Question 143724
{{{x+3*sqrt(x)-40=0}}} Start with the given equation



Let {{{y=sqrt(x)}}}. So {{{y^2=(sqrt(x))^2=x}}}. In other words, {{{y^2=x}}}



{{{y^2+3y-40=0}}} Replace {{{sqrt(x)}}} with y and replace x with {{{y^2}}}




{{{(y+8)(y-5)=0}}} Factor the left side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:

{{{y+8=0}}} or  {{{y-5=0}}} 


{{{y=-8}}} or  {{{y=5}}}    Now solve for y in each case



So the y values are 


 {{{y=-8}}} or  {{{y=5}}} 





Remember, we let {{{y=sqrt(x)}}}. 



{{{-8=sqrt(x)}}} Plug in {{{y=-8}}} 



{{{(-8)^2=x}}} Square both sides



{{{64=x}}} Square -8 to get 64



So the first possible answer is {{{x=64}}}. But we need to check our answer



Check:


{{{x+3*sqrt(x)-40=0}}} Start with the given equation



{{{64+3*sqrt(64)-40=0}}} Plug in {{{x=64}}}



{{{64+3*8-40=0}}} Take the square root of 64 to get 8



{{{64+24-40=0}}} Multiply



{{{48=0}}} Combine like terms. Since this is <b>not</b> true, this means that {{{x=64}}} is <b>not</b> a solution




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{{{5=sqrt(x)}}} Now plug in {{{y=5}}} (this is the other y-value we found previously)



{{{(5)^2=x}}} Square both sides



{{{25=x}}} Square 5 to get 25



So the second possible answer is {{{x=25}}}. But we need to check our answer




Check:


{{{x+3*sqrt(x)-40=0}}} Start with the given equation



{{{25+3*sqrt(25)-40=0}}} Plug in {{{x=25}}}



{{{25+3*5-40=0}}} Take the square root of 25 to get 5



{{{25+15-40=0}}} Multiply



{{{0=0}}} Combine like terms. Since this is true, this means that {{{x=25}}} is  a solution




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Answer:




So the solution is {{{x=25}}}