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

We have the quadratic equation: $x^2 + (a+3)x - (a+5) = 0$

We need to find the minimum possible value of $\alpha^2 + \beta^2$, where $\alpha$ and $\beta$ are the roots.

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For any quadratic equation $x^2 + px + q = 0$, we have:

Sum of roots: $\alpha + \beta = -p$

Product of roots: $\alpha \beta = q$

In our equation $x^2 + (a+3)x - (a+5) = 0$:

$\alpha + \beta = -(a+3)$

$\alpha \beta = -(a+5)$

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To find $\alpha^2 + \beta^2$, we use the identity:

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$

This works because when we expand $(\alpha + \beta)^2$, we get:

$(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2$

Therefore: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$

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Substituting our values:

$\alpha^2 + \beta^2 = (-(a+3))^2 - 2(-(a+5))$

Simplifying:

$(-(a+3))^2 = (a+3)^2 = a^2 + 6a + 9$

$-2(-(a+5)) = 2(a+5) = 2a + 10$

Therefore:

$\alpha^2 + \beta^2 = a^2 + 6a + 9 + 2a + 10 = a^2 + 8a + 19$

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We have $\alpha^2 + \beta^2 = a^2 + 8a + 19$

This is a quadratic expression in $a$. To find its minimum value, we complete the square:

$a^2 + 8a + 19 = (a + 4)^2 - 16 + 19 = (a + 4)^2 + 3$

We take half of the coefficient of $a$: $\tfrac{8}{2} = 4$

Square it: $4^2 = 16$

So $a^2 + 8a = (a + 4)^2 - 16$

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Since $(a + 4)^2 \geq 0$ for all real values of $a$, the minimum value of $(a + 4)^2 + 3$ is $3$.

This minimum occurs when $(a + 4)^2 = 0$, which means $a = -4$.

Therefore, the minimum possible value is $3$.

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When finding the sum of squares of roots, remember:

Use the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$

Apply sum and product of roots formulas

Complete the square to find the minimum value

This approach works for any quadratic equation!