1. Tardigrade
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  4. Let α, β are two real roots of equation x2+a x+b=0 (a, b ∈ R and b ≠ 0). If the quadratic equation g(x)=0 has two roots α+(1/α), β+(1/β) such that sum of roots is equal to product of roots, then the complete range of b is :

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Q. Let $\alpha, \beta$ are two real roots of equation $x^{2}+a x+b=0$ $(a, b \in R$ and $b \neq 0)$. If the quadratic equation $g(x)=0$ has two roots $\alpha+\frac{1}{\alpha}, \beta+\frac{1}{\beta}$ such that sum of roots is equal to product of roots, then the complete range of b is :

Complex Numbers and Quadratic Equations

Solution:

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$\alpha+\beta=-a ; \alpha \beta=b$
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sum of roots $=$ product of roots
$\alpha+\beta+\frac{1}{\alpha}+\frac{1}{\beta}=\left(\alpha+\frac{1}{\alpha}\right)\left(\beta+\frac{1}{\beta}\right)$
$\Rightarrow \alpha+\beta+\frac{(\alpha+\beta)}{\alpha \beta}=\alpha \beta=\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+\frac{1}{\alpha \beta}$
Using equation (1)
$\Rightarrow -a-\frac{a}{b}=b+\frac{a^{2}-2 b}{b}+\frac{1}{b}$
$\Rightarrow a^{2}+a(b+1)+\left(b^{2}-2 b+1\right)=0$
$D \geq 0$ (Quadratic in a)
$(b+1)^{2}-4\left(b^{2}-2 b+1\right) \geq 0$
$\Rightarrow (3 b-1)(b-3) \leq 0$
$ \Rightarrow b \in\left[\frac{1}{3}, 3\right]$