Roots of a Quadratic Equation

LCM stands for least common multiple. It is a multiple of 2 or more numbers. For example: 4 and 6 = 12 and the LCM of 10 and 15 = 30.

using differentiation, dx / dy will give you the derivative. from here, you can equate the answer to 0, and solve for x.

you could also factorise it, and using each brackets separately, equate them to 0, again finding values for x

you could also complete the square, and by doing so will give you a new layout to the equation. then you work backwards by making x the subject, leaving you with two x values.

The roots of a quadratic equation (that is when f(x) = 0) are given by the quadratic formula:

x = (-b +/- \sqrt(b^2 - 4 a c)) / (2 a)

Note that there are always two roots, as indicated by the +/-.

You can also find the roots of a quadratic equation by factorising it into the form (x + a)(x + b) = 0, which may be done via the FOIL method or similar. This should be covered in most textbooks.

What Are the Roots of a Quadratic Equation?

The roots (or solutions) of a quadratic equation are simply the values of x that make the equation equal to zero.

A Simple Example:

Take the equation:

x^2 - 4 = 0

When we are trying to find the roots of this equations, we’re asking: What values of x make this equation true?

If we rewrite it:

x^2 - 4 = 0

x^2 = 4

What numbers squared give 4?

x = sqrt(4)

x = 2, and x = -2

So the roots of this equation is x = 2 and x = -2

Visualizing It on a Graph

A quadratic equation is a parabola when graphed. The roots are simply the points where the parabola crosses the x-axis.

For example, in our equation x^2 − 4 = 0, the graph of y = x^2 − 4 is a U-shaped curve that touches the x-axis at x = −2 and x = 2.

So, finding the roots = finding where the graph hits the x-axis!

General Formula: The Quadratic Formula

For any quadratic equation in the form:

ax^2 + bx + c = 0

The roots are found using the quadratic equation:

Key Intuition:

• If the graph crosses the x-axis twice hence, two real roots.
• If it just touches the x-axis once hence, one root (a "double root").
• If it doesn’t cross at all hence, no real roots, only imaginary ones.

The most common and simple method to find the roots (or solutions) of a quadratic equation in the form ax² + bx + c = 0 is by using the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a. 

• A quadratic equation is typically written as ax² + bx + c = 0, where 'a', 'b', and 'c' are coefficients, and 'a' cannot be zero.

Quadratic Formula:

• The formula directly calculates the roots (x-values) of the equation:
• x = (-b ± √(b² - 4ac)) / 2a
• The ± symbol indicates that there are two possible solutions, one with a positive square root and one with a negative square root.

Discriminant:

• The expression inside the square root (b² - 4ac) is called the discriminant. It helps determine the nature of the roots:
• If b² - 4ac > 0, there are two distinct real roots.
• If b² - 4ac = 0, there is one real root (a repeated root).
• If b² - 4ac < 0, there are two complex (non-real) roots.

Steps to Solve:

1. Identify a, b, and c: Determine the coefficients of the quadratic equation.
2. Substitute into the formula: Plug the values of a, b, and c into the quadratic formula.
3. Simplify: Calculate the values of the two possible solutions (one with + and one with -).

Example:

• For the equation 2x² + 5x - 3 = 0:
• a = 2, b = 5, c = -3
• x = (-5 ± √(5² - 4 * 2 * -3)) / (2 * 2)
• x = (-5 ± √(49)) / 4
• x = (-5 ± 7) / 4
• x = 1/2 or x = -3 

I use this symbol ^ to show the presence of a power/exponent. You could complete the square so divide the coefficient of x by 2 to get -3/2. then rewrite the equation as (x-1.5)^2 which expands to x^2- 3x +2.25. So we need to get from 2.25 to -4 so what we do is either -2.25 and then -4 or we could do that in 1 step and -6.25. Hence we get x^2 - 3x - 4 = 0 written as (x-1.5)^2 -6.25. When this equals 0, we rearrange and then we get the square root of 6.25 (posotive OR negative which eventually helps to give us both the solutions of x=-1 and x=4) plus 1.5 = x. therefore x = +/- (square root of 6.25) and then +1.5. So we get x = 4 and x=-1

complete the square