Question

If alpha and beta are the zeros of 6 x square + x - 2 then what is the value of alpha^4
+ beta^4

Answer 1

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

$\green{\tt{\therefore{\alpha^{4}+\beta^{4}=\frac{337}{1296}}}}$

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

$\green{\underline \bold{Given:}} \\ \tt: \implies {6x}^{2} + x - 2 = 0 \\ \\ \red{\underline \bold{To \: Find:}}\\ \tt: \implies { \alpha }^{4} + { \beta }^{4} = ?$

• According to given question :

$\tt: \implies {6x}^{2} + x - 2 = 0 \\ \\ \text{Solving\:by\:middle\:term\:spiliting} \\ \tt: \implies {6x}^{2} - 3x + 4x - 2 = 0 \\ \\ \tt: \implies 3x(2x - 1) + 2(2x - 1) = 0 \\ \\ \tt: \implies (3x + 2)(2x - 1) = 0 \\ \\ \green{\tt: \implies x = \frac{ - 2}{3} \: and \: \frac{1}{2}} \\ \\ \green{\tt \circ \: \alpha = \frac{ - 2}{3} } \\ \\ \green{\tt \circ \: \beta = \frac{ 1}{2} } \\ \\ \bold{for \: finding \: value} \\ \tt: \implies { \alpha }^{4} + { \beta }^{4} \\ \\ \tt: \implies (\frac{ - 2}{3} )^{4} + (\frac{1}{2} )^{4} \\ \\ \tt: \implies \frac{16}{81} + \frac{1}{16} \\ \\ \tt: \implies \frac{256 + 81}{1296} \\ \\ \green{\tt: \implies \frac{337}{1296} }$

Answer 2

Answer:

337 / 1296

Step-by-step explanation:

Given :

6 x² + x - 2

= > Solving by splitting mid term :

= > 6 x² + 4 x - 3 x - 2

= > 2 x ( 3 x + 2 ) - ( 3 x + 2 )

= > ( 3 x + 2 ) ( 2 x - 1 )

Now :

α = > ( 2 x - 1 ) =  0

α = > 2 x = 1

α = > x = 1  /2

α = 1 / 2

= > α⁴ = 1 / 16 ... ( i )

Also :

β = > ( 3 x + 2 ) = 0

β = > 3 x = - 2

β = > x = - 2 / 3

β = - 2 / 3

= > β⁴ = 16 / 81 ... ( ii )

Adding equation ( i ) and ( ii ) we get :

α⁴ + β⁴ = 1 / 16 + 16 / 81

= > ( 256 + 81 ) / 1296

= > 337 / 1296

Therefore we get required answer.