Showing posts with label Precalculus. Show all posts
Showing posts with label Precalculus. Show all posts

Mar 11, 2010

Discriminant

We have already studied the quadratic formula,
Quadratic formula
The term which is inside the square root symbol is called the discriminant.
It is used to determine the nature of the roots of a quadratic equation. We can also determine the number of real roots for a quadratic equation with this number. The following table will give us the relation between the discriminant and the nature of the roots.


Discriminant


Nature of roots


Number of real roots


b2 - 4ac > 0


Two real roots


2


b2 - 4ac = 0


Double root


1


b2 - 4ac < 0


No real roots


0
Example 1:
Find the discriminant value for the equation x2 + 5x + 6 = 0 and determine the number of real roots.
Solution:
Step 1:
Given equation is x2 + 5x + 6 = 0.
It is in the form ax2 + bx + c = 0, where a = 1, b = 5, and c = 6.
Step 2:
The discriminant = b2 - 4ac
b2 - 4ac = (5)2 - 4(1)(6) = 25 - 24 = 1 > 0
=> The equation has two real roots.
=> The graph of this equation will touch the x-axis twice.
Step 3:
So, the given equation has two real roots.
Example: 2
Find the discriminant value of x2 - 12x + 36 = 0 and determine the number of real roots.
Solution:
Step 1:
Given equation is x2 - 12x + 36 = 0.
It is in the form ax2 + bx + c = 0 where, a = 1, b = -12, and c = 36.
Step 2:
Discriminant = b2 - 4ac = (-12)2 - 4(1)(36) = 144 - 144 = 0
=> The equation has a double root.
=> The graph of this equation touches the x-axis in only one point.
Step 3:
 So, the given equation has a double root.
Example: 3
Find the discriminant value 2x2 + x + 3 = 0 and determine the number of real roots.
Solution:
Step 1:
The given quadratic equation is 2x2 + x + 3 = 0.
It is in the form ax2 + bx + c = 0 where, a = 2, b = 1, and c = 3.
Step 2:
b2 - 4ac = (1)2 - 4(2)(3) = 1 - 24 = -23 < 0
=> The equation has no real roots.
=> The graph of this equation does not touch the x-axis.
Step 3:
So, the given quadratic equation has no real roots.
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Feb 20, 2010

Best method to solve quadratic equations

 So far, we have studied the four methods to solve a quadratic equation. How do we select the best method to solve among these four methods?
The following table will guide us to choose the best method to solve.


Method


When to choose




Use if we need only the appropriate solutions. However, we can use it always.




Use if we could easily determine the factors or the constant term is zero.




Use if the equation is of the form x2 + bx + c = 0 and b is even. However, we can use it always.




We can use this method to any type of quadratic equation. In some cases, the other methods will be easy to solve.
Example:
The product of two consecutive positive odd integers is 99. Find those two numbers.
Solution:
Step 1:
Given that the product of two consecutive positive odd integers is 99.
We are asked to find the two consecutive positive odd integers.
Let the first integer be x.
Then the consecutive positive odd integer is x + 2.
Given that the product of x and x + 2 = 99
=> x(x + 2) = 99
=> x2 + 2x - 99 = 0
So, we arrived at a quadratic equation x2 + 2x - 99 = 0.
Step 2:
We can solve the quadratic equation by using factoring.
x2 + 11x - 9x - 99 = 0
=> x(x + 11) - 9(x + 11) = 0
=> (x + 11)(x - 9) = 0
=> x + 11 = 0 or x - 9 = 0
=> x = -11 or x = 9
Step 3:
Since we are looking for a positive number, neglect -11.
So, x = 9.
If x = 9, then x + 2 = 9 + 2 = 11.
Step 4:
Hence, the two consecutive positive odd numbers are 9 and 11.
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Feb 16, 2010

Graphing Method: Quadratic Equations

In this method, we are going to find the solution or roots of a quadratic equation graphically. The graph of a quadratic equation is always a parabola. The points where the graph of the parabola cuts the x-axis are called the roots of the corresponding quadratic equation.
To graph a quadratic equation, we should graph the related quadratic function. For example, if the quadratic equation is x2 + 2x + 5 = 0, then the related quadratic function is f(x) = x2 + 2x + 5.
Let us solve few problems on quadratic equations by graphing.
Example 1:
Solve x2 + x - 6 = 0 by graphing.
Solution:
Step 1:
Given equation is x2 + x - 6 = 0.
Here a = 1, b = 1, and c = -6.
Graph the related function f(x) = x2 + x - 6.
The equation of the axis of symmetry is x = - b/2a
x = - b/2a = - 1/2(1)  = - 1/2 = -0.5
=> x = -0.5 is the axis of symmetry of the equation.
Step 2:
When x = -0.5,
f(x) = f(-0.5) = (-0.5) 2 + (-0.5) - 6 = 0
=> f(-0.5) = 0.25 - 0.5 - 6 = -6.25
So, the coordinates of the vertex are (-0.5, -6.25).
Step 3:
Now, make the table with all other points.
Put x = -4, -3, -2, 0, 1, 2, 3 and find f(x).
When x = -4,
f(x) = f(-4) = (-4)2 + (-4) - 6 = 16 - 4 - 6 = 6
When x = -3,
f(x) = f(-3) = (-3)2 + (-3) - 6 = 9 - 3 - 6 = 0
When x = -2,
f(x) = f(-2) = (-2)2 + (-2) - 6 = 4 - 2 - 6 = -4
When x = -1,
f(x) = f(-1) = (-1)2 + (-1) - 6 = 1 - 1 - 6 = -6
When x = 0,
f(x) = f(0) = (0)2 + (0) - 6 = 0 + 0 - 6 = -6
When x = 1,
f(x) = f(1) = (1)2 + (1) - 6 = 1 + 1 - 6 = -4
When x = 2,
f(x) = f(2) = (2)2 + 2 - 6 = 4 + 2 - 6 = 0
When x = 3,
f(x) = f(3) = (3)2 + 3 - 6 = 9 + 3 - 6 = 6


x


-4


-3


-2


-1


-0.5


0


1


2


3


f(x)


6


0


-4


-6


-6.25


-6


-4


0


6
Hence, the ordered pairs of the quadratic function are (-4, 6), (-3, 0), (-2, -4), (-1, -6), (-0.5, -6.25), (0, -6), (1, -4), (2, 0), and (3, 6).
Step 4:
Now, plot the ordered pairs in a coordinate plane and join the points.
Graph of Quadratic Equation
The point where the related function f(x) = 0 is the solution to this quadratic equation. This occurs at the x-intercepts. The x-intercepts are the points where the graph touches the x-axis. So, the graph has -3 and 2 as x-intercepts.
So, the solutions are -3 and 2.
Step 5:
Check:
We can also check our solutions by factoring the given equation.
x2 + x - 6 =0
=> x2 + 3x - 2x - 6 = 0
=> x(x + 3) - 2(x + 3) = 0
=> (x + 3)(x - 2) = 0
Using the zero product property,
x + 3 = 0 or x - 2 = 0
x = -3 or x = 2
Step 6:
So, the solutions of the equation are -3 and 2.
Example 2:
Solve x2 - 10x + 25 = 0 by graphing.
Solution:
Step 1:
Given equation is x2 - 10x + 25 = 0.
Here a = 1, b = -10, and c = 25.
Graph the related function f(x) = x2 - 10x + 25.
The equation of the axis of symmetry is x = - b/2a
x = - b/2a = - 10/2(1) = - b/2a = 5
=> x = 5 is the axis of symmetry of the given equation.
Step 2:
When x = 5,
f(x) = f(5) = (5)2 - 10(5) + 25 = 25 - 50 + 25 = 0
=> f(x) = 0
So, the coordinates of the vertex are (5, 0).
Step 3:
Now, make the table with all other points.
Put x = 2, 3, 4, 6, 7, 8 and find f(x).
When x = 2,
f(x) = f(2) = (2)2 - 10(2) + 25 = 4 - 20 + 25 = 9
When x = 3,
f(x) = f(3) = (3)2 - 10(3) + 25 = 9 - 30 + 25 = 4
When x = 4,
f(x) = f(4) = (4)2 - 10(4) + 25 = 16 - 40 + 25 = 1
When x = 6,
f(x) = f(6) = (6)2 - 10(6) + 25 = 36 - 60 + 25 = 1
When x = 7,
f(x) = f(7) = (7)2 - 10(7) + 25 = 49 - 70 + 25 = 4
When x = 8,
f(x) = f(8) = (8)2 - 10(8) + 25 = 64 - 80 + 25 = 9


x


2


3


4


5


6


7


8


f(x)


