algebra

2.  A quadratic equation is any equation that can be
rearranged in standard form as 𝒂𝒙𝟐
+ 𝒃𝒙 + 𝒄 = 𝟎 where
𝒂, 𝒃, and 𝒄 represent known numbers, and 𝒂 ≠ 𝟎.
 A quadratic equation is an equation of the second degree,
which contains at least one term that is squared.
 In the equation 𝒂𝒙𝟐
+ 𝒃𝒙 + 𝒄 = 𝟎, 𝒂𝒙𝟐
is the quadratic
term, 𝒃𝒙 is the linear term, and 𝒄 is the constant term.

3. 𝟑𝒙𝟐 + 𝟒𝒙 + 𝟔 = 𝟎
a = 3 b = 4 c = 6

4. 𝒙𝟐 − 𝟑𝒙 = −𝟏
Transfrom it to it’s STANDARD FORM
Transposition Method
𝒙𝟐
− 𝟑𝒙 + 𝟏 = 𝟎
𝒙𝟐 − 𝟑𝒙 = −𝟏

5. 1. 3𝑚 + 8 = 15
2. 𝑥2
− 5𝑥 + 10 = 0
3. 12 − 4𝑥 = 0
4.
1
2
(ℎ − 6) = 0
5. 25 − 𝑟2
= 4𝑟
6. (𝑥 + 2)2
= 0

6. 1. 𝟑𝒙 − 𝟐𝒙𝟐
= 𝟕
𝟐. 𝟓 − 𝟐𝟐
= 𝟔
𝟑. 𝟐𝒙(𝒙 − 𝟑) = 𝟏𝟓
4.(𝒙 + 𝟕)(𝒙 − 𝟕) = 𝟎
5. (𝒙 − 𝟒)𝟐
+𝟖 = 𝟎
Write each quadratic equation in standard form, ax2
+ bx + c = 0 then identify the values of a, b and c.

7. EQUATION STANDARD
FORM
VALUES OF a, b,
and c
𝟑𝒙 − 𝟐𝒙𝟐 = 𝟕 −𝟐𝒙𝟐 + 𝟑𝒙 − 𝟕 = 𝟎 𝒂 = −𝟐, 𝒃 = 𝟑, 𝒄 = −𝟕
𝟓 + 𝟐𝒙𝟐 = 𝟔𝒙 𝟐𝒙𝟐 − 𝒙 + 𝟓 = 𝟎 𝒂 = 𝟐, 𝒃 = −, 𝒄 = 𝟓
𝟐𝒙(𝒙 − 𝟑) = 𝟏𝟓 𝟐𝒙𝟐
− 𝟔𝒙 − 𝟏𝟓 = 𝟎 𝒂 = 𝟐, 𝒃 = −𝟔, 𝒄 = −𝟏𝟓
(𝒙 + 𝟕)(𝒙 − 𝟕) = 𝟎 𝒙𝟐 − 𝟒𝟗 = 𝟎 𝒂 = 𝟏, 𝒃 = 𝟎, 𝒄 = −𝟒𝟗
(𝒙 − 𝟒)𝟐
+𝟖 = 𝟎 𝒙𝟐
− 𝟖𝒙 + 𝟐𝟒 = 𝟎 𝒂 = 𝟏, 𝒃 = −𝟖, 𝒄 = 𝟐𝟒

9. Quadratic equations that can be written in the form 𝒙𝟐
= 𝒌 can
be solved by applying the following properties:
1. If 𝑘 > 0, then 𝑥2
= 𝑘, has two real solutions or roots: 𝑥 = ±√𝑘.
2. If 𝑘 = 0, then 𝑥2
= 𝑘, has one real solution or root: 𝑥 = 0.
3. If 𝑘 < 0, then 𝑥2
= 𝑘, has no real solutions or roots.

10. 1. Find the solutions of the equation 𝑥2
− 16 = 0
Solution:
Write the equation in the form 𝑥2
= 𝑘
𝑥2
− 16 = 0 𝑥2
− 16 + 16 = 0 + 16
𝑥2
= 16
𝑥2 = ± 16
𝑥 = ±4
𝑥 = 4; 𝑥 = −4
CHECKING:
𝒙 = 𝟒 𝒙 = −𝟒
𝑥2
− 16 = 0
42 − 16 = 0
16 − 16 = 0
0 = 0
𝑥2
− 16 = 0
−42 − 16 = 0
16 − 16 = 0
0 = 0
Answer: The equation 𝑥2
− 16 = 0 has two solutions: 𝒙 = 𝟒 𝒐𝒓 𝒙 = −𝟒

11. Extract Me!
1. 𝑡2 − 81 = 0
2. 2𝑠2 = 50
3. (𝑥 − 4)2 = 169
4. (2𝑠 − 1)2
= 225
5. 𝑥2
=
9
16

12. Steps on how to solve a quadratic equation by factoring:
1. Transform the quadratic equation into standard form if necessary.
2. Factor quadratic expression.
3. Apply the zero product property by setting each factor of the quadratic
expression equal to zero.
4. Solve each resulting equation.
5. Check the values of the variable obtained by substituting each in the
original equation.
ZERO PRODUCT PROPERTY- If the product of two real numbers is
zero, then either of the two is equal to zero or both numbers is equal to
zero.

