1. 6.4 - The Quadratic Formula
Factor the following 2 quadratic equations:
x2 + 6x – 16 = 0
2x2 + 4x – 10 = 0
The second one is more difficult, right?
That’s because the roots are not integers
We need to solve the relation using the Quadratic Formula

2. You can to solve using Vertex Form too:
2x2 + 4x – 10 = 0
2(x2 + 2x – 5) = 0
If we use (x + 1)2, we get x2 + 2x + 1
We need x2 + 2x – 5, so 6 less than above
2((x + 1)2 – 6) = 0
2(x + 1)2 – 12 = 0
2(x + 1)2 = 12
(x + 1)2 = 6
x + 1 = ± 6
x = -1 ± 6
Therefore the roots are x= & x =-1- 6

3. Solve Using Quadratic Formula
Based on form ax2 + bx + c = 0
2x2 + 4x – 10 = 0
a = 2; b = 4; c = -10
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
Quadratic Formula
Just substitute the values for a, b and c into the formula, and find the 2 values of x

4. Solve Using Quadratic Formula
2x2 + 4x – 10 = 0
a = 2; b = 4; c = -10
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
𝑥= −4± −4(2)(−10) 2(2)
𝑥= −4± 16−(−80) 4
𝑥= −4±
Therefore, the roots of 2x2 + 4x – 10 = 0 are x = − and x = x = −4−

5. Example #2
A rectangular field is going to be completely enclosed by 100m of fencing. Create a quadratic relation that shows how the area of the field will depend on its width. Then, determine the dimensions of the field that will results in an area of 575 m2. Round your answers to 2 decimal places.

6. Example #2 cont’d Let w be the width of the field
If total perimeter is 100m, and width is w, then length must be 100 −2𝑤 2 =50 −w
Area = length x width = lw
A = w(50 – w)
A = 50w – w2
575 = 50w – w2
0 = -w2 + 50w – 575
Now use quadratic formula to find roots. w
50 - w

7. Example #2 cont’d 0 = -w2 + 50w – 575 a = -1; b = 50; c = -575
𝑤= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
𝑤= −50± −4(−1)(−575) 2(−1)
𝑤= −50± 2500−2300 −2
𝑤= −50± −2
𝑤=17.9 𝑎𝑛𝑑 𝑤=32.1
Since 50 – w = 50 – 17.9 = 32.1, we can see
that w = 17.9m and l = 32.1 m

8. In Summary…
The roots of a quadratic equation of the form ax2 + bx + c = 0 can be determined using the quadratic formula:
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎

9. 6.4 Homework
pg. 343 # 4acf, 5acf, 6, 8–10 (cd), 12-16,

10. 6.5 Interpreting Quadratic Equation Roots
Quadratic relations can have two, one or no x-intercepts.
This graph shows the quadratic equation:
-x2 + x + 6 = 0, and has 2 solutions: x = -2 and x = 3

11. One/No Solutions/Root
This graph shows the quadratic equation:
x2 – 6x + 9 = 0, and has 1 solution: x = 3
This graph shows the quadratic equation:
2x2 – 4x + 5 = 0, and has no solutions.

12. Challenge Question
How can you determine the number of solutions to a quadratic equation without solving it?

13. Discriminant
The discriminant allows us to determine the number of real roots of each equation
𝐷= 𝑏 2 −4𝑎𝑐
If D > 0, there are 2 real solutions
If D = 0, there is 1 real solution
If D < 0, there are no real solutions.
EX. 3x2 + 4x + 5 = 0
𝐷= −4 3 5
𝐷=16−60=−44
D < 0, so no solutions.

14. Example #2 Determine the number of zeros for 𝑦=−2 𝑥 2 +16𝑥 −35.
Let’s try to solve by completing the square:
𝑦=−2 𝑥 2 −8𝑥+17.5
𝑥−4 2 = 𝑥 2 −8𝑥+16, but we need +17.5
𝑥− = 𝑥 2 −8𝑥+17.5
𝑦=−2 𝑥−
𝑦=−2 𝑥−4 2 −3
How can you tell the number to solutions by looking at the equation?

15. Example #2 cont’d 𝑦=−2 𝑥−4 2 −3
𝑦=−2 𝑥−4 2 −3
Sketch the graph. “a” is negative, so opening down, and the vertex is in the 3rd quadrant, so it never touches the x-axis, so no solutions.

16. Example #2 cont’d Another way to solve: use Quadratic Formula.
𝑦=−2 𝑥 2 +16𝑥 −35 (a = -2; b = 16; c = -35)
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
= −16± (16) 2 −4(−2)(−35) 2(−2)
= −16± 256−(280) −4
= −16± −24 −4
We cannot square root a negative number, so x does not exist. No solutions.

17. In Summary…
There are many ways to determine whether a quadratic equation has two, one or no real solutions, without solving the equation
A simple way is to use the determinant
𝐷= 𝑏 2 −4𝑎𝑐
If D > 0, there are 2 real solutions
If D = 0, there is 1 real solution
If D < 0, there are no real solutions.