What is root in equation x^2+5x=2+9square root(x^2+5x+2)?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should come up with the following substitution, such that:

`x^2 + 5x + 2 = t^2 => x^2 + 5x = t^2 - 2`

Replacing the variable in equation yields:

`t^2 - 2 = 2 + 9sqrt(t^2)`

Using the absolute value definition yields:

`sqrt(t^2) = |t| = {(t, t>=0),(-t, t<0):}`

Considering `t > 0` , yields:

`t^2 - 2 = 2 + 9t => t^2 - 9t - 4 = 0`

Using quadratic formula yields:

`t_(1,2) = (9 +- sqrt(81 + 16))/2 => t_(1,2) = (9 +- sqrt97)/2`

Since `t > 0 => t = (9 - sqrt97)/2` invalid

You need to solve for x theĀ  following equation, such that:

`x^2 + 5x + 2 = (9 + sqrt97)^2/4`

`4x^2 + 20x + 8 = 81 + 18sqrt97 + 97`

`4x^2 + 20x - 170 - 18sqrt97 = 0`

`x_(1,2) = (-20 +- sqrt(400 + 16(170 + 18sqrt97)))/8`

`x_(1,2) = (-20 +- 77.17)/4`

Considering `t < 0` , yields:

`t^2 - 2 = 2 - 9t => t^2 + 9t - 4 = 0`

`t_(1,2) = (-9 +- sqrt97)/2`

Since `t < 0 => t = (-9 + sqrt97)/2` invalid

You need to solve for x theĀ  following equation, such that:

`x^2 + 5x + 2 = (-9 - sqrt97)^2/4`

`4x^2 + 20x + 8 = 81 + 18sqrt97 + 97 => x_(1,2) = (-20 +- 77.17)/4`

Hence, evaluating the solutions to the given equation yields `x_(1,2) = (-20 +- 77.17)/4.`

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