Question

Hi. I need help with these questions.
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Answer 23 , 24,25,26

Answer 1

• 23) 4x² - 8k²x + k⁴ = 0
• 24) (i) 2, (ii) 7
• 25) (i) 11/8, (ii) -81/8
• 26) (a³ + b³ + c³ + 3abc)/a³

Explanation:

23..............................

Given

• Equation 2x(x - k) = k² with the roots α and β

To find

• The equation with roots α² and β²

Solution

• 2x(x - k) = k²
• 2x² - 2kx - k² = 0

The sum and the product of the roots

• α + β = - (-2k)/2 = k
• αβ = - k²/2

The equation with roots α² and β² is:

• (x - α²)(x - β²) = 0
• x² - (α² + β²)x + α²β²= 0
• x² - ((α + β)² - 2αβ)x + (αβ)² = 0
• x² - (k² - 2( -k²/2))x + (- k²/2)² = 0
• x² - (k² + k²)x + k⁴/4 = 0
• 4x² - 8k²x + k⁴ = 0

24..............................

Given

• Equation 3x² - 9x + 2 = 0 with the roots α and β

To find

The values of

• αβ² + α²β
• α² - αβ + β²

Solution

The sum and the product of the roots

• α + β = - (-9)/3 = 3
• αβ = 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..............................

Given

• Equation 2x² + 9x + 12 = 0 with the roots α and β

To 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 + β³

Solution

The sum and the product of the roots

• α + β = - 9/2
• αβ = 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)) = 0
• x² - (α - 1/α + β - 1/β)x + (α - 1/α)(β - 1/β) = 0
• x² - ((α + β) - (α + β)/αβ)x + αβ + 1/(αβ) - (α/β + β/α) = 0
• x² - ((α + β) - (α + β)/αβ)x + αβ + 1/(αβ) - ((α+β)² - 2αβ)/(αβ)= 0
• x² - (-9/2 - (-9/2)/6)x + 6 + 1/6 - ((-9/2)² - 2(6))/6 = 0
• x² - ( -9/2 + 3/4)x + 37/6 - (81/4 - 12)/6 = 0
• x² + 15/4x + 37/6 - 33/24 = 0
• x² + 90/24x + 148/24 - 33/24 = 0
• 24x² + 90x + 115 = 0
• Proven

(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..............................

Given

• Equation ax² + bx + c = 0 with the roots α and β

To find

• Express (1 - α³)(1 - β³) in terms of a, b and c

Solution

The sum and the product of the roots

• α + β = - b/a
• αβ = 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