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The minimum possible value of the sum of the squares of the roots of the equation $x^{2}+\left ( a+3 \right )x-\left ( a+5 \right )=0$ is

  1. $1$
  2. $2$
  3. $3$
  4. $4$
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Given that, $x^{2}+(a+3)x−(a+5)=0$

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

  • $\alpha+\beta= \frac{-(a+3)}{1}=-(a+3)$
  • $\alpha \cdot \beta= \frac{-(a+5)}{1}=-(a+5)$

Now, $\alpha^{2}+\beta^{2}= (\alpha+\beta)^{2}-2 \alpha \cdot \beta $

$\qquad \qquad  \quad = [-(a+3)]^{2}+2(a+5)$

$\qquad \qquad  \quad = a^{2}+9+6a+2a+10$

$\qquad \qquad  \quad = a^{2}+8a+19$

$\qquad \qquad  \quad = a^{2}+8a+16+3$

$\qquad \qquad  \quad = (a+4)^{2}+3$

The value is minimum when  $(a+4)^{2}=0 \quad [\because \text{The minimum value of any square term = 0}]$

The minimum value $=3$

$$\textbf{(or)}$$

Let, $f(a) = a^{2}+8a+19$

For the minimum value $f’(a) = 0$

$\Rightarrow 2a + 8 = 0$

$\Rightarrow a = -4$

The minimum value at $a = -4: f(-4) = (-4)^{2} + 8 (-4) + 19 = 16 – 32 + 19 = 3$

$\therefore$  The minimum possible value of $\alpha^{2}+\beta^{2}$ is $3.$

Correct Answer $:\text{C}$

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