Learning Task 3 . Solve for the variable of the following quadratic equations
C. by completing the square.
1. x2 + 5x + 6= 0
D. using quadratic formula.
1. x2 + 5x = 14
Answers
Answer:
A. By extracting square roots
1. x = 13
4. x = 6
2. b = 5/3
5. t = 9
3. y = 1/3
B. By factoring
1. x = -7
3. x = -7 or x = 2
2. m = -8 or m = -2
4. y = -5 or y = 1
C. By completing the square
1. x = 2 or x = 3
2. x = 10 or x = -8
3. x = 3 or x = -4
D. Using the quadratic formula
1. x = 2 or x = -7
2. x = 1 or x = -5
3. x = 3 or x = -4.5.
Step-by-step explanation:
A. By extracting square roots
1. x² = 169
Extracting square roots from both sides of the equation gives;
√(x²) = x and √169 = 13
∴ √(x²) = √169 = 13
x = 13
4. (x - 2)² = 16
Extracting square roots from both sides of the equation gives;
√(x - 2)²= √16 = 4
However;
√(x - 2)² = x - 2
∴ x - 2 = √16 = 4
x - 2 = 4
x = 4 + 2 = 6
x = 6
2. 9·b² = 25
√(9·b²) = √25 = 5
√(9·b²) = √9 × √(b²) = 3·b
∴ 3·b = √25 = 5
3·b = 5
b = 5/3
5. 2·(t - 3)² - 72 = 0
2·(t - 3)² = 72
(t - 3)² = 72/2 = 36
t - 3 = √36 = 6
t = 6 + 3 = 9
t = 9
3. (3·y - 1)² = 0
√(3·y - 1)² = √0
∴ 3·y - 1 = 0
3·y = 0 + 1 = 1
y = 1/3
B. By factoring
1. x² + 7·x = 0
By factoring, we have;
x·(x + 7) = 0
(x + 7) = 0/x = 0
x + 7 = 0
x = 0 - 7 = -7
x = -7
3. x² + 5·x - 14 = 0
Noting that 7 × (-2) = -14 and 7 + (-2) = 5, we get;
x² + 5·x - 14 = (x + 7)·(x - 2) = 0
∴ x = -7 or x = 2
2. m² + 8·m = -16
m² + 8·m + 16 = 0
Which gives;
m² + 8·m + 16 = (m + 8) × (m + 2) = 0
m = -8 or m = -2
4. 2·y² + 8·y - 10 = 0
Which gives;
2·(y² + 4·y - 5) = 0
y² + 4·y - 5 = 0/2 = 0
y² + 4·y - 5 = (y + 5) × (y - 1) = 0
y = -5 or y = 1
C. By completing the square
1. x² + 5·x + 6 = 0
x² + 5·x = -6
Adding (5/2)² to both sides of the equation gives;
x² + 5·x + (5/2)² = -6 + (5/2)²
Which gives;
(x + 5/2)² = -6 + (5/2)² = 1/4
x + 5/2 = ±√(-6 + (5/2)²) = ±√(1/4)
x + 5/2 = ±1/2
x = 5/2 - 1/2 or x = 5/2 + 1/2
∴ x = 2 or x = 3
2. x² + 2·x = 8
x² + 2·x + (2/2)² = 8 + (2/2)²
2/2 = 1 and (2/2)² = 1² = 1
∴ x² + 2·x + (2/2)² = 8 + (2/2)² gives;
x² + 2·x + 1 = 8 + 1 = 9
(x + 1)² = 9
x + 1 = ±√9
x = 1 + 9 or x = 1 - 9
x = 10 or x = -8
3. 2·x² + 2·x = 24
2·(x² + x) = 24
x² + x = 24/2 = 12
x² + x = 12
x² + x + (1/2)² = 12 + (1/2)²
(x + 1/2)² = 12 + (1/2)² = 49/4
√(x + 1/2)² = √(49/4) = ±7/2 = ± 3.5
x + 1/2 = ±3.5
x = -1/2 + 3.5 or x = -1/2 - 7/2
x = -0.5 + 3.5 or x = -0.5 - 3.5
x = 3 or x = -4
D. Using the quadratic formula
1. x² + 5·x = 14
Simplifying, gives;
x² + 5·x - 14 = 0
The quadratic formula for the equation, a·x² + b·x + c = 0, is given as follows;
Comparing with the equation in the question, we get;
x = 4/2 or x = -14/2
x = 2 or x = -7
2. 2·x² + 8·x - 10 = 0
quest
2. 2·x² + 8·x - 10 = 0
x = 4/4 or x = -20/4
x = 1 or x = -5
3. 2·x² + 3·x = 27
Rewriting the above equation in the form a·x² + b·x + c = 0, we get;
2·x² + 3·x - 27 = 0
Which gives;
x = 12/4 = 3 or x = -18/4 = -4.5
x = 3 or x = -4.5.
Step-by-step explanation:
Step-by-step explanation:
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