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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?

- $-2$
- $3$
- $6$
- $\text{None of these}$

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Quadratic equation $:x^2 – (A – 3)x – (A – 2)\rightarrow(1)$

Let given quadratic equation have roots $\alpha$ and $\beta.$

Now, $\alpha + \beta = \dfrac{A-3}{1} = A-3$

and $\alpha\cdot\beta = \dfrac{-(A-2)}{1} = -(A-2)$

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

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

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

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

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

$\implies A^{2} -5A+A+5$

$$\textbf{Not Complete Yet}$$

Let given quadratic equation have roots $\alpha$ and $\beta.$

Now, $\alpha + \beta = \dfrac{A-3}{1} = A-3$

and $\alpha\cdot\beta = \dfrac{-(A-2)}{1} = -(A-2)$

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

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

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

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

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

$\implies A^{2} -5A+A+5$

$$\textbf{Not Complete Yet}$$

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