Question

The function f(x)=x³+2x²+2x+1 can be rewritten as f(x)=(x+1)(x²+x+1). The equation f(x)=0 has only one real solution, -1.

Answer 1

The given quadratic equation can be solved using the quadratic formula. The solutions are x = 0.1354 and x = -0.1572.

Step-by-step explanation:

The given equation x² + 0.0211x - 0.0211 = 0 is a quadratic equation. We can solve this equation using the quadratic formula, which is:

x = (-b ± √(b² - 4ac)) / 2a

In this equation, a = 1, b = 0.0211, and c = -0.0211. Plugging in these values, we get:

x = (-0.0211 ± √(0.0211² - 4(1)(-0.0211))) / 2(1)

Simplifying further, we have:

x = (-0.0211 ± √(0.00044421 + 0.08484)) / 2

x = (-0.0211 ± √0.08528421) / 2

The square root of 0.08528421 is approximately 0.2919. Considering both the positive and negative roots, we get the solutions:

x = (-0.0211 + 0.2919) / 2 and x = (-0.0211 - 0.2919) / 2

Calculating further:

x = 0.1354 and x = -0.1572

So, the equation has two real solutions: x = 0.1354 and x = -0.1572.