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Question 26 of 36

Q.Find the algebraic equation whose roots are the translates of the roots of the equation x4−5x3+7x2−17x+11=0x^4 - 5x^3 + 7x^2 - 17x + 11 = 0 by −2-2.

Telangana TsbieTelangana Board of Intermediate Education 2019Subjective· 7mImportance★★★★★
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To shift roots by −2-2 (translate each root α\alpha to α−2\alpha-2), substitute x→x+2x \to x+2 in the original equation and simplify.

Given f(x)=x4−5x3+7x2−17x+11=0f(x) = x^4-5x^3+7x^2-17x+11 = 0 with roots αi\alpha_i. We want the equation whose roots are βi=αi+(−2)=αi−2\beta_i = \alpha_i + (-2) = \alpha_i - 2.

If β=α−2\beta = \alpha - 2, then α=β+2\alpha = \beta+2. So the new equation (in β\beta, renamed xx) is f(x+2)=0f(x+2) = 0.

Expand each power:

(x+2)4=x4+8x3+24x2+32x+16(x+2)^4 = x^4+8x^3+24x^2+32x+16

(x+2)3=x3+6x2+12x+8(x+2)^3 = x^3+6x^2+12x+8

(x+2)2=x2+4x+4(x+2)^2 = x^2+4x+4

Substitute into f(x+2)=(x+2)4−5(x+2)3+7(x+2)2−17(x+2)+11f(x+2) = (x+2)^4 - 5(x+2)^3 + 7(x+2)^2 - 17(x+2) + 11:

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