To solve x2 - 3x - 4, we need to identify a b and c, which are constants. 

A quadratic expression is written in the format ax2 + bx +c, where a b and c are constants (numbers)

In x2 -3x-4, a=1, b=-3 and c = -4 

Next, we find ac and b

Ac = 1 x -4 = -4 

b = -3

Next, we find two numbers that multiply to give ac and add to make b

So 1 x -4 = -4 and 1 + -4 = -3 

So the numbers are 1 and -4

We then put these in the brackets opposite x to factorise 

X2 = X x X so we have X in the brackets

(x+1)(x-4)

To get the roots, we need to solve the equations by making it =0 

(x+1)(x-4) =0 

x=-1 and x =4, because that makes the brackets = 0 and anything x 0 = 0 

The quadratic formula is given by the formula where it is applied to a quadratic equation of the form ax2+bx+c = 0. In this instance the values of a = 1, b = -3 and c = -4. Then you plug these values into the quadratic formula using a calculator (the formula is also given on formula sheet on a exam paper) x = -b+- sort(b2 - 4ac) / 2a. This will also give the values x = -1 and x = 4. You can check your answers by substitution these values into the original equation to see if the answer gives 0. This means the y-value is equal to zero as it touches the x-axis on the graph.

Another way to find the derivative of a quadratic equation is

Find the factors of the constant term (usually the number at the end of the equation). In this example its -4

The factors are

1 and 4

2 and 2

From these pairs you need to pick the pair that multiplies to get -4 and then adds to get the coefficient of x (in this example its -3)

-4 and 1 are going to be the correct pair as

-4 x 1 = -4 and -4 + 1 = -3

Then write these numbers in two brackets

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

To Solve take each bracket individually as an equation equal to 0

x-4=0 x+1=0

Solve each equation to find x

x=4 x=-1

The roots of a quadratic equation are the values of x where for that x value, the equation is equal to 0. So in the example x^2 - 3x -4=0, if you substitute x=-1 and x=-4 in, you get 0 out of the equation. So x=-1 and x=-4 are called the roots. To get the values of x you can factorise and solve the quadratic, or use the quadratic formula.

This can also be solved using factorisation.

The equation in question can be split into (x +/- ??)(x +/- ??) = 0.

The question marks in the brackets need to add to -3 and multiply to -4.

-1 and 4 multiply to -4, as does 1 and -4, as well as -2 and 2.

-1 + 4 = 3 - this is incorrect as we need the numbers to add to -3.

1 + -4 = -3 - this looks correct, as the numbers add to -3.

-2 + 2 = -4 - this is incorrect, as we need the numbers to add to -3.

The brackets can now be filled in as (x + 1)(x - 4) = 0.

The solution to the equation is taken by reversing the symbols before the numbers in the brackets, so the solution is x = -1 and x = 4.

These are the points where the plot of the equation crosses the x-axis.

Completing the Square is another method.

’the roots of a quadratic equation are found by setting the equation to 0. There are 3 ways to solve the equation: factorisation, using the quadratic formula and completing the square.’

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.

An easier solution is to find 2 numbers that multiply to give -3 and also add to give -4 (so this method can't be done). You can substitute values by trial and error (as above) but this will take a while. The best method here is to use the quadratic formula

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

Hi Robert,

I can see you have explained the roots of a quadratic equation beautifully here with an example. Kudos on that.

I feel that you can also use the generalised formula for finding the roots of a quadratic equation which goes as follows:

If your quadratic equation is of the form:

ax2+bx+c=0

Your roots will be X=(-b+/-sqrt(b^2-4ac))/2a

When using this for your problem, we get a= 1, b= -3, c= -4

Plugging these values in our quadratic formula we get: (3+/-5)/2

Which will give us the same roots as you have stated X=-1 and X=4.

Hope this helped your problem.

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

please check images

To derive the quadratic formula for solving equations of the form ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0, we can use the method of completing the square:

1. Start with the standard quadratic equation: ax^2+bx+c=0
2. Divide all terms by a (assuming a≠0) to normalize the coefficient of x^2: x^2+(b/a)x+c/a=0
3. Subtract c/a​ from both sides to isolate the quadratic and linear terms: x^2+(b/a)x=−(c/a)
4. Complete the square by adding (b/2a)^2 to both sides: x^2+(b/a)x+(b/2a)^2=−(c/a)+(b/2a)^2
5. Rewrite the left side as a perfect square and simplify the right side: (x+b/2a)^2=(b^2−4ac)/4a^2
6. Take the square root of both sides: x+b/2a=±sqrt(b^2−4ac)/2a​
7. Solve for x by isolating it: x=−b/2a ± sqrt(b^2−4ac)/2a​​
8. Combine the terms over a common denominator: x=(−b ± b^2−4ac)/2a

This is the quadratic formula, which provides the solutions to any quadratic equation where a≠0.

x = −b ± √(b2 − 4ac) / 2a