Solve for x in #2x + 20sqrt(x) - 42 = 0#?

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Answer 1 · 2018-04-14T13:09:47

Please look below.

This may look complicated but can be solved like a quadratic equation if we let #u = sqrtx#

#2x + 20sqrtx - 42 = 0#

#2u^2 + 20u - 42 = 0#

#u^2 + 10u - 21 = 0#

Using the quadratic equation:

#u = (-b +-sqrt(b^2 - 4ac))/(2a)#

#u = (-10 +-sqrt(10^2 - 4xx1 xx -21))/(2 xx 1)#

#u = (-10 +-sqrt(184))/(2)#

#u = (-10 +-2sqrt(46))/(2)#

#u = -5 +-sqrt(46)#

Therefore:

#sqrt(x) = sqrt(-5+-sqrt(46))#