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STEP III 1994 question 2 solution

From The Student Room

TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 1994 question 2 solution

(i)

\displaystyle \\x^4+10x^3+26x^2+10x+1=0\\x^2+10x+26+\frac{10}{x}+\frac{1}{x^2}=0\\ (x+\frac{1}{x})^2-2+10y+26=0\\y^2+10y+24=0\\(y+6)(y+4)=0\\x+\frac{1}{x}=-4\\x=-2\pm\sqrt{3}\\x+\frac{1}{x}=-6\\x=-3\pm\sqrt{10}


(ii)


\displaystyle \\x^4+x^3-10x^2-4x+16=0\\x^2+x-10-\frac{4}{x}+\frac{16}{x^2}=0\\ \text{ set } y=x-\frac{4}{x}\\(x-\frac{4}{x})^2+8+(x-\frac{4}{x})-10=0\\y^2+y-2=0\\(y-1)(y+2)=0\\x-\frac{4}{x}=1\\x=\frac{1\pm\sqrt{17}}{2}\\x-\frac{4}{x}=-2\\x=-1\pm\sqrt{5}

Solution by khaixiang

Retrieved from "http://thestudentroom.co.uk/wiki/STEP_III_1994_question_2_solution"
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