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2 years ago

HELLPP. ITS URGENT.50 PTS

Mathematics

2 answers:

Burka [1]2 years ago

5 0

Answer:

I replaced alpha by A and bitta by B. Sorry for this mistake

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ICE Princess25 [194]2 years ago

3 0

Answer:

4x² - 13x + 8 = 0
4x² - 11x + 5 = 0
16x² - 41x + 1 = 0
x² + 5x + 4 = 0
x² - 66x + 64 = 0

Step-by-step explanation:

Given

α and β are roots of 4x²-5x-1=0

Then the sum and product of the roots are:

α+b = -(-5)/4 = 5/4
αβ = -1/4

(i) Roots are α + 1 and β + 1, then we have:

(x - (α + 1))(x - (β + 1)) = 0
(x - α - 1)(x - β - 1) = 0
x² - (α+β+2)x + α+β+ αβ + 1 = 0
x² - (5/4+2)x +5/4 - 1/4 + 1 = 0
x² - 13/4x + 2= 0
4x² - 13x + 8 = 0

(ii) Roots are 2 - α and 2 - β, then we have:

(x + α - 2)(x + β - 2) = 0
x² + (a + β - 4)x - 2(α + β) + αβ + 4 = 0
x² + (5/4 - 4)x - 2(5/4) - 1/4 + 4 = 0
x² - 11/4x - 10/4 - 1/4 + 16/4 = 0
x² - 11/4x + 5/4x = 0
4x² - 11x + 5 = 0

(iii) Roots are α² and β², then:

(x - α²)(x-β²) = 0
x² -(α²+β²)x + (αβ)² = 0
x² - ((α+β)² - 2αβ)x + (-1/4)² = 0
x² - ((5/4)² -2(-1/4))x + 1/16 = 0
x² - ( 25/16 + 1/2)x + 1/16 = 0
x² - 33/16x + 1/16 = 0
16x² - 33x + 1 = 0

(iv) Roots are 1/α and 1/β, then:

(x - 1/α)(x - 1/β) = 0
x² - (1/α+1/β)x + 1/αβ = 0
x² - ((α+β)/αβ)x + 1/αβ = 0
x² - (5/4)/(-1/4)x - 1/(-1/4) = 0
x² + 5x + 4 = 0

(v) Roots are 2/α² and 2/β², then:

(x - 2/α²)(x - 2/β²) = 0
x² - (2/α² + 2/β²)x + 4/(αβ)² = 0
x² - 2((α+β)² - 2αβ)/(αβ)²)x + 4/(αβ)² = 0
x² - 2((5/4)² - 2(-1/4))/(-1/4)²x + 4/(-1/4)² = 0
x² - 2(25/16 + 8/16)/(1/16)x + 4(16) = 0
x² - 2(33)x + 64 = 0
x² - 66x + 64 = 0

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