# What is x*y if (squareroot2 -1)^5=x*squareroot2+y?

Asked on by gudeapp

shaznl1 | High School Teacher | (Level 1) Salutatorian

Posted on

The only way to solve this equation is to multiply (sqrt2 -1)^5 out

(sqrt2-1)(sqrt2-1) = 2+1-2sqrt2=3-2sqrt2

(3-2sqrt2)^2=(sqrt2-1)^4=(3-2sqrt2)(3-2sqrt2)=9+8-12sqrt2=17-12sqrt2

ok, the exponent here is five, not four, so lets multiply again a (sqrt2-1)

(17-12sqrt2)(sqrt2-1)=17sqrt2-24+12sqrt2-17=39sqrt2-41

by the form in the equation xsqrt2+y

we could say x=29 and y=41

another way is to one-stepped multiply the 5th exponent out by using the binomial theorem.

(x-y)^5=x^5-5x^4y+10x^3y^2-10x^2y^3+5xy^4-y^5

the coefficents are determined by the pascal triangle

since y=1 here, the y-terms are all one, so just omit it(. The formula becomes

x^5-5x^4+10x^3-10x^2+5x-1

substitue sqrt2 in there

4sqrt2-5*4+20sqrt2-10*2+5sqrt2-1=29sqrt2 - 41

Thus, x=29, and y=41

Therefore, the requested product x*y is 29*-41= -1189

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

(sqrt2 - 1)^5 = (sqrt2 - 1)^2*(sqrt2 - 1)^2*(sqrt2-1)

We'll raise to square the binomial (sqrt2 - 1)^2:

(sqrt2 - 1)^2 = 2 - 2sqrt2 + 1

(sqrt2 - 1)^2 = 3 - 2sqrt2

(sqrt2 - 1)^2*(sqrt2 - 1)^2 = (3 - 2sqrt2)^2

We'll raise to square the binomial (3 - 2sqrt2)^2:

(3 - 2sqrt2)^2 = 9 - 12sqrt2 + 8

(3 - 2sqrt2)^2 = 17 - 12sqrt2

We'll mutliply by (sqrt2 - 1):

(sqrt2 - 1)(3 - 2sqrt2)^2=(sqrt2 - 1)(17 - 12sqrt2)

We'll perform the multiplication:

(sqrt2 - 1)(17 - 12sqrt2) = 17sqrt2 - 24 - 17 + 12sqrt2

(sqrt2 - 1)^5 = (sqrt2 - 1)(17 - 12sqrt2) = 29sqrt2 - 41

But (sqrt2 - 1)^5 = xsqrt2 + y

Comparing, we'll get:

x = 29 and y = -41

Now, we'll perform the multiplication:

x*y = (29)*(-41) = -1189

The requested value of the product is x*y = -1189.

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