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Jlenok [28]
3 years ago
10

Hi. I need help with these questions. See image for question. Answer 23 , 24,25,26

Mathematics
1 answer:
diamong [38]3 years ago
6 0

Answer:

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

Step-by-step explanation:

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

<h3>Given</h3>
  • Equation 2x(x - k) = k² with the roots α and β
<h3>To find</h3>
  • The equation with roots α² and β²
<h3>Solution</h3>
  • 2x(x - k) = k²
  • 2x² - 2kx - k² = 0

<u>The sum and the product of the roots </u>

  • α + β = - (-2k)/2 = k
  • αβ = - k²/2

<u>The equation with roots α² and β² is:</u>

  • (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..............................

<h3>Given</h3>
  • Equation 3x² - 9x + 2 = 0 with the roots α and β
<h3>To find</h3>

<u>The values of </u>

  • αβ² + α²β
  • α² -  αβ + β²
<h3>Solution</h3>

<u>The sum and the product of the roots </u>

  • 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..............................

<h3>Given</h3>
  • Equation 2x² + 9x + 12 = 0 with the roots α and β
<h3>To find</h3>
  • (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
<h3>Solution</h3>

<u>The sum and the product of the roots </u>

  • 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..............................

<h3>Given</h3>
  • Equation ax² + bx + c = 0 with the roots α and β
<h3>To find</h3>
  • Express (1 - α³)(1 - β³) in terms of a, b and c
<h3>Solution</h3>

<u>The sum and the product of the roots </u>

  • α + β = - b/a
  • αβ = c/a

<u>The expression is evaluated as follows:</u>

  • (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
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