Question

If alpha beta are zeroes of x2-x-4 then find alpha4+beta4

Answer 1

Formulas :
(αβ) = c/a
α + β = -b/a
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Equation : x² - x - 4

To find : α⁴ + β⁴

Solution :

α⁴ + β⁴ = (α² + β²)² - 2(αβ)²
(Eq.1)

But, α² + β² = (α + β)² - 2(αβ)
= (1)² - 2(-4)
= 1 + 8
= 9

Substitution in Eq.1
→ α⁴ + β⁴ = (9)² - 2(-4)²
→ α⁴ + β⁴ = 81 - 32
→ α⁴ + β⁴ = 49

Answer 2

$\huge\tt{\bold{\underline{\underline{Question᎓}}}}$

If alpha beta are zeroes of x2-x-4 then find alpha4+beta4

$\huge\tt{\bold{\underline{\underline{Answer᎓}}}}$

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Formulas:-

$\bold{\boxed{ = > ( \alpha \beta ) = \frac{c}{a}}}$

$\bold{\boxed{ = > \alpha + \beta = - \frac{b}{a}}}$

$= > {eq}^{n} = {x}^{2} - x - 4$

$To \: Find \: = { \alpha }^{4} + { \beta }^{4}$

$= > { \alpha }^{4} + { \beta }^{4} = {( { \alpha }^{2} + { \beta }^{2}) }^{2} - 2( \alpha \beta )....(i)$

$= > But\: { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2( \alpha \beta )$

$= {(1)}^{2} - 2(4) = 1 + 8 = 9$

Substitute in Equation1:-

$= > { \alpha }^{4} + { \beta }^{4} = {(9)}^{2} - 2 { (- 4)}^{2}$

$= > { \alpha }^{4} + { \beta }^{4} = 81 - 32$

$= > { \alpha }^{4} + { \beta }^{4} = 49$

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