Question

Given that alpha and beta are the roots of the equation x^2+3x-4=0. Find alpha^3+beta^2

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\alpha^{3}+\beta^{2}=17\:or\:-63}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies {x}^{2} +3x-4 = 0 \\ \\ \tt: \implies \alpha \: and \: \beta \: are \: the \: zeores \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies { \alpha }^{3}+ { \beta }^{2} = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies {x}^{2} + 3x - 4 = 0 \\ \\ \tt: \implies {x}^{2} +4x-x-4 = 0 \\ \\ \tt: \implies x(x + 4) -1(x + 4) = 0 \\ \\ \tt: \implies (x + 4)(x -1) = 0 \\ \\ \green{\tt: \implies x = 1 \: and \: - 4} \\ \\ \tt \circ \: \alpha = 1 \\ \\ \tt \circ \: \beta = -4 \\ \\ \bold{For \: finding \: value : } \\ \tt : \implies \alpha^{3} + \beta^{2} \\ \\ \tt : \implies {1}^{3}+(-4)^{2}\\ \\ \green{\tt : \implies 17} \\ \\ \green{\tt \therefore \alpha^{3} + \beta^{2} = 17}$

$\bold{If \: \alpha = - 4 \: and \: \beta = 1} \\ \\ \tt: \implies { \alpha }^{3} + { \beta }^{2} \\ \\ \tt: \implies { (- 4)}^{2} + {1}^{2} \\ \\ \tt: \implies - 64 + 1 \\ \\ \green{\tt: \implies - 63 }\\ \\ \green{\tt \therefore { \alpha }^{3} +{ \beta }^{2} = - 63}$

Answer 2

QUESTION :

Given that alpha and beta are the roots of the equation x^2+3x-4=0. Find alpha^3+beta^2.

SOLUTION :

$\begin{lgathered}\bold{Factorising \::-} \\ \tt: \implies {x}^{2} + 3x - 4 = 0 \\ \\ \tt: \implies {x}^{2} +4x-x-4 = 0 \\ \\ \tt: \implies x(x + 4) -1(x + 4) = 0 \\ \\ \tt: \implies (x + 4)(x -1) = 0 \\ \\ \orange{\tt: \implies x = 1 \: and \: - 4} \\ \\ \tt \circ \: \alpha = 1 \\ \\ \tt \circ \: \beta = -4 \\ \\ \bold{For \: finding \: value : } \\ \tt : \implies \alpha^{3} + \beta^{2} \\ \\ \tt : \implies {1}^{3}+(-4)^{2}\\ \\ \green{\tt : \implies 17} \\ \\ \purple{\tt \therefore \alpha^{3} + \beta^{2} = 17}\end{lgathered} </p><p>$

$\begin{lgathered}\bold{If \: \alpha = - 4 \: and \: \beta = 1} \\ \\ \tt: \implies { \alpha }^{3} + { \beta }^{2} \\ \\ \tt: \implies { (- 4)}^{2} + {1}^{2} \\ \\ \tt: \implies - 64 + 1 \\ \\ \pink{\tt: \implies - 63 }\\ \\ \blue{\tt \therefore { \alpha }^{3} +{ \beta }^{2} = - 63}\end{lgathered}$