General Form of Quadratic Equations

is a form of mathematical model that states the relationship is equal to (=) and the highest rank is two. The quadratic equation has a general form:

ax2+bx+c=0 with a,b and c ϵ R and a≠0

example : 3x2+4x+1

Determining the Roots of a Quadratic Equation

Complete the quadratic equation, namely by looking for the value of x, there are several ways that can be done such as factoring, perfecting, and with the abc formula


Factoring the type ax2 +bx=0 can be done by separating x according to the distributive properties, namely:
x (ax + b) = 0
So x = 0 and ax + b = 0 Example:
Complete 3x2 +x=0
3x2 +x=0
x (3x + 1) = 0
x = 0 or
3x + 1 = 0
3x = -1
x = -1/3
so the solution is {-1/3, 0}

Factoring ax2+bx+c=0
For quadratic equations the type ax2 + bx + c = 0 can be factored in the form of:

with p and q integers or

It can be concluded#

b = p + q
c = pq / a or ac = pq

Factoring specifies a set of resolutions from 3x2-7x-6=0

With value a = 3, b = -7, c = -6
p + q = -7, p.q = 3. -6 = -18

to determine the value of p and q which when summed yielded -7 and when multiplying yielded -18 then got p = 2 and q = -9 so that:

(3x+2)(x-(9/3) )
x= -2/3

or x-3=0

so, the set of solutions {-2/3, 3}

Perfecting Quadratic

In completing the squares of the axillary equation ax2+bx+c=0 can be done with several stages:

1.     Move the constants to the right-hand segment
2.    If a ≠ 1, for both segments with a.

3. Add it to the right side and the left side quadratic of 1/2 times the coefficient x.

x2+ b/ax+ (b/2a)^2= c/a + (b/2a)^2

4. State it in the form of a perfect quadratic on the left side.

(x+b/2a)^2= c/a + (b/2a)^2

5. Then pull the root on the right side

x+b/2a=±√(c/a + (b/2a)^2 )


Messrs solution set for x2-6+2=0


x2-6+2=0 x2 - 6x= -2 x2 - 6x+ (-6/2)^2= -2 + (-6/2)^2 x2 - 6x+9=-2+9 (x-3)2=7 x-3=± √7 x=3± √7 x1 = 3+ √7 dan x2 = 3- √7

so the solution is {3- √7, 3 + √7}

Using the Abc Formula

Looking for the root values of the quadratic equation ax2+bx+c=0 is to use the following formula:

or can be written

so the roots are:


Determine the set of solutions for the 2x2 -5x -6=0 equation using the ABC formula


2x2 -5x -6=0

With the value a = 2, b = -5, c = -6 then

Determining the Root Types of Quadratic Equations with Discriminant.

In the abc formula obtained the formula:

in the formula above there is b2-4ac called discriminant (D). By using discriminan ( D= b2-4ac ), we can determine the root types from the quadratic equation, namely:

  1. If D> 0 then the quadratic equation ax2+bx+c=0 has 2 different real roots.
  2. If D <0 then the quadratic equation ax2+bx+c=0 has no real roots.
  3. If D = 0 then the quadratic equation ax2+bx+c=0 has the same 2 real roots.


Determine the type of root quadratic equation 3x2 -5x +2=0 , without first determining the roots.


With the values a = 3, b = -5 and c = 2 then
D = (-5)2-4.3.2
D = 1

Because D> 0 then the quadratic equation 3x2 -5x +2=0 has 2 different real roots.

Amounts and Results of the Roots of the Quadratic Equation

If you have obtained the roots of the quadratic equation, you can dance the results of the times and the sum of the roots of the quadratic equation. But there is an event where you are instructed to look for the results of the times or the number of root roots of the quadratic equation without getting the roots first.

Suppose the quadratic equation ax2+bx+c=0 has roots x1, x2 :

So, the formula for adding the roots of quadratic equations is:

x1 + x2 = -b / a

while for the root-root multiplication formula the quadratic equation is

So the formula for multiplying the roots of the quadratic equation is

x1.x2 = c / a

Symmetry forms of roots of quadratic equations

  1.     x12+ x22 =(x1+x2)2-2 x1x2
  2.     x13+ x23 =(x1+x2)3-3 x1x2(x1+x2)
  3.     x14+ x24 =( x12+ x22)3– 2(x1x2 )2


Foreheaded x1, x2 is the roots of the quadratic equation x2-3x+5=0 , specify the value of:

  1.      x1 + x2
  2.      x1x2
  3.    x12+ x22
  4.      (x1 / x2) + (x2 / x1)



a = 1, b = -3, and c = 5

menyusun persamaan kuadrat baru

Arrange new quadratic equations

Arrange equations when the roots are known

If a quadratic equation has the roots of x1 and x2, the quadratic equation can be expressed in the form of:

(x- x1) (x- x2) = 0


Known quadratic equations whose root roots are -2 and 3.


x1 = -2 and x2 = 3
(x – (- 2)) (x-3) = 0
(x + 2) (x + 3)

Arrange quadratic equations if the number and yield times of the roots are known.

If it is known that a quadratic equation has roots x1 and x2 and is known (x1 + x2) and (x1.x2), the quadratic equation can be formed as follows:

x2-( x1+ x2)x+(x1.x2)=0


Determine the quadratic equation whose roots are 3 and -1/2


x1 = 3 and x2 = -1/2

x1 + x2 = 3 -1/2 = 6/2 – 1/2 = 5/2

x1.x2 = 3 (-1/2) = -3/2

then the quadratic equation is

x2-( x1+ x2)x+(x1.x2)=0

x2– 5/2 x – 3/2=0 each segment multiplied by 2


That was the material friend from us about quadratic equations.

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