Quadratic Functions

Quadratic Functions

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Quadratic functions

A quadratic function has the general form

 

\(y=ax^2 + bx + c\)

 

The quadratic function is also known as a second degree polynomial.

 

1. Solving quadratic equation

There are three methods of solving quadratic equations.

 

2. Factorization

When the factors are multiplied that is expanded, you get back the quadratic equation. 

 

Example

 

\(x^2+5x+6=0 \)

\(x+3x+2=0\)

\(x=-3 \text { and } x=-2\)

 

Example

 

\(2x^2-x-1=0 \)

\(x-12x+1=0 \)

\(x=-12 \text { and } x=1 \)

 

3. Completing the square

For the quadratic expression  

 

\(y=ax^2 + bx + c\)

 

you complete the square by first factoring out a :

 

\(a(x^2+{b \over a} \times x +{c \over a})\)

 

and the using the formula :

 

\(a[(x+{b \over 2a}^2)+{c \over a}-{b^2 \over 4a^2}]\)

 

Example

For

 

\(x^2+5x+6=0\)

 

the coefficients are a=1, b=5 and c=6. Then, using the above formula, you have

 

\((x+{5 \over 2(1)})^2+6-{5^2 \over 4(1)^2}=0\)

\((x+{5 \over 2})^2+6-{25 \over 4}=0\)

\( (x+{5 \over2 })^2-{1 \over 4}=0\)

\((x+{5 \over 2})^2={1 \over 4}\)

\(x+{5 \over 2}-{1 \over 2} \text { and } x+{5 \over 2}={1 \over 2}\)

\(x=-{1 \over 2}-{5 \over 2} \text{ and } x={1 \over 2}-{5 \over 2}\)

\(x=-3 \text{ and } x=-2 \)

 

Example

For

 

\(2x^2-x-1=0\)

 

the coefficients are a=2, b=-1 and c= -1. Then, using the  completing square method, you have

 

\( 2(x^2-{1 \over 2}x-{1 \over 2})=0\)

\( 2[(x-{1 \over 4})^2-{1 \over 2}-{1 \over 16}]=0\)

\( 2[(x-{1 \over 4})^2-{9 \over 16}]=0\)

\( (x-{1 \over 4})^2-{9 \over 16}=0\)

\( (x-{1 \over 4})^2={9 \over 16}\)

\( x-{1 \over 4}=-{ 3 \over 4} \text { and } x-{1 \over 4}={3 \over 4} \)

\( x=-{3 \over 4}+{1 \over 4} \text { and } x={3 \over 4} +{1 \over 4}\)

\(x=-{1 \over 2} \text { and } x=1 \)

 

4. Quadratic formula

For the quadratic equation  

 

\(ax^2+bx+c=0 \)

 

the quadratic formula to find x is :

 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

 

Example

For

 

\(x^2+5x+6=0\)

 

where you have seen that the coefficients are a=1, b=5 and c=6,

 

\(x = {-5 \pm \sqrt{5^2-4(1)(6) }\over 2(1)}\)

\(x = {-5 \pm \sqrt{25-24} \over 2}\)

\(x = {- 5\pm \sqrt{1} \over 2}\)

\(=-{5-1 \over 2}, {-5+1 \over 2}\)

\(={-6 \over 2}, {-4 \over 2}\)

\(=-3,-2 \)

 

5. Roots of quadratic equation

The number of roots and the type of roots of the quadratic equation

 

\(ax^2+bx+c=0\)

 

can be determined by the discriminant D=b2-4ac.

  • If D>0, the quadratic equation has two different real roots
  • If D=0, the quadratic equation has one real root, repeated twice
  • If D<0, the quadratic equation has two different complex roots
 

Example

Notice that in 

 

\(x^2+5x+6=0\)

 

given above, 

 

\(D=5^2-4(1)(6)=1>0\)

 

and so you had two different real roots x=-3 and x=-2.

 

Tag Factorization Discriminant Roots

Prior knowledge

1.  What are quadratic functions?
2.  How do you solve quadratic equations?
3.  What are the types of roots of quadratic equations?

1. 

Which is not a quadratic equation?  

2. 

Which are the factors of 3x^2 - 14x - 5 ?

3. 

What does the discriminant of the quadratic equation 2x^2 + 4x - 5 = 0 tells you?

4. 

test

Reflection

1.  What are the solution methods of quadratic equations?
2.  Explain what a discriminant is and how the discriminant can be used to determine the types of roots of a quadratic euqation.
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