Alphabet Roots

Calculus Level 3

If α \alpha and β \beta are roots of the equation x 2 − p x + q = 0 x^{2}-px+q=0 then

( α + β ) x − 1 2 ( α 2 + β 2 ) x 2 + 1 3 ( α 3 + β 3 ) x 3 − … (\alpha+\beta)x-\frac{1}{2}(\alpha^{2}+\beta^{2})x^{2}+\frac{1}{3}(\alpha^{3}+\beta^{3})x^{3}- \ldots is equal to?

ln ⁡ ( q x 2 + p x + 1 ) \ln(qx^{2}+px+1) ln ⁡ ( p x 2 + q x + 1 ) \ln(px^{2}+qx+1) ln ⁡ ( p x 2 − q x + 1 ) \ln(px^{2}-qx+1) ln ⁡ ( q x 2 − p x + 1 ) \ln(qx^{2}-px+1) ln ⁡ ( p x 2 + q x − 1 ) \ln(px^{2}+qx-1)

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Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

Tanishq Varshney
Mar 1, 2015

we know l n ( 1 + x ) = x − x 2 2 + x 3 3 . . . . . . . . . . . ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}...........

α x − α 2 x 2 2 + . . . . . . . + β x − β 2 x 2 2 . . . \alpha x-\frac{\alpha^{2}x^{2}}{2}+.......+\beta x-\frac{\beta^{2}x^{2}}{2}...

= l n ( 1 + α x ) + l n ( 1 + β x ) ln(1+\alpha x)+ln(1+\beta x)

= l n ( α β x 2 + ( α + β ) x + 1 ) ln(\alpha \beta x^{2}+(\alpha+\beta)x+1)

= l n ( q x 2 + p x + 1 ) ln(qx^2+px+1)

17 Helpful 0 Interesting 2 Brilliant 0 Confused

Mind = blown

Jitender Sharma - 3 years ago

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