9


4


1


0


1


4


9
So, the ordered pairs of the quadratic function are (2, 9), (3, 4), (4, 1), (5, 0), (6, 1), (7, 4), and (8, 9).
Step 4:
Now, plot the ordered pairs in a coordinate plane and join the points.
The resultant graph will be as shown.
The point where the related function f(x) = 0 is the solution to this quadratic equation. This occurs at the x-intercepts. The x-intercepts are the points where the graph touches the x-axis.
Notice that the vertex of the parabola is the x-intercept.
Hence, the solution is 5.
Step 5:
Check:
Let us check it by factoring.
Factor the equation x2 - 10x + 25 = 0.
x2 - 5x - 5x + 25 = 0
=> x(x - 5) - 5(x - 5) = 0
=> (x - 5)(x - 5) = 0
=> x - 5 = 0 or x - 5 = 0
=> x = 5 or x = 5
So, there are two roots but they are identical.
So, this quadratic equation has only one root called 5 as double root.
Step 6:
So, the solution is 5.
Example 3:
Solve x2 + 2x + 5 = 0 by graphing.
Solution:
 Step 1:
Given equation is x2 + 2x + 5 = 0.
Here, a = 1, b = 2, and c = 5.
Graph the related function f(x) = x2 + 2x + 5.
The equation of the axis of symmetry is x = - b/2a
x = - b/2a= - 2/2(1) = -1
=> x = -1 is the axis of symmetry of the given equation.
Step 2:
When x = -1,
f(x) = f(-1) = (-1)2 + 2(-1) + 5 = 1 - 2 + 5 = 4
So, the coordinates of the vertex are (-1, 4).
Now, make the table with all other points.
Put x = -4, -3, -2, 0, 1, 2 and find f(x).
When x = -4,
f(x) = f(-4) = (-4)2 + 2(-4) + 5 = 16 - 8 + 5 = 13
=> f(-4) = 13
When x = -3,
f(x) = f(-3) = (-3)2 + 2(-3) + 5 = 9 - 6 + 5 = 8
=> f(-3) = 8
When x = -2,
f(x) = f(-2) = (-2)2 + 2(-2) + 5 = 4 - 4 +5 = 5
=> f(-2) = 5
When x = 0,
f(x) = f(0) = (0)2 + 2(0) + 5 = 5
=> f(0) = 5
When x = 1,
f(x) = f(1) = (1)2 + 2(1) + 5 = 1 + 2 + 5 = 8
=> f(1) = 8
When x = 2,
f(x) = f(2) = (2)2 + 2(2) + 5 = 13
=> f(2) = 13


x


-4


-3


-2


-1


0


1


2


f(x)


13


8


5


4


5


8


13
So, the ordered pairs of the quadratic function are (-4, 13), (-3, 8), (-2, 5), (-1, 4), (0, 5), (1, 8), and (2, 13).
Step 3:
Now, plot the ordered pairs in a coordinate plane and join the points.
The resultant graph will be as shown.
Quadratic equation by graphing method
The point where the related function f(x) = 0 is the solution to this quadratic equation. This occurs at the x-intercepts. The x-intercepts are the points where the graph touches the x-axis. So, the graph has no x-intercept.
Step 4:
So, there are no real number solutions for the given quadratic equation.
Here is some practice questions: Quadratic equations
To check your result: Quadratic equation solver
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Feb 15, 2010

Quadratic Formula

 
The solution of a quadratic equation in the standard form ax2 + bx + c = 0, where a, b, and c are real numbers and a 0, can be find using the quadratic formula given below.
 quadratic formula
Let us work out a few examples.
Example 1:
Use quadratic formula to solve x2 - 2x - 35 = 0.
Solution:
Step 1:
Given equation is x2 - 2x - 35 = 0.
Here, a = 1, b = -2, and c = -35.
The quadratic formula is
quadratic formula
Step 2:
Substitute the values of a, b, and c in the quadratic formula to find the value of x.
Solution for quadratic equation
Step 3:
So, the solution set is {5, -7}.
Example 2:
Use quadratic formula to solve 4x2 + 2x = 17.
Solution:
Step 1:
Given equation is 4x2 + 2x = 17.
It can be written as 4x2 + 2x - 17 = 0. The equation is in the form of ax2 + bx + c = 0, where a = 4, b = 2, and c = -17.
Step 2:
The quadratic formula is
quadratic formula
Substitute the values of a, b, and c in the quadratic formula to find the value of x.
Solution for quadratic equation
Step 3:
So, the solutions are 1.83 and -2.33.
Practice questions:
Solve the given quadratic equations using quadratic formula:
1. x2 – x – 12 = 0
2. 2x2 – 9x – 5 = 0
3. 6x2 + x – 48 = 0
4. x2 – 49 = 0
5. x2 + 3x – 18 = 0 
Quadratic formula is a best method to solve any quadratic equations. Practice with more quadratic equations using quadratic formula to solve it.
You can check your answer using quadratic equation calculator.
Note: Calculator shows answer in decimals not in fraction.
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