TS EAMCET · Maths · Quadratic Equation

If $α, β, γ$ are the roots of the equation $3 x^{3} - 26 x^{2} + 52 x - 24 = 0$ such that $α, β, γ$ are in geometric progression and $α < β < γ$ , then $3 α + 2 β + γ =$

A $\frac{68}{3}$
B $\frac{56}{3}$
C $12$
D $24$

Verified Solution

Answer & Solution

Correct Answer

(C) $12$

Step-by-step Solution

Detailed explanation

Let α=ar, β=a and γ=ar are the roots of 3x3-26x2+52x-24=0 Thus products of roots a3=243=8 ⇒a=2 And sum of roots 2r+2+2r=263 ⇒1r+r=103 ⇒3r2-10r+3=0 ⇒3r-1r-3=0 ⇒r=13, 3 But r<1,so r=13 Hence, the roots are…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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If α,β,γ are the roots of the equation 3x3-26x2+52x-24=0 such that α,β,γ are in geometric progression and α<β<γ, then 3α+2β+γ=

Step-by-step Solution

If $α, β, γ$ are the roots of the equation $3 x^{3} - 26 x^{2} + 52 x - 24 = 0$ such that $α, β, γ$ are in geometric progression and $α < β < γ$ , then $3 α + 2 β + γ =$