The cubic equation whose roots are the squares of the roots of the equation x³-2x²+3x-4=0 is [AP EAPCET 23 May 2025 S2]

Correct Answer is : x3+2x2−7x−16=0\; x³+2x²-7x-16=0x3+2x2−7x−16=0

Correct Answer is : x3+2x2−7x−16=0\; x³+2x²-7x-16=0x3+2x2−7x−16=0

Test Index

Quadratic Equations

Section: Mathematics

Question : 3 of 51

Marks: +1, -0

x^3-2x^2+3x-4=0

Solution:

\;x^3-2x^2+3x-4=0

\;\;\text{ So, }\; \alpha+\beta+\gamma=2 \alpha\beta+\beta\gamma+\gamma\alpha=+3

\; \alpha\beta\gamma=4

\;\;\text{ thus, }\; \alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2

\;\;\; -2(\alpha\beta+\beta\gamma+\gamma\alpha)

\;\;\; =4-2(3)=-2

\;\;\; \;\text{ and }\; \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2

\;\;\; =(\alpha\beta)^2+(\beta\gamma)^2+(\gamma\alpha)^2

\; \because \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2

\; =(\alpha\beta+\beta\gamma+\gamma\alpha)^2-2\alpha\beta\gamma(\alpha+\beta+\gamma)

\; =(3)^2-2(4)(2)=9-16=-7

and

\alpha^2\beta^2\gamma^2=(\alpha\beta\gamma)^2=4^2=16

Hence, the cubic equation with roots

\alpha^2, \beta^2, \gamma^2

\; x^3-(\alpha^2+\beta^2+\gamma^2)x^2+(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2)x-\alpha^2\beta^2\gamma^2

\; =x^3-(-2)x^2+(-7)x-16 \;\; x-\alpha^2\beta^2\gamma^2

\; =x^3+2x^2-7x-16=0

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