Roots of a Quadratic Equation

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)

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

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)

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: