Question

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

Answer 1

Answer:

• (i) (b² - 2ac)/c²

• (ii) (3abc - b³)/a³

Step-by-step 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

----------------

(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

Answer 2

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

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. $\tt \dfrac{1}{\alpha ^2} + \dfrac{1}{\beta ^2}$
2. α³ + β³

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

1. $\bf \dfrac{1}{\alpha ^2} + \dfrac{1}{\beta ^2} = \dfrac{b^2 - 2ac}{c^2 }$
2. $\bf \alpha ^3 + \beta ^3 = \dfrac{ - b^3 + 3abc}{a^3}$

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

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

We know that :

• $\underline{\boxed{\bf{Sum \: of \: roots = \dfrac{- coefficient \: of \: x}{coefficient \: of \: x^2}}}}$
• $\underline{\boxed{\bf{Product \: of \: roots = \dfrac{constant \: term}{coefficient \: of \: x^2}}}}$

$\tt : \implies \alpha + \beta = \dfrac{-b}{a}$

and

$\tt : \implies \alpha\beta = \dfrac{c}{a}$

Now, let's solve given values :

$\bf \: \: \: \: 1. \: \dfrac{1}{\alpha ^2} + \dfrac{1}{\beta ^2}$

$\tt : \implies \dfrac{\beta ^2 + \alpha ^2}{\alpha ^2 \beta ^2}$

$\tt : \implies \dfrac{\alpha ^2 + \beta ^2}{\alpha ^2 \beta ^2}$

Now, by using identity :

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

$\tt : \implies \dfrac{(\alpha + \beta)^2 - 2 \alpha\beta}{(\alpha\beta)^2}$

Now, by substituting values of :

• $\underline{\boxed{\bf{\alpha + \beta = \dfrac{-b}{a}}}}$
• $\underline{\boxed{\bf{\alpha\beta = \dfrac{c}{a}}}}$

$\tt : \implies \dfrac{\Bigg(\dfrac{-b}{a}\Bigg)^2 - 2 \times \dfrac{c}{a}}{\Bigg(\dfrac{c}{a}\Bigg)^2}$

$\tt : \implies \dfrac{\dfrac{b^2}{a^2} - \dfrac{2c}{a}}{\dfrac{c^2}{a^2}}$

$\tt : \implies \dfrac{\dfrac{b^2}{a^2} - \dfrac{2ac}{a^{2} }}{\dfrac{c^2}{a^2}}$

$\tt : \implies \dfrac{\dfrac{b^2 - 2ac}{a^2 }}{\dfrac{c^2}{a^2}}$

$\tt : \implies \dfrac{b^2 - 2ac}{\cancel{a^2} } \times \dfrac{ \cancel{a^2}}{c^2}$

$\tt : \implies \dfrac{b^2 - 2ac}{c^2 }$

$\underline{\bf Hence, \: \dfrac{1}{\alpha ^2} + \dfrac{1}{\beta ^2} = \dfrac{b^2 - 2ac}{c^2 }}$

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

By using identity :

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

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

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

By using identity :

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

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

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

Now, by substituting values of :

• $\underline{\boxed{\bf{\alpha + \beta = \dfrac{-b}{a}}}}$
• $\underline{\boxed{\bf{\alpha\beta = \dfrac{c}{a}}}}$

$\tt : \implies \Bigg(\dfrac{-b}{a}\Bigg)\Bigg( \bigg(\dfrac{-b}{a} \bigg)^2 -3 \times \dfrac{c}{a}\Bigg)$

$\tt : \implies \Bigg(\dfrac{-b}{a}\Bigg)\Bigg(\dfrac{b^2}{a^2} - \dfrac{3c}{a}\Bigg)$

$\tt : \implies \Bigg(\dfrac{-b}{a}\Bigg)\Bigg(\dfrac{b^2}{a^2} - \dfrac{3ac}{a^{2} }\Bigg)$

$\tt : \implies \Bigg(\dfrac{-b}{a}\Bigg)\Bigg(\dfrac{b^2 - 3ac}{a^2} \Bigg)$

$\tt : \implies \dfrac{-b}{a} \times \dfrac{b^2 - 3ac}{a^2}$

$\tt : \implies \dfrac{ - b(b^2 - 3ac)}{a \times a^2}$

$\tt : \implies \dfrac{ - b^3 + 3abc}{a^3}$

$\underline{\bf Hence, \: \alpha ^3 + \beta ^3 = \dfrac{ - b^3 + 3abc}{a^3}}$