Question 86200
a)

{{{5*sqrt(x)-x=6}}}


{{{5*sqrt(x)=6+x}}} Add x to both sides



{{{(5*sqrt(x))^2=(6+x)^2}}}Square both sides


{{{25x=(6+x)^2}}}


{{{25x=36+12x+x^2}}} Foil


{{{0=36+12x+x^2-25x}}} Subtract {{{25x}}} from both sides


{{{0=x^2-13x+36}}} Combine like terms and rearrange 



Now lets use the quadratic formula to solve for x:


Starting with the general quadratic


{{{ax^2+bx+c}}}


the general form of the quadratic equation is:


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


So lets solve {{{x^2-13*x+36}}}


{{{x = (13 +- sqrt( (-13)^2-4*1*36 ))/(2*1)}}} Plug in a=1, b=-13, and c=36




{{{x = (13 +- sqrt( 169-4*1*36 ))/(2*1)}}} Square -13 to get 169




{{{x = (13 +- sqrt( 169+-144 ))/(2*1)}}} Multiply {{{-4*36*1}}} to get {{{-144}}}




{{{x = (13 +- sqrt( 25 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (13 +- 5)/(2*1)}}} Simplify the square root




{{{x = (13 +- 5)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (13 + 5)/2}}} or {{{x = (13 - 5)/2}}}


Lets look at the first part:


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

{{{x=9}}} Divide


So one answer is

{{{x=9}}}

Now lets look at the second part:


{{{x=8/2}}} Subtract the terms in the numerator

{{{x=4}}} Divide


So another answer is

{{{x=4}}}


So our possible solutions are:

{{{x=9}}} or {{{x=4}}}


Now we must check to see if the solutions work:

Check:


Test the first solution


{{{5*sqrt(9)-9=6}}} Plug in x=9


{{{5*3-9=6}}} Take the square root


{{{15-9=6}}} Multiply


{{{6=6}}} Combine like terms. Sot the solution works.


Now lets test the other solution


{{{5*sqrt(4)-4=6}}} Plug in x=4


{{{5*2-4=6}}} Take the square root


{{{10-4=6}}} Multiply


{{{6=6}}} Combine like terms. Sot the solution works.


So our solutions are

{{{x=4}}} or {{{x=9}}}



Here is the graph of  {{{y=5*sqrt(x)-x-6}}} (just get all terms to one side)


{{{ graph( 500, 500, 0, 15, -5, 5, 5*sqrt(x)-x-6) }}} graph of {{{y=5*sqrt(x)-x-6}}} where the roots represent the solutions we found


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b)

{{{x^2-8=17}}}


{{{x^2-8-17=0}}}Subtract 17 from both sides


Now lets use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c}}}


the general form of the quadratic equation is:


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


So lets solve {{{x^2-8*x-17}}}


{{{x = (8 +- sqrt( (-8)^2-4*1*-17 ))/(2*1)}}} Plug in a=1, b=-8, and c=-17




{{{x = (8 +- sqrt( 64-4*1*-17 ))/(2*1)}}} Square -8 to get 64




{{{x = (8 +- sqrt( 64+68 ))/(2*1)}}} Multiply {{{-4*-17*1}}} to get {{{68}}}




{{{x = (8 +- sqrt( 132 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (8 +- 2*sqrt(33))/(2*1)}}} Simplify the square root




{{{x = (8 +- 2*sqrt(33))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (8 + 2*sqrt(33))/2}}} or {{{x = (8 - 2*sqrt(33))/2}}}



Which approximate to


{{{x=9.74456264653803}}} or {{{x=-1.74456264653803}}}



So our solutions are:

{{{x=9.74456264653803}}} or {{{x=-1.74456264653803}}}


Notice when we graph {{{x^2-8*x-17}}} we get:


{{{ graph( 500, 500, -11.744562646538, 19.744562646538, -11.744562646538, 19.744562646538,1*x^2+-8*x+-17) }}}


when we use the root finder feature on our calculator, we find that {{{x=9.74456264653803}}} and {{{x=-1.74456264653803}}}.So this verifies our answer