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How would solve equation (`sqrt(5+2sqrt6)` )^x+(`sqrt(5-2sqrt6)` )^x=98?

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lixalixa | Honors

Posted September 26, 2013 at 4:58 PM via web

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How would solve equation (`sqrt(5+2sqrt6)` )^x+(`sqrt(5-2sqrt6)` )^x=98?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted September 26, 2013 at 5:10 PM (Answer #1)

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You should notice that you may convert both radicands into two squares, such that:

`5 + 2sqrt6 = (sqrt3)^2 + 2*sqrt3*sqrt2 + (sqrt2)^2 = (sqrt3 + sqrt2)^2`

`5 - 2sqrt6 = (sqrt3)^2 - 2*sqrt3*sqrt2 + (sqrt2)^2 = (sqrt3 - sqrt2)^2`

Replacing `(sqrt3 + sqrt2)^2` for `5 + 2sqrt6` and `(sqrt3 - sqrt2)^2 ` for `5 - 2sqrt6` yields:

`(sqrt((sqrt3 + sqrt2)^2))^x + (sqrt((sqrt3 - sqrt2)^2))^x = 98`

Since `sqrt(a^2) = |a|` yields:

`|sqrt3 + sqrt2|^x + |sqrt3 - sqrt2|^x = 98`

`(sqrt3 + sqrt2)^x + (sqrt3 - sqrt2)^x = 98`

You need to notice that `sqrt 3 + sqrt 2 = 1/(sqrt 3 - sqrt 2),` hence, raising to `x` both sides, yields:

`(sqrt3 + sqrt2)^x = 1/(sqrt 3 - sqrt 2)^x`

You need to come up with the following substitution, such that:

`(sqrt3 + sqrt2)^x = y => (sqrt 3 - sqrt 2)^x = 1/y`

Replacing ` y` for `(sqrt3 + sqrt2)^x` and `1/y` for `(sqrt 3 - sqrt 2)^x` , yields:

`y + 1/y = 98 => y^2 - 98y + 1 = 0`

You need to use quadratic formula, such that:

`y_(1,2) = (98+-sqrt(98^2 - 4))/2`

`y_(1,2) = (98+-40sqrt6)/2 => y_(1,2) = 49 +- 20sqrt6`

You need to solve for x the following equations, such that:

`(sqrt3 + sqrt2)^x = 49 +- 20sqrt6`

Taking common logarithms both sides, yields:

`ln (sqrt3 + sqrt2)^x = ln(49 +- 20sqrt6)`

Using the power property of logarithms, yields:

`x*ln (sqrt3 + sqrt2) = ln(49 +- 20sqrt6) => x_(1,2) = ln(49 +- 20sqrt6)/ln (sqrt3 + sqrt2)`

Hence, evaluating the solutions, to the given equation, yields `x_(1,2) = ln(49 +- 20sqrt6)/ln (sqrt3 + sqrt2).`

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aruv | High School Teacher | Valedictorian

Posted September 26, 2013 at 7:34 PM (Answer #2)

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`5+2sqrt(6)=3+2+2sqrt(6)`

`=(sqrt(3))^2+(sqrt(2))^2+2xxsqrt(3)xxsqrt(2)`

`=(sqrt(3)+sqrt(2))^2`

Thus

`sqrt(5+2sqrt(6))=sqrt((sqrt(3)+sqrt(2))^2)=sqrt(3)+sqrt(2)`

`(sqrt(5+2sqrt(6))^x)=(sqrt(3)+sqrt(2))^x`     (i)

`5-2sqrt(6)=3+2-2sqrt(6)=(sqrt(3))^2+(sqrt(2))^2-2sqrt(3)sqrt(2)`

`=(sqrt(3)-sqrt(2))^2`

`Thus`

`(sqrt(5-2sqrt(6)))^x=(sqrt((sqrt(3)-sqrt(2))^2))^x`

`=(sqrt(3)-sqrt(2))^x`                         (ii) 

Thus from (i) and (ii)

`(sqrt(5+2sqrt(6))^x+(sqrt(5-2sqrt(6)))^x=(sqrt(3)+sqrt(2))^x+(sqrt(3)-sqrt(2))^x=98`

`` Let

`y=sqrt(3)+sqrt(2)=((sqrt(3)+sqrt(2))(sqrt(3)-sqrt(2)))/(sqrt(3)-sqrt(2))`

`=(3-2)/(sqrt(3)-sqrt(2))`

`=1/(sqrt(3)-sqrt(2))`

`sqrt(3)-sqrt(2)=1/y`

Thus given problem reduces to

`y^x+1/y^x=98`

`y^(2x)+1=98y^x`

`(y^x)^2-98y^x+1=0`

Let `y^x=t`

`t^2-98t+1=0`

`t=(98+-sqrt((-98)^2-4))/2`

`t=(98+-40sqrt(6))/2`

`t=(49+-20sqrt(6))`

i.e

`y^x=(49+-20sqrt(6))`

`y^x=97.9898 or .0102` 

If

`y^x=97.9898`

`(sqrt(3)+sqrt(2))^x=97.9898`

`ln((sqrt(3)+sqrt(2))^x)=ln(97.9898)`

`xln(sqrt(3)+sqrt(2))=ln(97.9898)`

`x=ln(97.9898)/(ln(sqrt(3)+sqrt(2)))`

`x=4`

`` If

`y^x=.0102`

`(sqrt(3)+sqrt(2))^x=.0102`

`ln((sqrt(3)+sqrt(2))^x)=ln(.0102)`

`xln(sqrt(3)+sqrt(2))=ln(.0102)`

`x=ln(.0102)/ln(sqrt(3)+sqrt(2))=-4.5854/1.1462`

`=-4`

Thus  x=-4 and 4.

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