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

Step-by-step explanation:5(2h+8) 6010h +40 6010h + 40-40 60-4010h 2010h/10 20/10h 2...

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

Ответ:

GGerardi7552

10.02.2022

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

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 :

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

$\\$

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

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

$\\$

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

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 :

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

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

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

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

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

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

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

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

$\\$

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

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