1. Solving Quadratic Equations Lesson 9-3

2. Standard form of a quadratic function
A quadratic equation is a equation that can be written in the form
𝒂 𝒙 𝟐 +𝒃𝒙+𝒄=𝟎
Where 𝑎≠0. This form is called the standard form of a quadratic equation

3. Quadratic equation can be solved by a variety of methods
Graphing
Squared Root property
Factoring
Completing the square
Quadratic formula

4. Solutions of a Quadratic Equation
A quadratic equation can have two, one or no real number solutions
The solutions of a quadratic equation are the x- intercepts of the related function
The solutions of a quadratic equation are often called
Roots of the equation or
Zeros of the function

5. Steps in Solving Quadratic Equations
If the equation is in the form (ax+b)2 = c, use the square root property to solve.
If not solved in step 1, write the equation in standard form.
Try to solve by factoring.
If you haven’t solved it yet, use the quadratic formula.

6. are the x-intercepts 2 and -2
Solving Quadratic Equation by graphing
Graph the parabola by: y x
Finding the equation of the axis of symmetry
The solution of
𝑥 2 −4=0
are the x-intercepts 2 and -2
𝑥= −𝑏 2𝑎 x= 0 2(1) 𝑥=0
Finding the vertex
(0) 2 −4=− Vertex = (0,-4)
Making a table using the x-values around the axis of symmetry 2 -2 x
𝒙 𝟐 −𝟒 y 2
(𝟐) 𝟐 −𝟒 -2
(−𝟐) 𝟐 −𝟒
Graphing each point on a coordinate plane
The roots of the equation or solution are the x-intercept or zeros of the related quadratic function

7. Examples of solving by graphing
What are the solution of each equation? Use a graph of the related function
𝑥 2 −1=0
𝑥 2 =0
𝑥 2 +1=0
There are two solutions
1 and -1
There is one solutions
There is no real number solution

8. Solving Quadratic Equation by using the square root
Square Root Property
If b is a real number and a2 = b, then
You can solve equations of the form
𝒙 𝟐 =𝒌
by finding the square root of each side. For example
𝑥 2 =81
𝑥=± 81
𝑥=±9

9. Solving Quadratic Equation by using the square root
Example
Solve:
𝐵) 3𝑥 2 −75=0
A) (y – 3)2 = 4
3𝑥 2 =75
𝑥 2 = 75 3
𝑦−3=±2
𝑥 2 = 25
𝑦=3±2
𝑦=1 𝑜𝑟 𝑦=5
𝑥=± 25
𝑥=±5

10. Solving Quadratic Equation by using
Quadratic formula
This method will work to solve ALL quadratic equations
For many equations it takes longer than some of the other methods.

11. Solving Quadratic Equation by using
Quadratic formula
A quadratic equation written in standard form, ax2 + bx + c = 0, has the solutions.
𝒙= −𝒃± 𝒃 𝟐 −𝟒𝒂𝒄 𝟐𝒂
When getting your solution, if the radicand in the quadratic formula is not a perfect square, you can use a calculator to approximate the solution

12. Solve x2 + x – = 0 by the quadratic formula.
Example
Solve x2 + x – = 0 by the quadratic formula.
x2 + 8x – 20 = (multiply both sides by 8)
a = 1, b = 8, c = 20
= − or −8−12 2
= −20 2 𝑜𝑟 4 2
𝑥=−10 𝑜𝑟 𝑥=2

13. So there is no real solution.
Example
Solve x(x + 6) = 30 by the quadratic formula.
(Simplify the polynomial and write it on standard form)
x2 + 6x + 30 = 0
a = 1, b = 6, c = 30
So there is no real solution.
Square root can’t be negative.

14. Solve 2x = x2 - 8. x2 – 2x – 8 = 0 Let a = 1, b = -2, c = -8 Example
write the equation on standard form
x2 – 2x – 8 = Let a = 1, b = -2, c = -8
𝑥= 𝑜𝑟
𝑥= 2−6 2
𝑥= 𝑜𝑟
𝑥=−2

15. The Discriminat
Before you solve a quadratic equation you can determine how many real-number solutions it has by using discriminant.
The expression under the radical sign in the formula (b2 – 4ac) is called the discriminant.
The discriminant will take on a value that is positive, 0, or negative.
The value of the discriminant indicates two distinct real solutions, one real solution, or no real solutions, respectively.

17. Example
Use the discriminant to determine the number and type of solutions for the following equation.
5 – 4x + 12x2 = 0
a = 12, b = –4, and c = 5
b2 – 4ac = (–4)2 – 4(12)(5)
= 16 – 240
= –224
Because the discriminant is negative, the equation has no-real number solution.

18. Because the discriminant is positive, the equation has two solution.
Example
Use the discriminant to determine the number and type of solutions for the following equation.
6 𝑥 2 −5𝑥=7
6 𝑥 2 −5𝑥−7=0
a = 6, b = –5, and c = -7
b2 – 4ac = (–6)2 – 4(6)(-7) =
= 232
Because the discriminant is positive, the equation has two solution.