If the roots of 10x3 - cx2 - 54x - 27 = 0 are in HP, find the value of c and all the roots.

Let the roots be α,β,γNow, α+β+γ=c10, αβ+αγ+βγ=−5410,αβγ=2710 Since α,β,γ are m HP, so 1α,1β,1γ are in AP. Thus, 1α+1γ=2β⇒ 1α+1β+1γ=3β⇒ 3β=βγ+αβ+αγαβγ=−5427=−2⇒ β=−32.Since β is a root of the given equation, so⇒ 10−323−c−322−54−32−27=0 −1354−9c4+81−27=0⇒ 9c4=−1354+81−27⇒ 9c4=−1354+54=−135+2164⇒ 9c4=814⇒ c=9Solving, we getα=3,β=−32,γ=−35

If the roots of 10x3 - cx2 - 54x - 27 = 0 are in HP, find the value of c and all the roots.

Let the roots be α,β,γNow, α+β+γ=c10, αβ+αγ+βγ=−5410,αβγ=2710 Since α,β,γ are m HP, so 1α,1β,1γ are in AP. Thus, 1α+1γ=2β⇒ 1α+1β+1γ=3β⇒ 3β=βγ+αβ+αγαβγ=−5427=−2⇒ β=−32.Since β is a root of the given equation, so⇒ 10−323−c−322−54−32−27=0 −1354−9c4+81−27=0⇒ 9c4=−1354+81−27⇒ 9c4=−1354+54=−135+2164⇒ 9c4=814⇒ c=9Solving, we getα=3,β=−32,γ=−35

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If the roots of 10x³ - cx² - 54x - 27 = 0 are in HP, find the value of c and all the roots.

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$α = 3, β = \frac{3}{2}, γ = \frac{3}{5}$

$α = 3, β = - \frac{3}{2}, γ = - \frac{3}{5}$

$α = 4, β = \frac{3}{2}, γ = - \frac{3}{5}$

answer is D.

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Detailed Solution

Let the roots be $α, β, γ$

Now, $α + β + γ = \frac{c}{10}$ , $α β + α γ + β γ = - \frac{54}{10}$ , $α β γ = \frac{27}{10}$ Since $α, β, γ$ are m HP, so $\frac{1}{α}, \frac{1}{β}, \frac{1}{γ}$ are in AP. Thus, $\frac{1}{α} + \frac{1}{γ} = \frac{2}{β}$

$\begin{array}{l} \Rightarrow \frac{1}{α} + \frac{1}{β} + \frac{1}{γ} = \frac{3}{β} \\ \Rightarrow \frac{3}{β} = \frac{β γ + α β + α γ}{α β γ} = - \frac{54}{27} = - 2 \\ \Rightarrow β = - \frac{3}{2} . \end{array}$

Since $β$ is a root of the given equation, so

$\begin{array}{l} \Rightarrow 10 {(- \frac{3}{2})}^{3} - c {(- \frac{3}{2})}^{2} - 54 (- \frac{3}{2}) - 27 = 0 \\ - \frac{135}{4} - \frac{9 c}{4} + 81 - 27 = 0 \\ \Rightarrow \frac{9 c}{4} = - \frac{135}{4} + 81 - 27 \\ \Rightarrow \frac{9 c}{4} = - \frac{135}{4} + 54 = \frac{- 135 + 216}{4} \\ \Rightarrow \frac{9 c}{4} = \frac{81}{4} \\ \Rightarrow c = 9 \end{array}$