There are 3 methods to solve a quadratic equation, these are

Completing square method

Factorisation

Using the quadratic formula

Here let's use the factorisation method,

The equation is x2-3x-4 = 0

So you have to find two numbers their sum is -3( the coefficient of X) and their product is -4( That is the power of the leading coefficient, in this case the 1 multiplied by the constant, in this case -4, therefore 1*-4= -4)

The two numbers are -4 and 1, because -4*1= -4 and -4+1= -3

Therefore, the equation becomes (x-4 ) (x+1)=0

so, x-4 =0 and x+1=0

x= 4 and x=-1

therefore the solution is x= 4 and x=-1

x = (-b ± √(b^2 - 4ac)) / (2a)

please check images

Yes I verified your values of x=4 and x=-1. These are called solutions of the equation rather than absolute values. They are the values of x for which the expression = zero. You can use quadratic equation here but you should also be able to factorise to get (x-4)(x+1).

The above equation can easily be solve by factorization

by looking at the pair factors of -4 that add up to -3

These include -4 and +1

So, the above quadratic equation becomes (x+1)(x-4)=0

either x+1=0 or x-4=0

when x+1=0 , x=-1

When X-4=0, then x=4

hence x=-1 or 4.

Two distinct real roots.

The other methods of finding the solutions of a quadratic equation are; completing the square and factorise

For a quadratic equation \( ax^2 + bx + c = 0 \), the roots are the values of \( x \) that make the equation equal to zero. These are not derivatives; they are solutions to the equation. For example, the roots of \( x^2 - 3x - 4 = 0 \) are \( x = -1 \) and \( x = 4 \), as shown in the calculations provided:

  - At \( x = -1 \): \( (-1)^2 - 3(-1) - 4 = 0 \).

  - At \( x = 4 \): \( (4)^2 - 3(4) - 4 = 0 \).

The derivative of a quadratic function, such as \( f(x) = ax^2 + bx + c \), is a linear expression representing the rate of change of the function. For \( f(x) = x^2 - 3x - 4 \), the derivative \( f'(x) = 2x - 3 \). This tells us the slope of the function at any point \( x \) but is not directly related to finding the roots of the original equation.

To find the roots of a quadratic equation, we use the quadratic formula:

  \[

  x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

  \]

This formula helps find the values of \( x \) that satisfy the equation \( ax^2 + bx + c = 0 \).

The standard method utilised to solve a quadratic equation is by considering the coefficients of the equation in this way:

ax^2 + bx + c = 0. "a" is the coefficient of the squared term, b of the non-squared x term and c of the term without the variable x. The formula is: x1,2 = (-b (+/-) sqrt(b^2-4ac))/2a

For instance: x^2-6x+5 = 0 can be solved in this way:

x1,2 (roots!) = (6 (+/-) sqrt(36-20))/2 = (6 (+/-) 4)/2 = 3 (+/-) 2 --> x1 = 5, x2 = 1

The second method can only be applied for equations of the form: x^2 +sx+ p = 0. This method consists in factorising the initial equation into two factors. To do that we have to consider the coefficient s as the sum of the two factors and p as their product. The two factors are called f1 and f2: x^2+sx+p = (x+f1) * (x+f2)

The values that cancel out one or the other parenthesis are the roots of the initial quadratic equation.

Same example: x^2-6x+5 = 0

Apply second method: choose f1 = -5 and f2 = -1.

-5 + (-1) = -6 = s, OK

-5 * (-1) = 5 = p OK

Rewrite the initial equation: x^2-6x+5 = (x-5) * (x-1) = 0. Hence, the solutions are: x1 = 5, x2 = 1, as before.

the Quadratic equations

A quadratic equation is an equation with an x^2. For example x^2 + 7x + 12 = 0

The root of a quadratic equation is the solution of the equation - what values of x in the equation above make the answer 0? In this case, we can factorise the equation to get (x+3)(x+4) = 0, so roots would be x = -3 and x = -4. Substituting these values back into the equation would give 0.

Following these rules and ensuring the quadratic always equate to zero before you apply the formula you will find all roots.

I'm not sure what your question is but I'm guessing you want to know roughly what the other methods are for solving quadratic equations, other than the quadratic formula.

1. Factorising
2. Completing the square
3. Factor theorem (for A level students, usually those over 16 years of age)

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.

This is a case where a minor error can result in a completely different solution, however, Robert has gotten away lucky this time!

The main misunderstanding here is the following: Derivative & Roots.

Derivative refers to a formula (commonly dy/dx or y') made from an original equation (commonly y) which can be used to find the slope of a tangent at a given point.

Roots (also called solutions) to the equation are (usually) x values which, when inputted, output y=0.

For the given formula: x2 - 3x -4 = 0, we know that the roots (x values) are likely to be our end goal, as the output is set to 0. Note that the format of a quadratic is: y = ax2 +bx + c, thus in this question a=1,b=-3,c=-4; the plotted graph can be seen below.

There are, at GCSE, 4 main methods for finding these x values: Graphing, Factoring, Quadratic and Completing the square; we will only cover Factoring and Quadratic for this example.

Factoring

Factoring refers to finding 2 values, which can add together to make b (-3) and multiply together to make c (-4) (note that in this example a=1, which greatly simplifies the process).

1 ) I've found that the best way of approaching this is by first listing the factors of |c| (absolute value of c) = 4.

This gives:

2,2

1,4

2 ) Now we have to consider that c is negative, so we can conclude that one of factors must also be negative, e.g. -1,4, 2,-2 etc. Next, we need to identify the values which can be added together to make -3.

-2 + 2 = 0

2 + -2 = 0

1 + -4 = -3

-1 + 4 = 3

3 ) We can clearly see that using 1 and -4, a value of -3 is acquired, satisfying our requirements. All that is left is to input them into brackets of the format (x + factor) and multiply them together:

(x+1)(x-4) = 0.

For the result to be 0, either x+1 = 0 or x-4 = 0. Both of these are 'events' which can occur so we must consider each of them.

if x+1=0, x=-1

if x-4=0, x=4

This gives our solutions to be x=4, x=-1 and can be confirmed by looking at the graph below.

Quadratic Formula

This is an alternative approach in which you simply input some values into a formula to get the roots, but what is the equation and what are the values?

Quadratic formula: x = ( -b ± sqrt ( b2 - 4ac ) ) / 2a

The values are taken from the quadratic formula, following the same format as mentioned in the Factoring solution.

quadratic equation format: ax2 + bx + c = 0(or y); our equation is x2 -3x -4 = 0, giving us a =1, b=-3, c=-4.

Our quadratic formula now becomes: x = ( ( -(-3) ± sqrt ( (-3)2 - 4(1)(-4) ) ) / 2(1) = ( ( 3 ± sqrt ( 9 + 16 ) ) / 2 = (3 ± 5)/2

The result has this plus-minus sign in it, which just means that the calculation can be taken as a plus or as a minus; in the same way as the Factoring solution, we must consider both.

(3 + 5) / 2 = 4

(3 - 5) / 2 = -1

Which, as expected, has given us the same solutions, x = 4, x = -1.

Robert You got the correct roots but you have to work more on the concept of derivative. The derivative of the formula is 2x-3 and will be used to determine the rate of change or slope of the function, not the values of the formula that satisfy the equation. The method to use is to find the roots of the derivative