Question 133080
Does your equation look like this: {{{sqrt(2x^2+5x+6)=x}}} ???



{{{sqrt(2x^2+5x+6)=x}}} Start with the given equation



{{{2x^2+5x+6=x^2}}} Square both sides



{{{2x^2+5x+6-x^2=0}}} Subtract {{{x^2}}} from both sides



{{{x^2+5x+6=0}}} Combine like terms




{{{(x+3)(x+2)=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:

{{{x+3=0}}} or  {{{x+2=0}}} 


{{{x=-3}}} or  {{{x=-2}}}    Now solve for x in each case



So our possible solutions are


 {{{x=-3}}} or  {{{x=-2}}} 




However, we must check our answers:


Let's check the first possible solution {{{x=-3}}}  


{{{sqrt(2x^2+5x+6)=x}}} Start with the given equation




{{{sqrt(2(-3)^2+5(-3)+6)=-3}}} Plug in {{{x=-3}}} 



{{{sqrt(18-15+6)=-3}}} Square and multiply 



{{{sqrt(9)=-3}}} Combine like terms



{{{3=-3}}} Take the square root of 9 to get 3



Since the two sides are clearly <b>not</b> equal, this means that  {{{x=-3}}}  is an extraneous solution. In other words,  {{{x=-3}}}  is not a real solution.



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Let's check the first possible solution {{{x=-2}}}  


{{{sqrt(2x^2+5x+6)=x}}} Start with the given equation




{{{sqrt(2(-2)^2+5(-2)+6)=-2}}} Plug in {{{x=-2}}} 



{{{sqrt(8-10+6)=-2}}} Square and multiply 



{{{sqrt(4)=-2}}} Combine like terms



{{{2=-3}}} Take the square root of 4 to get 2



Since the two sides are clearly <b>not</b> equal, this means that  {{{x=-2}}}  is an extraneous solution. In other words,  {{{x=-2}}}  is not a real solution.




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Answer:



So this shows us that {{{sqrt(2x^2+5x+6)=x}}}  does not have any solutions.