Algebra Electronic Presentation: Expert Voices

Pre Cal 30S January 22nd 2010

Emelda Iradukunda Haben Gabir Aruni Perera

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

Algebra

Algebra Absolute Values

Algebra Solving for Roots by Completeing the Square

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

Algebra

=> -(-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

Algebra Solving for “p” using Quadratic Formula

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

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

Algebra Generating Equations with given 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:

The Discriminant - The Nature of Roots Algebra

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

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.

Algebra The Discriminant - The Nature of Roots

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 )

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

Algebra Solving Rational Equations

The LCD is (x+1)(x-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

solve:

Step1:

Non-permissible Values are x=1,-1

Algebra Solving Rational 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

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

Algebra Solving Radical Equations

x = 16

(x) ( ) = 16

x =

x = 2³

x = 8

Algebra Equations Containing an Exponent

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.

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: ( )

Algebra Practice Questions

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

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

• http://www.youtube.com/user/yourteachermathhelp#p/u/574/GyCuj 1hx_zc • http://www.youtube.com/user/yourteachermathhelp#p/u/417/NxbLXwXXp7Y • http://www.youtube.com/watch?v=LY8VBsLf-4M&feature=related • http://www.youtube.com/user/yourteachermathhelp#p/u/427/FMfqBKfwkKc • http://www.purplemath.com/modules/absolute.htm • http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Dividing%20Rational%20Expressions.pdf • http://www.kutasoftware.com/FreeWorksheets/Alg1Worksheets/Simplifying%20Radicals.pdf

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