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Form the quadratic whose roots are given expressions in the roots of another

This type of question, asked 3 times: 27 Jul 2022, 28 Jul 2022, 4 Apr 2024. All 3 questions below, oldest first.

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  1. Question 1

    27 Jul 2022

    153
    If α,β\alpha,\beta are the roots of the equation x2−(5+3log⁡35−5log⁡53)x+3(3(log⁡35)13−5(log⁡53)23−1)=0x^{2}-\left( 5 +3^{\sqrt{\log_{3}5}}-5^{\sqrt{\log_{5}3}} \right)x+ 3\left( 3^{\left( \log_{3}5 \right)^{\frac{1}{3}}}-5^{\left( \log_{5}3 \right)^{\frac{2}{3}}}- 1 \right)= 0 then the equation, whose roots are α+1β\alpha +\frac{1}{\beta} and β+1α\beta +\frac{1}{\alpha}, is

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  2. Question 2

    28 Jul 2022

    152
    Let α,β\alpha,\beta be the roots of the equation x2−2x+6=0x^{2}-\sqrt{2}x +\sqrt{6}= 0 and 1α2+1,1β2+1\frac{1}{\alpha^{2}}+ 1,\frac{1}{\beta^{2}}+ 1 be the roots of the equation x2+ax+b=0x^{2}+ ax + b = 0. Then the roots of the equation x2−(a+b−2)x+(a+b+2)x^{2}- (a + b - 2)x + (a + b + 2) =0= 0 are:

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  3. Question 3

    4 Apr 2024

    74
    If 2 and 6 are the roots of the equation ax2+bx+1=0ax^{2}+ bx + 1 = 0, then the quadratic equation, whose roots are 12a+b\frac{1}{2a + b} and 16a+b\frac{1}{6a + b}, is :

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