JEE Advanced 2022 Paper 1 · Mathematics · Quadratic Equations · medium

Find the value of (1)/(α^2+1)+(1)/(β^2+1)+(1)/(γ^2+1).

Numerical answer

Let

\alpha,\beta,\gamma

be the roots of

x^3-5x^2+8x-3=0

. Find the value of

\dfrac{1}{\alpha^2+1}+\dfrac{1}{\beta^2+1}+\dfrac{1}{\gamma^2+1}

. Give the answer correct to two decimal places.

Show answer and solution

Answer: 1.04

Let

p(x)=x^3-5x^2+8x-3

. Then

\alpha^2+1=(\alpha-i)(\alpha+i)

, so

\prod(\alpha^2+1)=p(i)p(-i)=|p(i)|^2

. Here

p(i)=-i+5+8i-3=2+7i

, hence the product is

53

. For the numerator write

\sum_{cyc}(\beta^2+1)(\gamma^2+1)=3+2\sum\alpha^2+\sum\beta^2\gamma^2

. With

e_1=5,\ e_2=8,\ e_3=3

\sum\alpha^2=25-16=9

and

\sum\beta^2\gamma^2=e_2^2-2e_1e_3=64-30=34

. So the numerator is

3+18+34=55

and the required sum is

55/53\approx1.0377

, i.e.

1.04

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