Maths Q13 – AP-EAMCET 2026

Home Maths Complex & QuadraticQuestion #61

α, β, γ

are the roots of the equation

3 x^{3} - 26 x^{2} + 52 x - 24 = 0

such that

α, β, γ

are in GP and

α < β < γ

, then

3 α + 2 β + γ =

Clue 1 of 3: Properties of Roots

Recall Vieta's formulas relating the coefficients of a polynomial equation to the sum and products of its roots. Also, remember the general form of three terms in a Geometric Progression (GP).

🧠 Full Solution Path

Step 1: Identify Equation and GP Properties✦ Active

The given cubic equation is $3 x^{3} - 26 x^{2} + 52 x - 24 = 0$ . Its roots $α, β, γ$ are in Geometric Progression (GP) and satisfy the condition $α < β < γ$ . We can represent the roots in GP as $α = a / r$ , $β = a$ , and $γ = a r$ , where $a$ is the middle term and $r$ is the common ratio. The condition $α < β < γ$ implies that the common ratio $r$ must be greater than 1 (assuming $a$ is positive).

💡 Teacher's Secret Hint

Always check the order of roots in GP. If the sequence is increasing, the common ratio $r$ must be greater than 1. If it's decreasing, $r$ must be between 0 and 1.

Step 2: Apply Vieta's Formulas for Product of Roots○ Expand

For a cubic equation $A x^{3} + B x^{2} + C x + D = 0$ , Vieta's formula for the product of roots is $α β γ = - D / A$ . In the given equation $3 x^{3} - 26 x^{2} + 52 x - 24 = 0$ , we have $A = 3$ , $B = - 26$ , $C = 52$ , and $D = - 24$ . Therefore, the product of roots is:

α β γ = \frac{- (- 24)}{3} = \frac{24}{3} = 8

Substituting the GP terms for the roots, we get $(a / r) (a) (a r) = a^{3}$ . Thus, we have:

a^{3} = 8 ⟹ a = 2

Since $β = a$ , the middle root is $β = 2$ .

💡 Teacher's Secret Hint

The product of roots formula for GP terms $(a / r) (a) (a r) = a^{3}$ is very useful as it directly gives the middle term $a$ without $r$ . This simplifies subsequent steps.

Step 3: Apply Vieta's Formulas for Sum of Roots to Find Common Ratio○ Expand

Vieta's formula for the sum of roots is $α + β + γ = - B / A$ . For our equation, this gives:

α + β + γ = \frac{- (- 26)}{3} = \frac{26}{3}

Now, substitute the GP terms $a / r, a, a r$ and the value $a = 2$ into this sum:

\frac{2}{r} + 2 + 2 r = \frac{26}{3}

To solve for $r$ , multiply the entire equation by $3 r$ to eliminate denominators:

6 + 6 r + 6 r^{2} = 26 r

Rearrange the terms into a standard quadratic equation:

6 r^{2} - 20 r + 6 = 0

Divide by 2 to simplify:

3 r^{2} - 10 r + 3 = 0

Solve this quadratic equation for $r$ using the quadratic formula $r = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ :

r = \frac{10 \pm \sqrt{(- 10)^{2} - 4 (3) (3)}}{2 (3)} = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm \sqrt{64}}{6} = \frac{10 \pm 8}{6}

This yields two possible values for $r$ : $r_{1} = \frac{10 + 8}{6} = \frac{18}{6} = 3$ and $r_{2} = \frac{10 - 8}{6} = \frac{2}{6} = \frac{1}{3}$ . Since the condition $α < β < γ$ requires $r > 1$ , we select $r = 3$ .

💡 Teacher's Secret Hint

Remember to simplify the quadratic equation before applying the quadratic formula if possible (e.g., dividing by a common factor like 2 here). Also, always use the conditions given in the problem (like $α < β < γ$ ) to correctly choose between multiple solutions for $r$ .

Step 4: Determine the Roots and Calculate the Final Expression○ Expand

With $a = 2$ and $r = 3$ , the roots are:

α = a / r = 2 / 3

β = a = 2

γ = a r = 2 \times 3 = 6

Let's verify the condition $α < β < γ$ : $2 / 3 < 2 < 6$ , which is true. Now, we calculate the required expression $3 α + 2 β + γ$ :

3 (\frac{2}{3}) + 2 (2) + 6

= 2 + 4 + 6

= 12

The value of the expression is 12, which corresponds to option 3.

💡 Teacher's Secret Hint

Always double-check your arithmetic in the final calculation step to avoid minor errors. It's also good practice to quickly verify the roots with all Vieta's formulas if time permits.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Maths Question 13 – AP-EAMCET 2026

Clue 1 of 3: Properties of Roots

🧠 Full Solution Path