Understanding Quadratic Equations: Standard Form, Solution Methods, and Graphs

Quadratic Equation:

A quadratic equation is an equation in which the highest power of the variable is two (also known as a second-degree polynomial equation). A general form of equation is "ax² + bx + c = 0", where "a", "b", and "c" are constants and "x" is the variable being solved for. 

Quadratic equations are interesting because they can be graphed as a parabola. A parabola is a curved line that, depending on the values of a, b, and c in the equation, looks like a smile or a frown. We can graph the equation to see what x values make the equation true, as well as where the parabola intersects the x-axis (the horizontal line where y = 0).

Standard form of equation:

ax² + bx + c = 0

Quadratic Equation how to solve:

To find the solutions (or roots) of the equation, we can use the quadratic formula and factor method.

Quadratic Formula:

x = (-b ± √(b² - 4ac)) / 2a

Discriminant of a quadratic equation:

The term inside the square root, "b² - 4ac" is known as the discriminant. The equation has two real solutions if the discriminant is positive. 

If the discriminant is zero, there is only one real solution to the equation. 

If the discriminant is negative, there are two complex solutions to the equation (which are conjugates of each other).

Discriminant Value	Number of Real Solutions	Nature of Solutions
Positive (b2 - 4ac > 0)	Two	Real and distinct
Zero (b2 - 4ac = 0)	One	Real and repeated
Negative (b2 - 4ac < 0)	Two	Complex conjugates (non-real)

Example 1: Solving equation ( 5x² + 6x + 1 = 0) using quadratic formula: 

To solve the equation 5x² + 6x + 1 = 0, we can use the quadratic formula:

x = [-b ± √(b² - 4ac)] / 2a

where a = 5, b = 6, and c = 1.

Substituting these values into the formula, we get:

x = [-6 ± √(6² - 4(5)(1))] / 2(5)

x = [-6 ± √(36 - 20)] / 10

x = [-6 ± √16] / 10

x = [-6 ± 4] / 10

So we have two possible solutions:

x = (-6 + 4) / 10 = -1/5

x = (-6 - 4) / 10 = -1

Therefore, the solutions to the equation 5x^2 + 6x + 1 = 0 are x = -1/5 and x = -1.

Example 2: Solving equation ( 5x² + 6x + 1 = 0) using factoring:

To factor this quadratic equation, we need to find two numbers that multiply to give 5 and add to give 6. These numbers are 1 and 5. So we can rewrite the equation as:

5x² + 5x + x + 1 = 0

Now we can factor by grouping:

(5x² + 5x) + (x + 1) = 0

5x(x + 1) + 1(x + 1) = 0

(5x + 1)(x + 1) = 0

So the solutions to the equation 5x² + 6x + 1 = 0 are:

5x + 1 = 0, which gives x = -1/5

x + 1 = 0, which gives x = -1

Therefore, the solutions to the equation 5x² + 6x + 1 = 0 are x = -1/5 and x = -1, which is the same answer we got using the quadratic formula.

Now lets plot graph for ( f(x) = 5x² + 6x + 1 = 0 ):

Quadratic Formula Calculator