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Exercise 3.1 · Q3

Q.If α,β\alpha, \beta and γ\gamma are the roots of the cubic equation x3+2x2+3x+4=0x^3+2x^2+3x+4=0, form a cubic equation whose roots are

(i) 2α,2β,2γ2\alpha, 2\beta, 2\gamma
(ii) 1α,1β,1γ\dfrac1\alpha, \dfrac1\beta, \dfrac1\gamma
(iii) −α,−β,−γ-\alpha, -\beta, -\gamma
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Step 1. Read off Vieta's relations. For x3+2x2+3x+4=0x^3+2x^2+3x+4=0: α+β+γ=−2\alpha+\beta+\gamma=-2, αβ+βγ+γα=3\alpha\beta+\beta\gamma+\gamma\alpha=3, αβγ=−4\alpha\beta\gamma=-4.

Step 2. Part (i), roots 2α,2β,2γ2\alpha,2\beta,2\gamma. Sum =2(−2)=−4=2(-2)=-4; pairwise sum =4(3)=12=4(3)=12; product =8(−4)=−32=8(-4)=-32. Equation: x3−(−4)x2+12x−(−32)=0x^3-(-4)x^2+12x-(-32)=0, i.e. x3+4x2+12x+32=0x^3+4x^2+12x+32=0.

Step 3. Part (ii), roots 1α,1β,1γ\tfrac1\alpha,\tfrac1\beta,\tfrac1\gamma. Sum =Σαβαβγ=3−4=−34=\dfrac{\Sigma\alpha\beta}{\alpha\beta\gamma}=\dfrac3{-4}=-\tfrac34; pairwise sum =Σααβγ=−2−4=12=\dfrac{\Sigma\alpha}{\alpha\beta\gamma}=\dfrac{-2}{-4}=\tfrac12; product =1αβγ=−14=\dfrac1{\alpha\beta\gamma}=-\tfrac14. Equation: x3+34x2+12x+14=0x^3+\tfrac34x^2+\tfrac12x+\tfrac14=0; multiplying by 44: 4x3+3x2+2x+1=04x^3+3x^2+2x+1=0.

Step 4. Part (iii), roots −α,−β,−γ-\alpha,-\beta,-\gamma. Sum =2=2; pairwise sum =(−α)(−β)+⋯=Σαβ=3=(-\alpha)(-\beta)+\cdots=\Sigma\alpha\beta=3; product =(−α)(−β)(−γ)=−αβγ=4=(-\alpha)(-\beta)(-\gamma)=-\alpha\beta\gamma=4. Equation: x3−2x2+3x−4=0x^3-2x^2+3x-4=0.

✓Final answer

(i) x3+4x2+12x+32=0x^3+4x^2+12x+32=0 (ii) 4x3+3x2+2x+1=04x^3+3x^2+2x+1=0 (iii) x3−2x2+3x−4=0x^3-2x^2+3x-4=0

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