Answer:
Step-by-step explanation:
13.
Given
Quadratic equation
- 4x² - 3x - 4 = 0
- With the roots of α and β
To Find
- The quadratic equation with roots of 1/(3α) and 1/(3β)
Solution
The sum and the product of the roots of the given equation:
- α + β = -b/a ⇒ α + β = -(-3)/4 = 3/4
- αβ = c/a ⇒ αβ = -4/4 = - 1
New equation is:
- (x - 1/(3α))(x - 1/(3β)) = 0
- x² - (1/(3α) + 1/(3αβ))x + 1/(3α3β) = 0
- x² - ((3α + 3 β)/(3α3β))x + 1/(3α3β) = 0
- x² - ((α + β)/(3αβ))x + 1/(9αβ) = 0
- x² - (3/4)/(3(-1))x + 1/(9(-1)) = 0
- x² + 1/4x - 1/9 = 0
- 36x² + 9x - 4 = 0
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14.
Given
Quadratic equation
- 3x² +2x + 7 = 0
- With the roots of α and β
To Find
- The quadratic equation with roots of α + 1/β and β + 1/α
Solution
The sum and the product of the roots of the given equation:
- α + β = -b/a ⇒ α + β = -2/3
- αβ = c/a ⇒ αβ = 7/3
New equation is:
- (x - (α + 1/β))(x - (β + 1/α)) = 0
- x² - (α + 1/β + β + 1/α)x + (α + 1/β) (β + 1/α) = 0
- x² - (α + β + (α + β)/αβ )x + αβ + 1/αβ + 2 = 0
- x² - (-2/3 - (2/3)/(7/3))x + 7/3 + 1/(7/3) + 2 = 0
- x² - (-2/3 - 2/7)x + 7/3 + 3/7 + 2 = 0
- x² + (14 + 6)/21x + (49 + 9 + 42/21) = 0
- x² + 20/21x + 100/21 = 0
- 21x² + 20x + 100 = 0
