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MCQMediumJEE 2024Nature of Roots & Formation of Equations

JEE Mathematics 2024 Question with Solution

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,rp, q, r are 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)\left((\alpha - \beta)^2\right) is:

  • A

    809\frac{80}{9}

  • B

    99

  • C

    203\frac{20}{3}

  • D

    88

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: α\alpha and β\beta are the roots of px2+qx−r=0px^2 + qx - r = 0, where p≠0p \neq 0. Also, p,q,rp, q, r are consecutive terms of a non-constant G.P. and 1α+1β=34\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{4}.

Find: The value of (α−β)2\left(\alpha - \beta\right)^2.

Using the relations between roots and coefficients,

α+β=−qp,αβ=−rp\alpha + \beta = -\frac{q}{p}, \qquad \alpha\beta = -\frac{r}{p}

Now,

1α+1β=α+βαβ=34\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{3}{4}

Substituting the values of α+β\alpha + \beta and αβ\alpha\beta,

−qp−rp=qr=34\frac{-\frac{q}{p}}{-\frac{r}{p}} = \frac{q}{r} = \frac{3}{4}

Since p,q,rp, q, r are consecutive terms of a G.P., let

p=A,q=AR,r=AR2p = A, \qquad q = AR, \qquad r = AR^2

Then,

qr=ARAR2=1R\frac{q}{r} = \frac{AR}{AR^2} = \frac{1}{R}

So,

1R=34\frac{1}{R} = \frac{3}{4}

Hence,

R=43R = \frac{4}{3}

For the quadratic equation,

α+β=−R,αβ=−R2\alpha + \beta = -R, \qquad \alpha\beta = -R^2

Now use

(α−β)2=(α+β)2−4αβ(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta

Therefore,

(α−β)2=(−R)2−4(−R2)(\alpha - \beta)^2 = (-R)^2 - 4(-R^2) =R2+4R2=5R2= R^2 + 4R^2 = 5R^2

Substituting R=43R = \frac{4}{3},

(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}

Therefore, the value of (α−β)2\left(\alpha - \beta\right)^2 is 809\frac{80}{9}. The correct option is A.

Using coefficient ratio directly

Given: 1α+1β=34\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{4} and the quadratic equation is px2+qx−r=0px^2 + qx - r = 0.

Find: (α−β)2\left(\alpha - \beta\right)^2.

From the quadratic,

α+β=−qp,αβ=−rp\alpha + \beta = -\frac{q}{p}, \qquad \alpha\beta = -\frac{r}{p}

Hence,

1α+1β=α+βαβ=−qp−rp=qr\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-\frac{q}{p}}{-\frac{r}{p}} = \frac{q}{r}

So,

qr=34\frac{q}{r} = \frac{3}{4}

Because p,q,rp, q, r are consecutive terms of a G.P.,

qp=rq=R\frac{q}{p} = \frac{r}{q} = R

Thus,

q=pR,r=pR2q = pR, \qquad r = pR^2

Then,

qr=pRpR2=1R=34\frac{q}{r} = \frac{pR}{pR^2} = \frac{1}{R} = \frac{3}{4}

Therefore,

R=43R = \frac{4}{3}

Now,

α+β=−qp=−R=−43\alpha + \beta = -\frac{q}{p} = -R = -\frac{4}{3}

and

αβ=−rp=−R2=−169\alpha\beta = -\frac{r}{p} = -R^2 = -\frac{16}{9}

So,

(α−β)2=(α+β)2−4αβ(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta =(−43)2−4(−169)= \left(-\frac{4}{3}\right)^2 - 4\left(-\frac{16}{9}\right) =169+649=809= \frac{16}{9} + \frac{64}{9} = \frac{80}{9}

Therefore, the correct option is A, and the required value is 809\frac{80}{9}.

Common mistakes

  • Using 1α+1β=1α+β\frac{1}{\alpha} + \frac{1}{\beta} = \frac{1}{\alpha + \beta}. This identity is incorrect. The correct relation is 1α+1β=α+βαβ\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}.

  • Forgetting that the product of roots for px2+qx−r=0px^2 + qx - r = 0 is −rp-\frac{r}{p}, not rp\frac{r}{p}. The negative sign comes from the constant term being −r-r.

  • Treating consecutive terms of a G.P. as having common difference instead of common ratio. Here one must write p=A,q=AR,r=AR2p = A, q = AR, r = AR^2 or equivalently use qp=rq\frac{q}{p} = \frac{r}{q}.

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