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.

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 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.

Two values of x in quadratic equation are called roots of quadratic equation. We can solve quadratic equation by four methods, factorising, by completing square, Using quadratic formula and by graphical method. Roots of quadratic equation always satisfies 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.

A quadratic equation is an equation of the form:

ax² + bx + c = 0

where 'a', 'b', and 'c' are constants, and 'a' is not equal to zero.

I use this symbol ^ to show the presence of a power/exponent. You could complete the square so divide the coefficient of x by 2 to get -3/2. then rewrite the equation as (x-1.5)^2 which expands to x^2- 3x +2.25. So we need to get from 2.25 to -4 so what we do is either -2.25 and then -4 or we could do that in 1 step and -6.25. Hence we get x^2 - 3x - 4 = 0 written as (x-1.5)^2 -6.25. When this equals 0, we rearrange and then we get the square root of 6.25 (posotive OR negative which eventually helps to give us both the solutions of x=-1 and x=4) plus 1.5 = x. therefore x = +/- (square root of 6.25) and then +1.5. So we get x = 4 and x=-1

Completing the square method is another, iterative method too is also another and factorisation method.

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

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 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

The quadratic equation is given as

x= (-b plus or minus square root b^2-4ac) / 2a

a is the coefficient of x^2

b is the coefficient of x

and c is the number without x.

You insert the values for a, b and c into the quadratic equation and you find the value for x which is the root.

If (b^2 - 4ac) is greater than 0, there are two distinct roots.

If (b^2 - 4ac) is equal to 0, there is only 1 root.

If (b^2 - 4ac) is less than 0, there are no real roots (only complex roots).