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

There isn't a question here but the information provided all looks accurate.

Hi Robert! I think there’s a little confusion here between derivatives and roots of a quadratic equation. Let me clear that up:

The quadratic formula is used to find the roots (or solutions) of a quadratic equation — values of x that make the equation equal zero. For example, with the equation:

x² - 3x - 4 = 0

The roots are x = -1 and x = 4, because plugging either value into the equation makes it true (equals 0).

On the other hand, the derivative of a quadratic expression (like f(x) = x² - 3x - 4) is from calculus. The derivative tells you the slope of the curve at different points, and in this case, it’s:

f'(x) = 2x - 3

This doesn’t give you the roots — it tells you how steep the graph is at any value of x.

Hope that clears things up! Let me know if you’d like help visualising it — always happy to explain

The roots of a quadratic equation are the values of  that satisfy the equation, meaning they make the expression equal to zero. A general quadratic equation has the form:

The roots (also called solutions or zeros) can be found using the quadratic formula:

This formula gives the two values of  where the graph of the quadratic equation (a parabola) intersects the x-axis.

In the example mentioned in the image, the equation given is:

To solve this, we apply the quadratic formula:

• , ,

So, the two solutions are:

These are the correct roots of the equation. The term "derivative" is misused in the original post. The derivative of a function refers to its rate of change or slope, not the roots. What the post refers to as "derivatives" are simply the roots or solutions.

In summary:

• Roots are the values that make the quadratic expression zero.
• They are found using factorisation, completing the square, or the quadratic formula.
• The term derivative should not be used when referring to the roots.

The roots of a quadratic equation are the values of  that satisfy the equation, meaning they make the expression equal to zero. A general quadratic equation has the form:

The roots (also called solutions or zeros) can be found using the quadratic formula:

This formula gives the two values of  where the graph of the quadratic equation (a parabola) intersects the x-axis.

In the example mentioned in the image, the equation given is:

To solve this, we apply the quadratic formula:

• , ,

So, the two solutions are:

These are the correct roots of the equation. The term "derivative" is misused in the original post. The derivative of a function refers to its rate of change or slope, not the roots. What the post refers to as "derivatives" are simply the roots or solutions.

In summary:

• Roots are the values that make the quadratic expression zero.
• They are found using factorisation, completing the square, or the quadratic formula.
• The term derivative should not be used when referring to the roots.

Hi Robert,

You've correctly identified that x = -1 and x = 4 are the values that satisfy the quadratic equation x² - 3x - 4 = 0, and your verification by substituting them back into the equation is spot on!

Just a small clarification on the terminology:

1. The values of x (like -1 and 4) that make the quadratic equation equal to zero are called the roots or solutions of the equation.
2. The term derivative is a concept from calculus. The derivative of the function f(x) = x² - 3x - 4 tells us about its rate of change or slope (which is f'(x) = 2x - 3). It's a different concept from the roots where the function's value is zero.
3. The quadratic formula is a specific method used precisely to find the roots (or solutions) of a quadratic equation of the form ax² + bx + c = 0.
4. Therefore, methods like factoring, completing the square, or using the quadratic formula (or a calculator that implements it) are all ways to find the roots/solutions, not the derivative.

Great job on finding and checking the roots for that example! Hope this helps clarify the difference between roots and derivatives.

complete the square