Question

if alpha and beta are the zeros of the polynomial P of X is equals to x square - 5 x minus 6 then find the value of Alpha ^4beta^2+alpha^2beta^4​

Answer 1

||✪✪ CORRECT QUESTION ✪✪||

if ɑ and β are the zeros of the polynomial P = x² - 5x - 6 = 0 , Than find the value of ɑ⁴β² + ɑ²β⁴ ?

|| ✰✰ ANSWER ✰✰ ||

we know that :-

→ The sum of the roots of the Equation ax² + bx + c = 0 , is given by = (-b/a)

and ,

→ Product of roots of the Equation is given by = c/a.

So, we can say that, :-

→ ɑ + β = (-b/a) = -(-5)/1 = 5 ----- Equation (1)

→ ɑ * β = c/a = (-6)/1 = (-6) ------ Equation (2)

Now, we know that, a²+b² = (a+b)² - 2ab

or,

→ ɑ² + β² = (ɑ + β)² - 2ɑ * β

Putting values of Equation (1) and Equation(2) , we get,

→ ɑ² + β² = 5² - 2 * (-6)

→ ɑ² + β² = 25 + 12

→ ɑ² + β² = 37. ---------- Equation (3).

_____________________

Now, we have to Find , ɑ⁴β² + ɑ²β⁴ = ?

→ ɑ⁴β² + ɑ²β⁴

→ ɑ²β²(ɑ² + β²)

→ (ɑβ)² * (ɑ² + β²)

Putting values of Equation (3) and Equation (2) now, we get,

→ (-6)² [ 37]

→ 36 * 37

→ 1332 (Ans).

Hence, The value of ɑ⁴β² + ɑ²β⁴ is 1332.

Answer 2

$\huge \mathfrak{solution}$

Given :

$\bold{p(x) = {x}^{2} - 5x - 6}$

To find :

$\bold{{ \alpha }^{4} { \beta }^{2} + { \alpha }^{2} { \beta }^{4} }$

Here :

a= 1

b= -5

and

c= -6

As we know that ,

$\alpha + \beta = \frac{ - b}{a} = - (\frac{- 5}{1} ) = 5$

and

$\alpha \beta = \frac{c}{a} = - \frac{6}{1} = - 6$

Now ,

$\star \: { \alpha }^{4} { \beta }^{2} + { \alpha }^{2} { \beta }^{4} \\ \\ \star \: { \alpha }^{2} { \beta }^{2} ( { \alpha }^{2} + { \beta }^{2} ) \\ \\ \star \: ( \alpha \beta ) ^{2} ( \: ( \alpha + \beta ) ^{2} - 2\alpha \beta ) \\ \\ \star \: ( - 6 )^{2} ( {5}^{2} - 2( - 6)) \\ \\ \star \: 36(25 + 12) \\ \\ \star \: (36)(37) \\ \\ \star1332$