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Example of Quadratic Equation Problem – After previously we discussed about Examples of Inverse Function Problems. This time, together with the material we will discuss the material about the quadratic equation formula, we will describe in detail and complete the meaning of quadratic and its solution, understanding quadratic equations, various roots of quadratic equations and the properties of roots of quadratic equations along with examples of problems. Alright, here’s the explanation.

## Definition of Square

In mathematics, Square is a root of the number x equal to the number r such that r2 = x, or, in other words, the number r which when squared will yield the product of the multiplication of the number itself) equal to x. Equality is a square of variables and has the highest level of two. The general form is: Where a, b, are coefficients, and c is a constant, and a 0. The solution or solution of an equation is called the roots of the quadratic equation.

## Kinds of Roots of Quadratic Equations

In order to determine the roots of a quadratic equation, we can use the formula D = b2 – 4ac. if the value of D has been formed, of course it will be easier to find the roots. Here are some general types of quadratic equations:

### At Real Root (D ≥ 0):

Example:

Determine the type of root of the following equation:

The solution :
From the equation = x2 + 4x + 2 = 0

Is known :

• D = b2 – 4ac
• D = 42 – 4(1)(2)
• D = 16 – 8
• D = 8 ( D> 8, so the conclusion is that the roots are the same as real roots but different )

»At the real root is equal to x1 = x2 when D = 0

Example :
Prove when in this equation has twin real roots:

The solution :
From = 2 × 2 + 4x + 2 = 0

Is known :

• D = b2 – 4ac
• D = 42 – 4(2)(2)
• D = 16 – 16
• D = 0 ( D = 0, then it is proved that the real roots are twins )

### Imaginary or Unreal Root ( D < 0 )

Example:
Determine the type of root of the following equation:

Solution:
In the equation at = x2 + 2x + 4 = 0

Is known :

• D = b2 – 4ac
• D = 22 – 4(1)(4)
• D = 4 – 16
• D = -12 ( D<0, then the roots are not real )

### Rational Root ( D = k2 )

Example :
Determine the type of root of the following equation:

Solution:

In the Equation results at = x2 + 4x + 3 = 0

Is known :

• D = b2 – 4ac
• D = 42 – 4(1)(3)
• D = 16 – 12
• D = 4 = 22 = k2 (From D=k2= 4 So the conclusion of the root of the equation is rational)

## Properties of the Roots of a Quadratic Equation

The following are types of Quadratic Equations:

In determining which quadratic equation is determined from the results of the discriminant value (D = b2 – 4ac) which distinguishes the types of roots – the roots of a quadratic equation into 3, namely:

• If D> 0, then the conclusion is that this equation has two different real roots.
• When D is a perfect square, then both are rational roots.
• If D is not a perfect square, then it can be concluded that both are irrational roots.
• When D = 0, so it can be concluded that the equation has two roots which are twin roots, real, and rational.
• If D
• Extension form for real roots:

x1 + x2 > 0

x1 x2 > 0

x1 + x2 < 0

x1 x2 > 0

x1 x2 < 0

x1 x2 > 0

### The Two Roots Are Opposite

x1 + x2 = 0 (b = 0)

x1 x2 < 0

### The Two Roots Are Opposite of Each Other

x1 + x2 = 1 (c = a)

## Example of Quadratic Equation Problem

Example No1:

Show that x1=4 and x2=-4 are the roots of the equation x²-16=0 !

Discussion :

We substitute the value of x1=4 in the equation x²-16=0, then

4²-16=16-16=0 (true)

We substitute the value of x2=-4 in the equation x²-16=0, then

(-4)²-16=16-16=0 (true)

because based on the substitution above, the sentence is correct, so x1=4 and x2=-4 are the roots of the equation x²-16=0.

Contoh No2:

Investigate whether x = 3 is the root or solution of the equation 5x²-13x+6 = 0?

Discussion :

We substitute the value of x=3 in the equation 5x²-13x+6=0, then

5(3)²-13(3)+6=5(9)-39+6=45-39+6=12 (salah)

Because it produces an incorrect sentence, then x=3 is not the root of the equation 5x²-13x+6=0.

ContohNo3 :

One of the roots of the equation y²-6y+2p = 0 is y = -2. Determine the value of p!

Discussion :

we substitute y=-2 into the equation y²-6y+2p=0, then

(-2)²-6(-2)+2p= 0

4   +   12   + 2p = 0

16   +  2p  = 0

2p  = -16

p = -8

So, p value = -8

Example NO4:

Find the roots of the following equation!

a. 2x (x-5) = 0

b. (3x-4)(x+2)=0

Discussion

a. 2x (x-5) = 0

⇔ 2x = 0

⇔ x = 0

or

⇔ x-5 = 0

⇔     x = 5

The roots are x1 = 0 and x2 = 5

b. (3x-4)(x+2)=0

⇔ 3x-4 = 0

⇔ 3x = 4

⇔        x = 4/3

or

⇔ x+2 = 0

⇔      x  = -2

the roots are x1 = 4/3 and x2 = -2

Example No. 5:

Find the roots of the following equation!

a. 4x² =25

b. (x+5)² = 36

Discussion :

a.  4x²  = 25

⇔(2x)²= ±√25

⇔ 2x = ± 5

⇔      x   = ± 2½

the roots are x1 = 2½ and x2 = -2½

b. (x+5)² = 36

⇔  x+5    = ±√36

⇔  x+5    = ± 6

⇔       x    = -5 ± 6

x1 = -5+6 and x2 = -5-6

⇔ x1 = 1 x2 = -11

the roots are x1 = 1 and x2 = -11.

Example No. 6:

Determine the solution of the following equations by factoring!

a. 2x²+10x = 0

b. 4x²-9 = 0

c. x²-6x-40 = 0

Discussion :

a. 2x²+10x = 0

⇔ 2x(x+5) = 0

2×1 = 0 and x2+5 = 0

⇔ x1 = 0 x2 = -5

the solutions are x1 = 0 and x2 = -5

b.      4x²  –    9    = 0

⇔ (2x + 3) (2x-3) = 0

2 x1 + 3 = 0 and 2 x2 – 3 = 0

⇔ 2 x1 = -3 2 x2 = 3

⇔ x1 = -3/4 x2 = 3/2

the solutions are x1 = -3/4 and x2 = 3/2

c.  x² – 6x – 40 = 0

⇔ (x-10)(x+4) = 0

x1-10 = 0 and x2+4 = 0

⇔ x1 = 0 x2 = -4

the solutions are x1 = 0 and x2 = -4 