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Given the quadratic equation $x^2 – (A – 3)x – (A – 2)$, for what value of $A$ will the sum of the squares of the roots be zero?

  1. $-2$
  2. $3$
  3. $6$
  4. $\text{None of these}$
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Given that,  $x^{2}-(A-3)x-(A-2) = 0 \quad \longrightarrow (1)$

Let the roots of a quadratic equation be $\alpha$ and $\beta$ respectively.

Now,  

  • Sum of roots $ = \alpha+\beta = A-3$
  • Product of roots $ = \alpha\beta = -(A-2) = 2-A$

Given that,  $\alpha^{2} + \beta^{2} = 0$

$\Rightarrow (\alpha+\beta)^{2}-2\alpha\beta = 0$

$\Rightarrow (A-3)^{2}-2(2-A) = 0$

$\Rightarrow A^{2}+9-6A-4+2A = 0$

$\Rightarrow A^{2}-4A+5 = 0$

$\Rightarrow A = \frac{4\pm\sqrt{16-20}}{2}$

$\Rightarrow A = \frac{4\pm\sqrt{-4}}{2}$

$\Rightarrow A = \frac{4\pm2i}{2} \quad [\because \sqrt{-1} = i]$

$\Rightarrow A = \frac{4+2i}{2},\frac{4-2i}{2}$

$\Rightarrow \boxed{A = 2+i, 2-i}$  

Correct Answer $: \text{D}$

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