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

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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.
  • 5.
  • 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