Roots of a Quadratic Equation

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.

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

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.

There are several ways to solve the roots of a quadratic equation. Some of them include: factorisation method, completing the square, quadratic formular and graphical method. Here is an example:

Solve the quadratic equation 2x^2 - 5x + 3 = 0.

Using the quadratic formula

1. Identify the coefficients: a = 2, b = -5, c = 3.
2. Apply the formula: x = (-b ± √(b^2 - 4ac)) / 2a.
3. Calculate: x = (-(-5) ± √((-5)^2 - 4 * 2 * 3)) / (2 * 2).
4. Simplify: x = (5 ± √(1)) / 4.
5. Solve for two possible solutions:

x1 = (5 + 1) / 4 = 3/2

x2 = (5 - 1) / 4 = 1

Using the Factorisation method:

1. Observe that the equation doesn't factor easily into two linear expressions.
2. Consider trial and error to find two numbers that multiply to 3 (ac) and add up to -5 (b). In this case, the numbers are -1 and -3.
3. Rewrite the equation: 2x^2 - x - 4x + 3 = 0.
4. Group the terms: (2x^2 - x) + (-4x + 3) = 0.
5. Factor out common factors: x(2x - 1) - 2(2x - 1) = 0.
6. Combine like terms: (2x - 1)(x - 2) = 0.
7. Set each factor equal to zero and solve for x:

2x - 1 = 0 => x = 1/2

x - 2 = 0 => x = 2

There are also easy approachable factorisation methods to solve the same equation.

There are 4 different ways of finding the roots of the quadratic equation.

The first is using a factoring method

This method involves expressing the quadratic equation in a factored form as the product of two binomials. For x2−3x−4=0x2−3x−4=0, you look for two numbers that multiply to give the constant term (-4) and add to give the coefficient of xx (-3). The numbers -4 and 1 satisfy this condition, so you can write:

x2−3x−4=(x−4)(x+1)=0

x2−3x−4=(x−4)(x+1)=0

Setting each factor equal to zero gives the roots:

x−4=0⇒x=4

x−4=0⇒x=4

x+1=0⇒x=−1

x+1=0⇒x=−1

The second method is completing the square

This method involves manipulating the equation to form a perfect square trinomial, which can then be solved by taking the square root of both sides. For the given equation:

1. Move the constant term to the other side: x2−3x=4x2−3x=4.
2. Divide the coefficient of xx by 2, square it, and add it to both sides: x2−3x+(−32)2=4+(−32)2x2−3x+(2−3​)2=4+(2−3​)2.
3. This gives x2−3x+94=4+94x2−3x+49​=4+49​, or x2−3x+94=254x2−3x+49​=425​.
4. Now, you have (x−32)2=254(x−23​)2=425​.
5. Taking the square root of both sides gives x−32=±52x−23​=±25​.
6. Solving for xx gives x=32±52x=23​±25​, resulting in x=4x=4 or x=−1x=−1.

The third method is using the quadratic formula

The quadratic formula is a universal method that can solve any quadratic equation. The roots of the equation ax2+bx+c=0ax2+bx+c=0 can be found using:

x=−b±b2−4ac2a

x=2a−b±b2−4ac

​​

For x2−3x−4=0x2−3x−4=0, a=1a=1, b=−3b=−3, and c=−4c=−4, so:

x=−(−3)±(−3)2−4(1)(−4)2(1)=3±9+162=3±252=3±52

x=2(1)−(−3)±(−3)2−4(1)(−4)

​​=23±9+16​​=23±25

​​=23±5​

Thus, x=4x=4 or x=−1x=−1.

The last method is using a graphical method

The roots of the quadratic equation are the x-coordinates of the points where the graph of the equation intersects the x-axis. You can graph the quadratic function y=x2−3x−4y=x2−3x−4 and find the points where y=0y=0. This method visually represents the solutions but requires graphing tools or software.

factorisation in an equation aX^2+bX+c, where a=1, the two factors will add up to b and multiply to make c then making (X+_)(X+_)=0. Causing the roots to be the opposite signs of the factors given.

The quadratic formula can be used to solve quadratic equations that you are unable to factorise. Using the formula (-b ± sqrt(b^2 - 4ac)) / 2a. Where a quadratic is in the form ax^2 + bx + c = 0

substituting the values of a=1 b=-3 and c=-4

x = (-(-3) ± √((-3)² - 4(1)(-4))) / (2(1))

when simplified results in

x = (3 ± √(9 + 16)) / 2

x = (3 ± √25) / 2

Hence

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

x = -1/4

x = (-b ± √(b^2 - 4ac)) / (2a)

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

What's the question

Use the quadratic formula to get answer by direct substitution

The use of the discriminant.

Factorisation is also key as there are 2 numbers that multiply to -4 and that add to -3.

Completing the square method, Factorisation method, Quadratic Formula method are other different method for finding the roots or zeros of quadratic equations.