Roots of a Quadratic Equation

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 roots of a quadratic equation ax2 + bx + c = 0 can be found using the quadratic formula that says x = (-b ± √ (b2 - 4ac)) /2a.

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)

The roots of a quadratic equation are the values of 'x' that make the equation true. They're like the solutions to a math puzzle, where the equation is in the form 'ax² + bx + c = 0'. These roots are the 'x' values that satisfy the equation and make both sides equal.

let's solve a quadratic equation using factorisation, a method to find the roots without using the quadratic formula:

Suppose we have the quadratic equation: x² - 5x + 6 = 0

Step 1: Write down the equation: x² - 5x + 6 = 0

Step 2: Try to factorise the quadratic expression on the left side of the equation. We need two numbers that multiply to the last term (6) and add up to the middle term (-5). Those numbers are -2 and -3 because (-2) * (-3) = 6 and (-2) + (-3) = -5.

Step 3: Rewrite the middle term (-5x) using the two numbers we found in the previous step: x² - 2x - 3x + 6 = 0

Step 4: Group the terms and factorise by grouping: x(x - 2) - 3(x - 2) = 0

Step 5: Factorise out the common factor, which is (x - 2): (x - 2)(x - 3) = 0

Step 6: Set each factor equal to zero and solve for 'x': x - 2 = 0 --> x = 2 x - 3 = 0 --> x = 3

So, the roots of the quadratic equation x² - 5x + 6 = 0 are x = 2 and x = 3.

In this example, we factorised the quadratic expression and then set each factor equal to zero to find the roots of the equation.

Alternatively, we may find the roots of the above quadratic equation by using the quadratic formula, as follows:

The quadratic formula is: x = (-b ± √(b² - 4ac)) / 2a

In our equation, a = 1, b = -5, and c = 6.

Substitute these values into the formula:

x = (-(-5) ± √((-5)² - 4(1)(6))) / (2 * 1)   i.e.  x = (5 ± √(25 - 24)) / 2   i.e. x = (5 ± √1) / 2  i.e.   x = (5 ± 1) / 2

So, the two possible solutions are: x = (5 + 1) / 2 = 6 / 2 = 3    and x = (5 - 1) / 2 = 4 / 2 = 2

The roots of the quadratic equation x² - 5x + 6 = 0 are x = 3 and x = 2, which matches the solution we found earlier using factorisation.

You can also use completing the square

My explanation is shown below:

X^2 - 3x - 4 = 0

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

x=4 or x=-1

I don't think you are refering to the derivative, which is the resultant equation when differentiating an equation.

I think you are talking about the roots of the equation, which are -1 and 4, as you have correctly stated.

The formula for solving a quadratic equation is:

x = (-b +/-sq root (b2 -4ac))/2a

If we compare the equation, x2 - 3x - 4 = 0  to ax2 + bx + c

we see that a = 1, b= -3, c=-4

So, substituting into the quadratic equation formula, we get:

-b=3, b2= -3 x -3 = 9, 4ac= 4x1 x -4 = -16, 2a = 2

x = (-b +/-sq root (b2 -4ac))/2a

x = (3 +/- sq root (9+16))/2

x =(3 +/- sq root 25)/2

x =(3 +/- 5)/2

So the solution to the equation is either x = (3 +5)/2 = 8/2 = 4 

or x = (3-5)/2 = -2/2 = -1

Here is a quick method:

ax^2+b+c=0

Find two numbers that add together to equal b and times together to make c. These 2 numbers y & z will be in the quadratic form here:

(x +y)*(x+z)=0

solve x from this equation.

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 derivative of a quadratic equation (not the quadratic formula) results in a linear equation. For instance, for the quadratic equation ��2+��+�=0

ax2

+bx+c=0, its derivative is 2��+�

2ax+b.

The roots of a quadratic equation are the values of �

x that make the equation equal to zero. In your example, for the quadratic equation �2−3�−4=0

x2

−3x−4=0, the roots are �=−1

x=−1 and �=4

x=4

The quadratic formula x=2a

−b±b2

−4ac

If you have further questions or specific points you'd like clarification on, feel free to ask!

You can find the roots of a quadratic equation with many different methods:

• Quadratic equation; namely x = [-b+-sqrt(b^2-4ac)] / 2a
• Factorising; e.g. x^2+7x+12=0 therefore (x+4)(x+3) = 0, so either x+4=0 or x+3=0, therefore x=-4 or -3
• Completing the square; e.g. x^2+2x-5 = 0, therefore (x+1)^2-5-1 = 0; (x+1)^2-6 = 0; (x+1)^2=6; (x+1) = +-sqrt(6); x = -1 +- sqrt(6)
• Graphically; e.g. if ax^2 + bx + c = 0, you can graph f(x) = ax^2 + bx + c, and find solutions for x where f(x) = 0.