13. 1. 𝟒𝒙𝟐
= 𝟔𝒙
Solution:
4𝑥2
= 6𝑥
4𝑥2
− 6𝑥 = 0
2𝑥(2𝑥 − 3) = 0
ZERO PRODUCT PROPERTY
𝟐𝒙 = 𝟎
𝟐𝒙
𝟐
=
𝟎
𝟐
𝒙 = 𝟎
𝟐𝒙 − 𝟑 = 𝟎
𝟐𝒙 = 𝟑
𝟐𝒙
𝟐
=
𝟑
𝟐
𝒙 =
𝟑
𝟐
Answer: The solutions are 0 and
𝟑
𝟐

14. FACTOR THEN SOLVE!
1. 𝑥2 + 7𝑥 = 0
2. 𝑡2 + 8𝑡 + 16 = 0
3. 𝑥2 − 14𝑥 = 5𝑥
4. 4𝑑2
+ 9 = 12𝑑
5. 81 − 4𝑥2 = 0

15. To solve quadratic equation by completing the square, the following steps
can be followed:
1. All the terms with unknowns are on the left side.
2.Divide each term of the equation with the numerial coeffcient of 𝒙𝟐 if
necessary.
3. To get the constant term needed to complete the square, get the numerical
coeffucuent of x, divide it by 2 and square it. Add the result to both sides of the
equation.
4. Factor the perfect square trinomial. Values will be obtained for the right side.
5. Extract the square root from both sides. Two values will be obtained for the
right side.
6. Equate the linear expressions to each of the two values.
7. Solve each of the resulting linear equations.
8. Check your answer by substituting to the original equation.

16. Solve 𝒙𝟐
− 𝟖𝒙 − 𝟗 = 𝟎 by completing th esquare.
Solution:
𝒙𝟐 − 𝟖𝒙 − 𝟗 = 𝟎
𝒙𝟐 − 𝟖𝒙 − 𝟗 + 𝟗 = 𝟎 + 𝟗
𝒙𝟐 − 𝟖𝒙 = 𝟗
𝒙𝟐 − 𝟖𝒙 + (𝟒)𝟐 = 𝟗 + (𝟒)𝟐
𝒙𝟐 − 𝟖𝒙 + 𝟏𝟔 = 𝟗 + 𝟏𝟔
𝒙𝟐
− 𝟖𝒙 + 𝟏𝟔 = 𝟐𝟓
(𝒙 − 𝟒)𝟐= 𝟐𝟓
(𝒙 − 𝟒)𝟐= 𝟐𝟓
𝑥 − 4 = ±5
𝑥 − 4 = 5
𝑥 = 5 + 4
𝑥 = 9
𝑥 − 4 = −5
𝑥 = −5 + 4
𝑥 = −1
THE SOLUTIONS ARE 9 AND -1.
CHECKING:
For 𝒙 = 𝟗 For 𝒙 = −𝟏
𝒙𝟐 − 𝟖𝒙 − 𝟗 = 𝟎 𝒙𝟐 − 𝟖𝒙 − 𝟗 = 𝟎
𝟗𝟐 − 𝟖(𝟗) − 𝟗 = 𝟎 (−𝟏)𝟐−𝟖(−𝟏) − 𝟗 = 𝟎
𝟖𝟏 − 𝟕𝟐 − 𝟗 = 𝟎 𝟏 + 𝟖 − 𝟗 = 𝟎
𝟖𝟏 − 𝟖𝟏 = 𝟎 𝟗 − 𝟗 = 𝟎
𝟎 = 𝟎 𝟎 = 𝟎

17. In solving any quadratic equation 𝒂𝒙𝟐
+ 𝒃𝒙 + 𝒄 = 𝟎, 𝑎 ≠ 0
using quadratic formula, determine the values of 𝑎, 𝑏, 𝑎𝑛𝑑 𝑐
then substitute this equation
𝒙 =
−𝒃 ± 𝒃𝟐 − 𝟒𝒂𝒄
𝟐𝒂
Simply the result if possible, then check the solutions
obtained against the original equation.

18. Find the solution of the equation 𝑥2
+ 2𝑥 − 8 = 0
𝒙 =
−𝟐 + 𝟔
𝟐
𝒙 =
𝟒
𝟐
𝒙 = 𝟐
𝒙 =
−𝟐−𝟔
𝟐
𝒙 =
−𝟖
𝟐
𝒙 = −𝟒
𝒙 =
−𝟐 ± 𝟐𝟐 − 𝟒(𝟏)(−𝟖)
𝟐(𝟏)
𝒙 =
−𝟐 ± 𝟒 + 𝟑𝟐
𝟐
𝒙 =
−𝟐 ± 𝟑𝟔
𝟐
𝒙 =
−𝟐 ± 𝟔
𝟐