Question

X²+6x+16<0x²+x+20>0​

Answer 1

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Answer 2

To solve these inequalities, we can first find the critical points by setting each quadratic expression equal to zero and then determine the intervals where the expressions are positive or negative.

1. For the first inequality: x² + 6x + 16 < 0

Let's find the critical points by solving for x: x² + 6x + 16 = 0

This is a quadratic equation, and you can solve it using the quadratic formula: x = (-b ± √(b² - 4ac)) / (2a)

In this case, a = 1, b = 6, and c = 16. Plugging these values into the formula:

x = (-6 ± √(6² - 4(1)(16))) / (2(1)) x = (-6 ± √(36 - 64)) / 2 x = (-6 ± √(-28)) / 2

Since the discriminant (the value inside the square root) is negative, there are no real solutions to the equation x² + 6x + 16 = 0. This means the quadratic is always positive.

So, for the first inequality, x² + 6x + 16 < 0 has no real solutions, which means there are no values of x that make this expression less than zero. Therefore, it is never less than zero.

2. For the second inequality: x² + x + 20 > 0

Let's find the critical points by solving for x: x² + x + 20 = 0

This is a quadratic equation, and you can solve it using the quadratic formula as we did before. However, you'll find that the discriminant is negative again, which means there are no real solutions to this equation.

Therefore, just like the first inequality, there are no values of x that make x² + x + 20 greater than zero. So, the second inequality is also never greater than zero.

In summary, neither of the inequalities has real solutions. The first one is never less than zero, and the second one is never greater than zero.