The document discusses solving quadratic equations by finding the roots or solutions of the equation. It explains that a quadratic equation is of the form ax^2 + bx + c = 0, where a ≠ 0. The roots are the values of x that make the equation equal to 0. To solve the equation, it is set equal to 0 and the square root property, that if x^2 = k then x = ±√k, is applied to find the two roots of the quadratic equation. Several examples are shown step-by-step to demonstrate solving quadratic equations to find their two roots.

1. EXTRACTING THE ROOTS

2. An equation of the form ax2 + bx + c = 0, where a, b, c are real
numbers and a≠ 0 is called quadratic equation.
If b = 0, then we have an equation of the form ax2 + c = 0.
To solve quadratic equation is to find the roots, zeros, or solution of the
quadratic equation. The roots, zeros, or solution of a quadtatic equation
are the values of x that will make the equation ax2 + c = 0 true.
To solve a quadratic equation of the form ax2 + c = 0, we use the
square root property.
Square Root Property
If x2 = k, then x = ± 𝑘, where k is a non-negative integer.

3. Example 1. Solve the quadratic equation x2 – 25 = 0
Steps Solution
1. Write the equation in the form of
ax2 = c
x2 – 25 = 0
x2 – 25 + 25 = 0 + 25
x2 = 25
2. Apply the square root property x2 – 25 = 0
x2 = 25
X = ± 25 ; x = 5 and x = -5
So, the roots are 5 and -5

4. Example 2. Solve the quadratic equation x2 – 16 = 0.
Steps Solution
1. Write the equation in the form of
ax2 = c
x2 – 16 = 0
x2 – 16 + 16 = 0 + 16
x2 = 16
2. Apply the square root property
x2 = 16
x = ± 16 ; x = 4, and x= -4
So, the roots are 4 and -4.

6. Example 3. Solve the quadratic equation 3x2 = 27.
Steps Solution
1. Write the equation in the form of
ax2 = c
3x2 = 27
3 3
Multiply the quadratic equation by the
reciprocal of a.
1
3
(3x2) =
1
3
(27)
x2 = 9
2. Apply the square root property x2 = 9
= ± 9
x = ±3 ; roots = 3 and -3

7. Checking:
3x2 = 27
3(3)2 = 27
3(9) = 27
27 = 27
Or
3(-3)2 = 27
3(9) = 27
27 = 27
Checking:
x2 – 16 = 0
(4)2 – 16 = 0
16 -16 = 0
0 = 0
Or
x2 – 16 = 0
(-4)2 – 16 = 0
16 – 16 = 0
0= 0

8. Therefore the roots are
9
2
and -
9
2
Example 4. Solve for the
roots of 4x2 = 81.

9. Solution:
x + 13 = ±5
x + 13 = 5
x = 5 – 13
x = -8
X + 13 = -5
x = -5 – 13
x = -18
Example 5. Solve for the
roots of (x + 13)2 = 25.

10. Example 6. Find the roots of 5x2 – 125 = 0
1. ax2 = c
5x2 = 125
5x2 = 125
5 5
x2 = 25
2. Square root property
= Squared both sides
x2 = 25
𝑥2 = 25
x = ±25
x = 5, x = -5

11. Example 7. (x + 7)2 = 16
(𝑥 + 7)2 = 16
x + 7 = ±4
x + 7 = 4 and x + 7 = -4
x = 4 – 7
x = -3
x = -4 – 7
x = -11
Checking :
(x + 7)2 = 16
(-3 + 7)2 = 16
(4)2 = 16
16 = 16
(-11 + 7)2 = 16
(4)2 = 16
16 = 16

12. Example 8. Find the roots of (x – 4)2 = 9.
(𝑥 − 4)2 = 9
x – 4= ±3
x – 4 = 3
x = 3 + 4
x = 7
x – 4 = - 3
x = -3 + 4
x = 1
Checking:
(x – 4)2 = 9
(7 – 4)2 = 9
(3)2 = 9
9 = 9
(x – 4)2 = 9
(1 – 4)2 = 9
(-3)2 = 9
9 = 9