1. Tardigrade
  2. Question
  3. Mathematics
  4. Let α and β are the roots of equation ax2+bx+c=0 (a ≠ 0). If 1, α +β , α β are in arithmetic progression and α , 2, β are in harmonic progression, then the value of ((α )2 + (β )2 - 2 (α )2 (β )2/2 ((α )2 + (β )2)) is equal to

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Q. Let $\alpha $ and $\beta $ are the roots of equation $ax^{2}+bx+c=0 \, \left(a \neq 0\right).$ If $1, \, \alpha +\beta , \, \alpha \beta $ are in arithmetic progression and $\alpha , \, 2, \, \beta $ are in harmonic progression, then the value of $\frac{\left(\alpha \right)^{2} + \left(\beta \right)^{2} - 2 \left(\alpha \right)^{2} \left(\beta \right)^{2}}{2 \left(\left(\alpha \right)^{2} + \left(\beta \right)^{2}\right)}$ is equal to

NTA AbhyasNTA Abhyas 2020Complex Numbers and Quadratic Equations

Solution:

$1,$ $\alpha +\beta ,\alpha \beta $ are in A.P.
$\Rightarrow 1,\frac{- b}{a},\frac{c}{a}$ are in A.P.
$\Rightarrow 1+\frac{c}{a}=\frac{- 2 b}{a}$
$\Rightarrow a+c+2b=0$ … $\left(1\right)$
$\frac{1}{\alpha },\frac{1}{2},\frac{1}{\beta }$ are in A.P.
$\Rightarrow \frac{1}{\alpha }+\frac{1}{\beta }=1$
$\Rightarrow \alpha +\beta =\alpha \beta $
$\Rightarrow \frac{- b}{a}=\frac{c}{a}$
$\Rightarrow b+c=0$ … $\left(2\right)$
From $\left(1\right)$ & $\left(2\right)$ we get,
$a=-b=c$
$\Rightarrow \alpha ,\beta $ are roots of equation $x^{2}-x+1=0$
Now, $\frac{\alpha^{2}+\beta^{2}-2 \alpha^{2} \beta^{2}}{2\left(\alpha^{2}+\beta^{2}\right)}=\frac{1}{2}-\frac{(\alpha \beta)^{2}}{(\alpha+\beta)^{2}-2 \alpha \beta}$
$=\frac{1}{2}-\frac{\left(1\right)^{2}}{\left(1\right)^{2} - 2 \left(1\right)}=\frac{1}{2}+1=1.5$