1.
Algebra
Electronic Presentation: Expert Voices
Pre Cal 30S
January 22nd 2010
Emelda Iradukunda
Haben Gabir
Aruni Perera
2.
Algebra
Table of Contents
• Absolute Values
• Solving for Roots by Completing the Square
• Solving for “p” using the Quadratic Formula
• Generating Equations with given Roots
• The Discriminant - The Nature of Roots
• Solving Rational Equations
• Solving Radical Equations
4.
Algebra
Solving for Roots by Completeing the Square
5.
Algebra
Solving for “p” using Quadratic Formula
Equation:
4x - 10x² + 4 = 0
1) Substitute ‘p’ in place of x², therefore:
p = x²
=> 4p² - 10p + 4 = 0
2) Solve for ‘p’ using the quadratic formula
Quadratic Formula:
1x² + 1x + 1 = 0
a b c
6.
Algebra
Solving for “p” using Quadratic Formula
=> -(-10) ± √(-10)² - 4(4)(4)
2(4)
=> 10 ± √ 100 - 64
8
=> 10 ± √ 36
8
=> 10 ± 6
8
Now we have 2 possible solutions:
p1 = 16 = 2 p2 = 4 = 1
8 8 2
x1 = ± √2 x2 = √1 = 1 = ± √2
√2 √2 2
7.
Algebra
Generating Equations with given Roots
Formula for Equation:
x² - (sum of roots)x + (product of roots) = 0
Roots:
4 + √6 and 4 - √6
1) Given the roots, we need to find the sum of the roots.
4 + √6 + 4 - √6 = 8
2) Given the roots, we need to find the product of the roots.
4 + √6 * 4 - √6 = 16 - 4√6 + 4√6 - 6 = 16 - 6 = 10
8.
Algebra
Generating Equations with given Roots
3) Substitute the sum of the roots and product of the roots into the
formula.
x² - (sum of roots)x + (product of roots) = 0
=> x² - 8x + 10 = 0
9.
Algebra
The Discriminant - The Nature of Roots
To find the value of the discriminant, we must use the formula:
b² - 4ac
Equation:
2x²- 2x - 6 = 0
1) Substitute the equation into the formula.
=> (-2)² - 4 (2)(-6)
=> 4 - (-48) = 52
2) Determine the nature of the roots of this value using the following
rules:
10.
Algebra
The Discriminant - The Nature of Roots
b² - 4ac > 0
- there are two roots
- if the value is a perfect square, the roots are rational
- if the value is not a perfect square, the roots are
irrational
b² - 4ac = 0
- there is only one root
- the function only crosses the x-axis at the vertex of
the parabola
b² - 4ac < 0
- the roots of the quadratic function are imaginary
- the parabola does not cross the x-axis at any point
11.
Algebra
The Discriminant - The Nature of Roots
=> 52 = 2 irrational roots
3) Find the exact value of the roots using this formula:
- b ± √(value of discriminant)
2a
=> - (-1) ± √52 = 1 ± √52
2 (2) 4
Now we have 2 roots:
r1 = 1 + √52 r2 = 1 - √52
4 4
4) Draw the quadratic on a graph.
12.
Algebra
Solving Rational Equations
Solving rational equations steps:
Completely factorize the equation
List all impossible values of x ( values that will make the
denominator equal to 0 )
Get rid of any factors that cancel each other out
Find the LCD and multiply it by both sides of equation ( this is
done to get rid of the denominators )
13.
Algebra
Solving Rational Equations
Solve:
The non-permissible values are: 2 (x can’t equal to 2)
Nothing to factor, it is already factored.
So now we multiply by the LCD, which in this case is x-2.
3x = 2x - 4 + 6
3x - 2x - 2 = 0
X-2=0
X=2
14.
Algebra
Solving Rational Equations
solve:
Step1:
The LCD is (x+1)(x-1)
Non-permissible Values are x=1,-1
now we multiply both side by this
4x + 1 = 2x - 2 - x² - 1
x² + 4x + 4 = 0
(x+2)(x+2) = 0
x = -2
15.
Algebra
Solving Radical Equations
Radical equation is an equation that contains radicals or rational
exponents.
Solve by:
• Eliminating the radicals and obtain a linear or quadratic
equation
• Solve the linear or quadratic using the method of quadratic
and linear equations
Important thing to remember when eliminating radicals:
• If a = b then a^n =b^n
• If you raise one side of an equation to a power , then you must
keep the other side of the equation balanced by raising it to the
same power
16.
Algebra
Solving Radical Equations
For Ex. =3
Square each side to get rid of square root sign
(√x)² = (3)²
x=9
solve: ³√x-5 = 0
Before raising both side of an equation to the nth power, you
need to isolate the radical expression on one side of the
equation
³√x = 5
(³√x)³ = (5)³
x = 125
17.
Algebra
Equations Containing an Exponent
x = 16
(x) ( ) = 16
x=
x = 2³
x=8
18.
Algebra
Practice Questions
Solving for Roots by Completing the Square:
x² + 2x + 3 = 0
Absolute Values
3x – 5 = 10
Solving for “p” using the Quadratic Formula:
x - 5x² + 4 = 0
Generating Equations with given Roots:
Given the roots 4 ± (5) ½, find the original quadratic equation.
19.
Algebra
Practice Questions
The Discriminant - The Nature of Roots
x² - 8x + 16 = 0
Solving Rational Equations
x = -2
x-3
Solving Radical Equations
Simplify this radical equation:
( )
20.
Algebra
Solutions to Practice Questions
Solving for Roots by Completing the Square:
y = (x + 1) ² + 2
Absolute Values
x= 5, x = 5
3
Solving for “p” using the Quadratic Formula
±4, ±1
Generating Equations with given Roots
x² - 8x + 21 = 0
21.
Algebra
Solutions to Practice Questions
The Discriminant - The Nature of Roots
Discriminant = 0; one real root
Solving Rational Equations
x = 1, x = 2
Solving Radical Equations
x=1
2
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