Mathematics 9 Lesson 1-D: System of Equations Involving Quadratic EquationsJuan Miguel Palero
This powerpoint presentation discusses or talks about the topic or lesson System of Equations involving Quadratic Equations. It also discusses and explains the rules, steps and examples of System of Equations involving Quadratic Equations
Mathematics 9 Lesson 1-D: System of Equations Involving Quadratic EquationsJuan Miguel Palero
This powerpoint presentation discusses or talks about the topic or lesson System of Equations involving Quadratic Equations. It also discusses and explains the rules, steps and examples of System of Equations involving Quadratic Equations
Quadratic Equations
In One Variable
1. Quadratic Equation
an equation of the form
ax2 + bx + c = 0
where a, b, and c are real numbers
2.Types of Quadratic Equations
Complete Quadratic
3x2 + 5x + 6 = 0
Incomplete/Pure Quadratic Equation
3x2 - 6 = 0
3.Solving an Incomplete Quadratic
4.Example 1. Solve: x2 – 4 = 0
Solution:
x2 – 4 = 0
x2 = 4
√x² = √4
x = ± 2
5.Example 2. Solve: 5x² - 11 = 49
Solution:
5x² - 11 = 49
5x² = 49 + 11
5x² = 60
x² = 12
x = ±√12
x = ±2√3
6.Solving Quadratic Equation
7.By Factoring
Place all terms in the left member of the equation, so that the right member is zero.
Factor the left member.
Set each factor that contains the unknown equal to zero.
Solve each of the simple equations thus formed.
Check the answers by substituting them in the original equation.
8.Example: x² = 6x - 8
Solution:
x² = 6x – 8
x² - 6x + 8 = 0
(x – 4)(x – 2) = 0
x – 4 = 0 | x – 2 = 0
x = 4 x = 2
9.By Completing the Square
Write the equation with the variable terms in the left member and the constant term in the right member.
If the coefficient of x² is not 1, divide every term by this coefficient so as to make the coefficient of x² equal to 1.
Take one-half the coefficient of x, square this quantity, and add the result to both members.
Find the square root of both members, placing a ± sign before the square root of the right member.
Solve the resulting equation for x.
10.Example: x² - 8x + 7 = 0
11.By Quadratic Formula
Example: 3x² - 2x - 7 = 0
12.Solve the following:
1. x² - 15x – 56 = 0
2. 7x² = 2x + 6
3. 9x² - 3x + 8 = 0
4. 8x² + 9x -144 = 0
5. 2x² - 3 + 12x
13.Activity:
Solve the following quadratic formula.
By Factoring By Quadratic Formula
1. x² - 5x + 6 = 0 1. x² - 7x + 6 = 0
2. 3 x² = x + 2 2. 10 x² - 13x – 3 = 0
3. 2 x² - 11x + 12 = 0 3. x (5x – 4) = 2
By Completing the Square
1. x² + 6x + 5 = 0
2. x² - 8x + 3 = 0
3. 2 x² + 3x – 5 = 0
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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
