Question

If α&β are the zeros of quadratic polynomial 4x^2 - 5x - 1 then find the value of α^4 + β^4

Answer 1

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

$\green{\tt{\therefore{\alpha^{4}+\beta^{4}=\frac{5}{4}}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies {4x}^{2} - 5x - 1 = 0 \\ \\ \tt: \implies \alpha \: and \: \beta \: are \: the \: zeores \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies { \alpha }^{4} + { \beta }^{4} = ?$

• According to given question :

$\tt \circ \: 4 {x}^{2} - 5x - 1 \\ \\ \tt \circ \: a = 4 \\ \\ \tt\circ \:b = - 5 \\ \\ \tt \circ \: c = - 1 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ \tt: \implies x = \frac{ - ( - 5) \pm \sqrt{( - 5)^{2} - 4 \times 4\times( - 1) } }{2 \times 4} \\ \\ \tt: \implies x = \frac{5 \pm \sqrt{25 + 16} }{8} \\ \\ \green{\tt: \implies x = \frac{5 \pm \sqrt{41} }{8} } \\ \\ \tt \circ \: \alpha = \frac{5 + \sqrt{41} }{8} \\ \\ \tt \circ \: \beta = \frac{5 + \sqrt{41} }{8} \\ \\ \bold{For \: finding \: value : } \\ \tt: \implies { \alpha }^{4} + { \beta }^{4} \\ \\ \tt: \implies (\frac{5 + \sqrt{41} }{8})^{4} + (\frac{5 - \sqrt{41} }{8} )^{4} \\ \\ \tt: \implies \frac{5 + 4 \sqrt{41} }{8} + \frac{5 - 4 \sqrt{41} }{8} \\ \\ \tt: \implies \frac{5 + 4 \sqrt{41} + 5 - 4 \sqrt{41} }{8} \\ \\ \tt: \implies \frac{10}{8} \\ \\ \green{\tt: \implies \frac{5}{4} } \\ \\ \green{\tt \therefore { \alpha }^{4} + { \beta }^{4} = \frac{5}{4} }$

Answer 2

QUESTION :

If α&β are the zeros of quadratic polynomial 4x^2 - 5x - 1 then find the value of α^4 + β^4

SOLUTION :

$\begin{lgathered}\tt \circ \: f(x) = 4 {x}^{2} - 5x - 1 \\ \\ \tt \circ \: a = 4 \\ \\ \tt\circ \:b = - 5 \\ \\ \tt \circ \: c = - 1 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \leadsto x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ \tt: \leadsto x = \dfrac{ - ( - 5) \pm \sqrt{( - 5)^{2} - 4 \times 4\times( - 1) } }{2 \times 4} \\ \\ \tt: \mapsto x = \dfrac{5 \pm \sqrt{25 + 16} }{8} \\ \\ \purple{\tt: \implies x = \frac{5 \pm \sqrt{41} }{8} } \\ \\ \tt \circ \: \alpha = \frac{5 + \sqrt{41} }{8} \\ \\ \tt \circ \: \beta = \dfrac{5 + \sqrt{41} }{8} \\ \\ \bold{For \: finding \: value : } \\ \tt: \implies { \alpha }^{4} + { \beta }^{4} \\ \\ \tt: \implies (\frac{5 + \sqrt{41} }{8})^{4} + (\frac{5 - \sqrt{41} }{8} )^{4} \\ \\ \tt: \implies \frac{5 + 4 \sqrt{41} }{8} + \frac{5 - 4 \sqrt{41} }{8} \\ \\ \tt: \implies \frac{5 + 4 \sqrt{41} + 5 - 4 \sqrt{41} }{8} \\ \\ \tt: \implies \frac{10}{8} \\ \\ \green{\tt: \implies \frac{5}{4} } \\ \\ \blue{\tt \therefore { \alpha }^{4} + { \beta }^{4} = \frac{5}{4} }\end{lgathered}$