23) 4x² - 8k²x + k⁴ = 024) (i) 2, (ii) 725) (i) 11/8, (ii) -81/826) (a³ + b³ + c³ + 3abc)/a³
Step-by-step explanation:
23
GivenEquation 2x(x - k) = k² with the roots α and βTo findThe equation with roots α² and β²Solution2x(x - k) = k²2x² - 2kx - k² = 0
The sum and the product of the roots
α + β = - (-2k)/2 = kαβ = - k²/2
The equation with roots α² and β² is:
(x - α²)(x - β²) = 0x² - (α² + β²)x + α²β²= 0x² - ((α + β)² - 2αβ)x + (αβ)² = 0x² - (k² - 2( -k²/2))x + (- k²/2)² = 0x² - (k² + k²)x + k⁴/4 = 04x² - 8k²x + k⁴ = 0
24
GivenEquation 3x² - 9x + 2 = 0 with the roots α and βTo find
The values of
αβ² + α²βα² - αβ + β² Solution
The sum and the product of the roots
α + β = - (-9)/3 = 3αβ = 2/3
(i)
αβ² + α²β =αβ(α + β) =2/3(3) =2
(ii)
α² - αβ + β² = α² + 2αβ + β² - 3αβ =(α + β)² - 3αβ =3² - 3(2/3) =9 - 2 =7
25
GivenEquation 2x² + 9x + 12 = 0 with the roots α and βTo find(a) show that the quadratic equation whose roots are (α - 1/α) and (β - 1/β) is 24x² + 90x + 115 = 0
The values of
(i) αβ (1/α² + 1/β²)(ii) α³ + β³Solution
The sum and the product of the roots
α + β = - 9/2αβ = 12/2 = 6
a) The quadratic equation whose roots are (α - 1/α) and (β - 1/β) is:
(x - (α - 1/α))(x - (β - 1/β)) = 0x² - (α - 1/α + β - 1/β)x + (α - 1/α)(β - 1/β) = 0x² - ((α + β) - (α + β)/αβ)x + αβ + 1/(αβ) - (α/β + β/α) = 0x² - ((α + β) - (α + β)/αβ)x + αβ + 1/(αβ) - ((α+β)² - 2αβ)/(αβ)= 0x² - (-9/2 - (-9/2)/6)x + 6 + 1/6 - ((-9/2)² - 2(6))/6 = 0x² - ( -9/2 + 3/4)x + 37/6 - (81/4 - 12)/6 = 0x² + 15/4x + 37/6 - 33/24 = 0x² + 90/24x + 148/24 - 33/24 = 024x² + 90x + 115 = 0Proven
(i)
αβ (1/α² + 1/β²) =αβ(α² + β²)/(α²β²) =((α + β)² -2αβ)/(αβ) =((-9/2)² - 2(6))/6 =(81/4 - 12)/6 =81/24 - 2 =33/24 =11/8
(ii)
α³ + β³ =(α + β)(α² - αβ + β²) = (α + β)(α² + 2αβ + β² - 3αβ) =(α + β)((α + β)² - 3αβ) =(-9/2)((-9/2)² - 3(6)) =-9/2(81/4 - 18) =-9/2(9/4) =-81/8
26
GivenEquation ax² + bx + c = 0 with the roots α and βTo findExpress (1 - α³)(1 - β³) in terms of a, b and cSolution
The sum and the product of the roots
α + β = - b/aαβ = c/a
The expression is evaluated as follows:
(1 - α³)(1 - β³) = 1 - (α³ + β³) + α³β³ =1 - (α + β)((α + β)² - 3αβ) + (αβ)³ =1 - (-b/a)((-b/a)² - 3c/a) + (c/a)³ =1 + (b/a)³ + 3bc/a² + c³/a³ =1 + (b³ + c³ + 3abc)/a³ =(a³ + b³ + c³ + 3abc)/a³
