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NDA 2020
Exam
NDA
Year
2020
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2020 · Mathematics · Quadratic Equation

The roots α and β of a quadratic equation, satisfy the relations...

The roots α\alpha and β\beta of a quadratic equation, satisfy the relations α+β=α2+β2\alpha+\beta=\alpha^2+\beta^2 and αβ=α2β2\alpha \beta=\alpha^2 \beta^2. What is the number of such quadratic equations?

Options

  1. A0
  2. B2
  3. C3
  4. D4

Correct answer

B. 2

Step-by-step solution

αβ=α2β2⇒αβ(1−αβ)=0∴αβ=0 or  1\begin{aligned} & \alpha \beta=\alpha^2 \beta^2 \Rightarrow \alpha \beta(1-\alpha \beta)=0 \\ & \therefore \alpha \beta=0 \text{\text{ or } } 1 \end{aligned} When αβ=1\alpha \beta=1 then α=1β\alpha=\frac{1}{\beta} Again from  α+β=α2+β2⇒1β+β=1β2+β2⇒β2−β=1β−1β2⇒β(β−1)=(β−1)β2⇒(β−1)(β−1β2)=0\begin{aligned} & \text{ Again\text{ from } } \alpha+\beta=\alpha^2+\beta^2 \\ & \Rightarrow \frac{1}{\beta}+\beta=\frac{1}{\beta^2}+\beta^2 \Rightarrow \beta^2-\beta=\frac{1}{\beta}-\frac{1}{\beta^2} \\ & \Rightarrow \beta(\beta-1)=\frac{(\beta-1)}{\beta^2} \Rightarrow(\beta-1)\left(\beta-\frac{1}{\beta^2}\right)=0\end{aligned}$\begin{aligned} & \Rightarrow(\beta-1)\left(\beta^3-1\right)=0 \Rightarrow(\beta-1)^2\left(\beta^2+\beta+1\right)=0 \\ & \therefore \beta=1 \text{ and } \beta=\frac{-1 \pm \sqrt{3} i}{2}\end{aligned} Again, when β=1\beta=1, then α=1β=1\alpha=\frac{1}{\beta}=1, roots (α,β)=(1,1)(\alpha, \beta)=(1,1) When β=−1+3i2\beta=\frac{-1+\sqrt{3} i}{2}, then α=−1−3i2\alpha=\frac{-1-\sqrt{3} i}{2} Roots (α,β)=(−1±3i2,−1∓3i2)Thus,number(\alpha, \beta)= \left(\frac{-1 \pm \sqrt{3} i}{2}, \frac{-1 \mp \sqrt{3} i}{2}\right) Thus, number of different quadratic equations=2..
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