Question

In Exercise 5.3.1 problem #2 you solved the general equation of a quadratic, ax² + bx+c = 0, resulting in:

AM
HI
b± √b² +4ac
2a
+√b²-4ac-b²
2a
-b± √b²-4ac
2a
A=Ib±
NI
b
6²-4ac
2a
62-4ac
a

Answer 1

The question is about the quadratic formula, which is used to solve equations of the form ax² + bx + c = 0. To find the solutions, substitute the constants a, b, and c into the formula -b ± √(b² - 4ac) / 2a, and calculate accordingly.

Step-by-step explanation:

The subject in question relates to solving a quadratic equation, which is a second-degree polynomial equation in one variable of the form ax² + bx + c = 0, where a, b, and c are constants, and a ≠ 0.

The solution to such an equation can be found using the quadratic formula, which is -b ± √(b² - 4ac) / 2a.

When substituting the values of a, b, and c from the provided equation into the quadratic formula, you can solve for the variable, often denoted as x or t.

For example, if we have a given quadratic equation with constants a = 1, b = 10.0, and c = -200, we would substitute these into the quadratic formula to find the solutions to the equation.

Step-by-step, first you would calculate the discriminant by squaring b and subtracting 4 times a times c from it.

Then, you apply the formula to find the two possible solutions for the variable.