Math, asked by kanikamalhans, 1 year ago

solve for 20 pts by explanation .. (will Mark the brainliest)

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Answered by siddhartharao77

Given Equation is x^4 - 4x^3 + 6x^2 - 4x = 2008.

It can be written as

x^4 - 4x^3 + 6x^2 - 4x + 1 = 2009

We know that (a - b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4.

(x - 1)^4 = (2009)

x - 1 = 2009^1/4

On Squaring both sides, we get

$(x - 1)^2 = (2009^ \frac{1}{4} )^{2}$

$(x - 1)^2 = - \sqrt{2009}$

$(x - 1)^2 = - 7 * 41^ \frac{1}{2}$

$x^2 - 2x + 1 = - 7 \sqrt{41}$

$x^2 - 2x + 1 + 7 \sqrt{41} = 0$

Therefore:

The product of the roots = $1 + 7 \sqrt{41}$

Hope this helps!

siddhartharao77: :-)

neosingh: good ans bro,i was not able to do it,u made it look easy,going by options i think this method should work, but in hindsight i think this ques is wrong, as technically there are 4 roots of this equation, but we are reducing it to quadratic form, isn't it wrong?

kanikamalhans: thank u bro!!!! u made it so much simpler!

kanikamalhans: plzz can u do other question i hv asked?

kanikamalhans: plzz

kanikamalhans: that cube root a .. qstn

Answered by cskooo7

Answer:

hey mate here is ur answer

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