Question

Find the real solutions of the equation.

a) x = 0
b) x = 3
c) x = -3/4
d) x = 5/4

Answer 1

To find the real solutions of the given quadratic equation x²+0.0211x-0.0211=0, we can use the quadratic formula. The formula is x = (-b ± √(b²-4ac)) / (2a). By plugging the values of a, b, and c into the formula, we can determine the real solutions for x.

Step-by-step explanation:

The given equation is a quadratic equation. To solve it, we can use the quadratic formula. The quadratic formula states that for an equation of the form ax²+bx+c=0, the solutions for x can be found using the formula:

x = (-b ± √(b²-4ac)) / (2a)

In this case, the equation is x²+0.0211x-0.0211=0. The values of a, b, and c are 1, 0.0211, and -0.0211 respectively. Plugging these values into the quadratic formula, we can find the solutions for x:

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

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