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Solve equation `x^2` +`x` =2`sqrtx` *`sqrt(x+1)`

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greenbel | (Level 2) Honors

Posted September 10, 2013 at 3:01 PM via web

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Solve equation `x^2` +`x` =2`sqrtx` *`sqrt(x+1)`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted September 10, 2013 at 3:27 PM (Answer #1)

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You need to factor out x, to the left side, such that:

`x(x + 1) = 2sqrt(x(x+1))`

You may use the following notation, such that:

`x(x+1) = t`

Replacing `t` for `x(x+1)` yields:

`t = 2sqrt(t)`

Squaring both sides yields:

`t^2 = 4t => t^2 - 4t = 0`

Factoring out t yields:

`t(t - 4) = 0 `

Using zero product rule, yields:

`t = 0`

`t - 4 = 0=> t = 4`

You need to solve for x the following equations:

`{(x(x+1) = 0),(x(x+1) = 4):}`

Using zero product rule yields:

`x = 0`

`x + 1 = 0= > x = -1`

You need to solve for ` x` the second equation, such that:

`x^2 + x - 4 = 0`

`x_(1,2) = (-1+-sqrt(1 + 16))/2 => x_(1,2) = (-1+-sqrt17)/2`

Since radicals in the given equation hold for x >= 0, you cannot accept the negative values `x = -1` and `x = (-1-sqrt17)/2` .

Hence, evaluating the solutions to the given radical equation, yields `x = 0, x = (-1+sqrt17)/2` .

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