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Question 30 of 36

Q.Solve the following equation x4−10x3+26x2−10x+1=0x^4 - 10x^3 + 26x^2 - 10x + 1 = 0.

Telangana TsbieTelangana Board of Intermediate Education 2024Subjective· 7mImportance★★★★★
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This is a reciprocal (palindromic) equation — divide by x2x^2 and substitute y=x+1xy=x+\frac1x to reduce it to a quadratic.

The coefficients 1,−10,26,−10,11,-10,26,-10,1 are palindromic, so x4−10x3+26x2−10x+1=0x^4-10x^3+26x^2-10x+1=0 is a reciprocal equation of even degree (type I).

Divide throughout by x2x^2 (note x=0x=0 is not a root):

x2−10x+26−10x+1x2=0x^2-10x+26-\dfrac{10}{x}+\dfrac{1}{x^2}=0

(x2+1x2)−10(x+1x)+26=0\left(x^2+\dfrac{1}{x^2}\right) - 10\left(x+\dfrac{1}{x}\right)+26=0

Let y=x+1xy=x+\dfrac1x, so x2+1x2=y2−2x^2+\dfrac{1}{x^2}=y^2-2:

(y2−2)−10y+26=0⇒y2−10y+24=0⇒(y−4)(y−6)=0(y^2-2)-10y+26=0 \Rightarrow y^2-10y+24=0 \Rightarrow (y-4)(y-6)=0

So y=4y=4 or y=6y=6.

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