Question

let alpha and beta are the roots of x2-6x-2=0,with alpha beta . if an=alphan-betan for n= (greater than or equal to one) 1,then value of a10-2a8/ 2a9 is??

Answer 1

Answer:

answer you will get is 3

Step-by-step explanation:

you can substitute and solve them

Answer 2

Therefore, the value of $a10-2a8/2a9$  is$-3 - 2\sqrt{11}$

The sum and product of the roots alpha and beta of the quadratic equation$x^2 - 6x - 2 = 0$  are:

sum = alpha + beta = 6

product = alpha * beta = -2

We can use these values to find expressions for $alpha^n$ and $beta^n$ using the binomial theorem:

$alpha^n = (alpha + beta)^n + (-beta)^n$

$beta^n = (alpha + beta)^n - alpha^n$

Using these expressions, we can find an expression for an:

$an = alpha^n - beta^n$

$= (alpha + beta)^n + (-beta)^n - ((alpha + beta)^n - alpha^n)$

$= alpha^n + (-beta)^n - alpha^n$

$= (-beta)^n$

Therefore, an is equal to $(-beta)^n.$

Now we can find the value of $a10-2a8/2a9$ using the formula we just derived:

$a10 = (-beta)^10 = beta^10$

$a8 = (-beta)^8 = beta^8$

$a9 = (-beta)^9 = -beta^9$

Substituting these values into the expression, we get:

$a10 - 2a8 / 2a9 = (beta^10 - 2beta^8) / (-2beta^9)$

$= (beta^8(beta^2 - 2)) / (-2beta^9)$

$= (2 - beta^2) / 2beta$

Now we need to find the value of beta. To do this, we can use the quadratic formula:

$beta = (6 + \sqrt{44} ) / 2 = 3 +\sqrt{11}$

Substituting this into the expression we just derived, we get:

$a10 - 2a8 / 2a9 = (2 - (3 +\sqrt{11} )^2) / (2(3 + \sqrt{11}))$

$= (2 - 11 - 6\sqrt{11} ) / (2(3 + \sqrt{11} ))$

$= (-9 - 6\sqrt{11} )) / (6 + 2\sqrt{11})$

$= -(9 + 6\sqrt{11} ) / (6 + 2\sqrt{11} )$

$= -3 - 2\sqrt{11}$

Therefore, the value of $a10-2a8/2a9$

is  $-3 - 2\sqrt{11} .$

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