Bahan Ulangan harian bab 2

MAIN MATERIAL TEST CHAPTER 2

2.1 Function, Quadratic Equation and Quadratic Inequalities
2.1.2 Quadratic Equation
2. Solving Quadratic Equation
Examples :
Find the solution set of Quadratic Equation/ find the root of Quadratic Equation these following!
a. x^2 – 2x – 8 = 0
b. 2x^2 + 3x – 5 = 0
Answer :
a. x^2 – 2x – 8 = 0, by factoring, we get :
(x – 4)(x + 2) = 0
x = 4 or x = -2
So, the roots of QE are 4 or -2
So, the solution set is {4,-2}
2x^2 + 3x – 5 = 0, by factoring, we get :
(2x -5)(x+1) = 0
x= 5/2 or x = -1
So, the roots are 5/2 or -1
So, the solution set are {5/2 ,-1}
Composing quadratic equation when the roots are connected to the roots of other QE
Example :
Suppose x1 and x2 are the roots QE x2 + 2x – 2 = 0. Find the new QE which roots as follows : 〖x_1〗^2 and 〖x_2〗^2
Answer :
QE x^2 + 2x – 2 = 0, value of a = 1, b = 2, and c = -2
x1 + x2 = (-b)/a = -2 and x1x2 = c/a = -2
α and β are the roots of asked quadratic equation, then :
α = x1^2 and β = x2^2 , so :
α + β = x1^2 + x2^2 = (x1 + x2)^2 – 2x1x2
= ( -2)2 – 2(-2)
= 4 + 4 = 8
αβ = x1^2.x2^2 = (x1x2)^2 = (-2)^2 = 4
so, the new quadratic equation is x^2 – (α + β)x + αβ = 0
x^2 – 8x + 4 = 0
Roots quadratic equations
Example :
1)Without solving the QE find the type of roots of the following QE :
4x^2 + 4x + 1 = 0
Answer :
value of a = 4, b = 4 and c = 1
Mean that a > 0 and D = b^2 – 4ac = 4^2 – 4.4.1
D = 16 – 16 = 0
So, the roots of the quadratic equation are two similar, real and Rational Numbers

2) Find the value m so that equation mx^2 – 4mx + 2 = 0 has roots which
Are : real and different
Answer :
QE has two different roots and real if D > 0,hence :
Value of a = m, b = 4m and c = 2
D = b^2 – 4ac = (4m)^2 – 4.m.2= 16m^2 – 8m > 0
16m^2 – 8m = 0, by factoring, we get : 8m(2m – 1) = 0
8m = 0 or 2m – 1 = 0
m = 0 or m = ½
+ _ +


0 ½ 1

x= 1, so that 8.1(2.1 – 1) = 8, mean that (+)
So, the value m are m <> ½

2. 1.3 Quadratic Inequalities

Examples :
Find the solution sets of these inequalities using number line!
2x^2 – x – 1 ≤0
2x^2 – 5x – 3 ≤ 0
x^2 – 4x – 5 > 0
2x^2 – 3x – 4 ≥ x2 + x + 1
Answers :

2x^2 – x – 1 ≤ 0
Change the above equation into the corresponding quadratic equation (step 1)
Factorise the equation to find the zero value makers (step 2)
(2x +1)(x -1) = 0
x= - ½ or x = 1
Place the zero makers on a number line (step 3)




-1/2 0 1

Substitute any numbers to decide the sign of intervals.
Say we take x = 0 ( in interval -1/2 ≤ x ≤ 1
X = 0, then 2.02 – 0 – 1 = -1 ( -) (step 4)
Because the asked inequality has sign “≤” mean that ( - )


+++ _ _ _ _ _ _ _ _ + + + +

-1/2 0 1

So, the solution set is {x "" -1/2 ≤x ≤1,x "" R}
Notes and remember!!!

The graph crosses x-axis on two different points if D > 0
The graph touches x-axis on one point if D = 0
The graph does not cross or touch x-axis if D < 0
The graph is in top x-axis, if a> 0 and D < 0
The graph is under x-axis, if a < 0 and D < 0

D = b^2 – 4ac
The next material :
Drawing the graph with steps
The aplication of quadratic equations and functions
Find the solution set of quadratic inequalities by using the numbers line or by using the test point method



Good luck
Practice make perfect. Ok,....


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