Question 97129
{{{sqrt(2y+7)+4=y}}} Your problem looks like this right?



{{{sqrt(2y+7)=y-4}}} Subtract 4 from both sides



{{{2y+7=(y-4)^2}}} Square both sides



{{{2y+7=y^2-8y+16}}} Foil



{{{0=y^2-10y+9}}} Get all terms to one side





Let's use the quadratic formula to solve for y:



Starting with the general quadratic


{{{ay^2+by+c=0}}}


the general solution using the quadratic equation is:


{{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}


So lets solve {{{y^2-10*y+9=0}}} ( notice {{{a=1}}}, {{{b=-10}}}, and {{{c=9}}})


{{{y = (--10 +- sqrt( (-10)^2-4*1*9 ))/(2*1)}}} Plug in a=1, b=-10, and c=9




{{{y = (10 +- sqrt( (-10)^2-4*1*9 ))/(2*1)}}} Negate -10 to get 10




{{{y = (10 +- sqrt( 100-4*1*9 ))/(2*1)}}} Square -10 to get 100  (note: remember when you square -10, you must square the negative as well. This is because {{{(-10)^2=-10*-10=100}}}.)




{{{y = (10 +- sqrt( 100+-36 ))/(2*1)}}} Multiply {{{-4*9*1}}} to get {{{-36}}}




{{{y = (10 +- sqrt( 64 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{y = (10 +- 8)/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{y = (10 +- 8)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{y = (10 + 8)/2}}} or {{{y = (10 - 8)/2}}}


Lets look at the first part:


{{{x=(10 + 8)/2}}}


{{{y=18/2}}} Add the terms in the numerator

{{{y=9}}} Divide


So one answer is

{{{y=9}}}




Now lets look at the second part:


{{{x=(10 - 8)/2}}}


{{{y=2/2}}} Subtract the terms in the numerator

{{{y=1}}} Divide


So another answer is

{{{y=1}}}


So our possible solutions are:

{{{y=9}}} or {{{y=1}}}



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Now we have to check our answers:


Lets check y=9



{{{sqrt(2y+7)+4=y}}} Start with the given equation



{{{sqrt(2(9)+7)+4=9}}} Plug in {{{y=9}}}


{{{sqrt(18+7)+4=9}}} Multiply



{{{sqrt(25)+4=9}}} Add



{{{5+4=9}}} Take the square root



{{{9=9}}} Add. Since the equation is true, y=9 is a solution





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Lets check y=1



{{{sqrt(2y+7)+4=y}}} Start with the given equation



{{{sqrt(2(1)+7)+4=1}}} Plug in {{{y=1}}}


{{{sqrt(2+7)+4=1}}} Multiply



{{{sqrt(9)+4=9}}} Add



{{{3+4=9}}} Take the square root



{{{7=9}}} Add. Since the equation is not true, y=1 is a not a solution


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Answer:


So our only solution is {{{y=9}}}