we know: 1α+1β=34.\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{4}.

Using the relationship between the roots and the coefficients of a quadratic equation: 1α+1β=α+βαβ.\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta}.
Substituting the values for α+β=−R\alpha + \beta = -R and αβ=−R2\alpha \beta = -R^2, we get: α+βαβ=−R−R2=RR2=1R.\frac{\alpha + \beta}{\alpha \beta} = \frac{-R}{-R^2} = \frac{R}{R^2} = \frac{1}{R}.
Equating this to the given value: 1R=34.\frac{1}{R} = \frac{3}{4}.
From this, we find: R=43.R = \frac{4}{3}.

Now, we calculate (α−β)2(\alpha - \beta)^2 using the formula: (α−β)2=(α+β)2−4αβ.(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta.
Substituting the known values: α+β=−Randαβ=−R2,\alpha + \beta = -R \quad \text{and} \quad \alpha \beta = -R^2,
we get: (α−β)2=(−R)2−4(−R2).(\alpha - \beta)^2 = (-R)^2 - 4(-R^2).

Simplify this expression: (α−β)2=R2−4(−R2)=R2+4R2=5R2.(\alpha - \beta)^2 = R^2 - 4(-R^2) = R^2 + 4R^2 = 5R^2.
Substituting R=43R = \frac{4}{3} into the equation: (α−β)2=5(43)2=5⋅169=809.(\alpha - \beta)^2 = 5 \left(\frac{4}{3}\right)^2 = 5 \cdot \frac{16}{9} = \frac{80}{9}.
Thus, the final answer is: (α−β)2=809.(\alpha - \beta)^2 = \frac{80}{9}.