Question

HELP IT'S URGENT.
Please show workings.
No 4 (see image)

Answer 1

`\huge \underline{\tt Question} :`

If α and β are the roots of the equation ax² + bx + c = 0, where a, b and c are constants such that a ≠ 0, find in terms of a, b and c expressions for :

1. 
   `\tt (1)/(\alpha ^2) + (1)/(\beta ^2)`
2. α³ + β³

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`\huge \underline{\tt Answer} :`

1. 
   `\bf (1)/(\alpha ^2) + (1)/(\beta ^2) = (b^2 - 2ac)/(c^2 )`
2. 
   `\bf \alpha ^3 + \beta ^3 = ( - b^3 + 3abc)/(a^3)`

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`\huge \underline{\tt Explanation} :`

As, α and β are the roots of the equation ax² + bx + c = 0

We know that :

• 
  `\underline{\boxed{\bf{Sum \: of \: roots = (- coefficient \: of \: x)/(coefficient \: of \: x^2)}}}`
• 
  `\underline{\boxed{\bf{Product \: of \: roots = (constant \: term)/(coefficient \: of \: x^2)}}}`

`\tt : \implies \alpha + \beta = (-b)/(a)`

and

`\tt : \implies \alpha\beta = (c)/(a)`

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Now, let's solve given values :

`\bf \: \: \: \: 1. \: (1)/(\alpha ^2) + (1)/(\beta ^2)`

`\tt : \implies (\beta ^2 + \alpha ^2)/(\alpha ^2 \beta ^2)`

`\tt : \implies (\alpha ^2 + \beta ^2)/(\alpha ^2 \beta ^2)`

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Now, by using identity :

• 
  `\underline{\boxed{\bf{a^2+ b^2 = (a+b)^2 -2ab}}}`

`\tt : \implies ((\alpha + \beta)^2 - 2 \alpha\beta)/((\alpha\beta)^2)`

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Now, by substituting values of :

• 
  `\underline{\boxed{\bf{\alpha + \beta = (-b)/(a)}}}`
• 
  `\underline{\boxed{\bf{\alpha\beta = (c)/(a)}}}`

`\tt : \implies (\Bigg((-b)/(a)\Bigg)^2 - 2 * (c)/(a))/(\Bigg((c)/(a)\Bigg)^2)`

`\tt : \implies ((b^2)/(a^2) - (2c)/(a))/((c^2)/(a^2))`

`\tt : \implies ((b^2)/(a^2) - (2ac)/(a^(2) ))/((c^2)/(a^2))`

`\tt : \implies ((b^2 - 2ac)/(a^2 ))/((c^2)/(a^2))`

`\tt : \implies \frac{b^2 - 2ac}{\cancel{a^2} } * \frac{ \cancel{a^2}}{c^2}`

`\tt : \implies (b^2 - 2ac)/(c^2 )`

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`\underline{\bf Hence, \: (1)/(\alpha ^2) + (1)/(\beta ^2) = (b^2 - 2ac)/(c^2 )}`

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`\bf \: \: \: \: 2. \: \alpha ^3 + \beta ^3`

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By using identity :

• 
  `\underline{\boxed{\bf{a^3+ b^3 = (a+b)(a^2 -ab + b^2)}}}`

`\tt : \implies (\alpha + \beta)(\alpha ^2 - \alpha\beta + \beta ^2)`

`\tt : \implies (\alpha + \beta)(\alpha ^2 + \beta ^2 - \alpha\beta)`

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By using identity :

• 
  `\underline{\boxed{\bf{a^2+ b^2 = (a+b)^2 -2ab}}}`

`\tt : \implies (\alpha + \beta)(\alpha + \beta)^2 -2 \alpha\beta - \alpha\beta)`

`\tt : \implies (\alpha + \beta)((\alpha + \beta)^2 -3 \alpha\beta)`

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Now, by substituting values of :

• 
  `\underline{\boxed{\bf{\alpha + \beta = (-b)/(a)}}}`
• 
  `\underline{\boxed{\bf{\alpha\beta = (c)/(a)}}}`

`\tt : \implies \Bigg((-b)/(a)\Bigg)\Bigg( \bigg((-b)/(a) \bigg)^2 -3 * (c)/(a)\Bigg)`

`\tt : \implies \Bigg((-b)/(a)\Bigg)\Bigg((b^2)/(a^2) - (3c)/(a)\Bigg)`

`\tt : \implies \Bigg((-b)/(a)\Bigg)\Bigg((b^2)/(a^2) - (3ac)/(a^(2) )\Bigg)`

`\tt : \implies \Bigg((-b)/(a)\Bigg)\Bigg((b^2 - 3ac)/(a^2) \Bigg)`

`\tt : \implies (-b)/(a) * (b^2 - 3ac)/(a^2)`

`\tt : \implies ( - b(b^2 - 3ac))/(a * a^2)`

`\tt : \implies ( - b^3 + 3abc)/(a^3)`

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`\underline{\bf Hence, \: \alpha ^3 + \beta ^3 = ( - b^3 + 3abc)/(a^3)}`

Answer 2

• (i) (b² - 2ac)/c²

• (ii) (3abc - b³)/a³

Explanation:

α and β are the roots of the equation:

• ax² + bx + c = 0

Sum of the roots is:

• α + b = -b/a

Product of the roots is:

• αβ = c/a

Solving the following expressions:

(i)

• 1/α² + 1/β² =
• (α² + β²) / α²β² =
• ((α + β)² - 2αβ) / (αβ)² =
• ((-b/a)² - 2c/a) / (c/a)² =
• (b²/a² - 2c/a) * a²/c² =
• b²/c² - 2ac/c² =
• (b² - 2ac)/c²

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(ii)

• α³ + β³ =
• (α + β)(α² - αβ + β²) =
• (α + β)((α + β)² - 3αβ) =
• (α + β)³ - 3αβ(α + β) =
• (-b/a)³ - 3(c/a)(-b/a) =
• -b³/a³ + 3bc/a²=
• 3abc/a³ - b³/a³=
• (3abc - b³)/a³