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Question

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + 2ax + (3a + 10) = 0$ such that $\alpha < 1 < \beta$. Then the set of all possible values of $a$ is :

The correct answer is
$(-\infty,-2) \cup (5,\infty)$

We are given the quadratic equation $x^2 + 2ax + (3a + 10) = 0$. Let the roots be $\alpha$ and $\beta$. We are given the condition $\alpha < 1 < \beta$.

Analyzing the Roots Condition

The condition $\alpha < 1 < \beta$ means that the number 1 lies strictly between the two roots of the quadratic equation. This implies that the quadratic equation must have two distinct real roots.

The condition for a quadratic equation $Ax^2 + Bx + C = 0$ to have two distinct real roots is that its discriminant, $\Delta = B^2 - 4AC$, must be positive ($\Delta > 0$).

Calculating the Discriminant

For the given equation $x^2 + 2ax + (3a + 10) = 0$, we have:

  • $A = 1$
  • $B = 2a$
  • $C = 3a + 10$

The discriminant is:

$\Delta = (2a)^2 - 4(1)(3a + 10)$

$\Delta = 4a^2 - 12a - 40$

Solving the Discriminant Inequality

We require $\Delta > 0$ for distinct real roots:

$4a^2 - 12a - 40 > 0$

Divide the inequality by 4:

$a^2 - 3a - 10 > 0$

Factor the quadratic expression:

$(a - 5)(a + 2) > 0$

This inequality holds true when $a$ is less than the smaller root ($-2$) or greater than the larger root ($5$).

Therefore, the possible values of $a$ are:

$a < -2$ or $a > 5$

Conclusion

The set of all possible values of $a$ that satisfy the condition is the union of the two intervals:

$(-\infty, -2) \cup (5, \infty)$

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Similar Questions

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Important Questions from Algebra

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  2. The number of $3 \times 2$ matrices A, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^T A$ is 5, is ______
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