Question 102555
{{{x-5/x= 4}}} Start with the given equation



{{{x^2-5= 4x}}} Multiply both sides by the LCD x



{{{x^2-4x-5=0 }}} Subtract 4x from both sides



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



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


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




So lets solve {{{x^2-4*x-5=0}}} ( notice {{{a=1}}}, {{{b=-4}}}, and {{{c=-5}}})





{{{x = (--4 +- sqrt( (-4)^2-4*1*-5 ))/(2*1)}}} Plug in a=1, b=-4, and c=-5




{{{x = (4 +- sqrt( (-4)^2-4*1*-5 ))/(2*1)}}} Negate -4 to get 4




{{{x = (4 +- sqrt( 16-4*1*-5 ))/(2*1)}}} Square -4 to get 16  (note: remember when you square -4, you must square the negative as well. This is because {{{(-4)^2=-4*-4=16}}}.)




{{{x = (4 +- sqrt( 16+20 ))/(2*1)}}} Multiply {{{-4*-5*1}}} to get {{{20}}}




{{{x = (4 +- sqrt( 36 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (4 +- 6)/(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>)




{{{x = (4 +- 6)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (4 + 6)/2}}} or {{{x = (4 - 6)/2}}}


Lets look at the first part:


{{{x=(4 + 6)/2}}}


{{{x=10/2}}} Add the terms in the numerator

{{{x=5}}} Divide


So one answer is

{{{x=5}}}




Now lets look at the second part:


{{{x=(4 - 6)/2}}}


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

{{{x=-1}}} Divide


So another answer is

{{{x=-1}}}


So our solutions are:

{{{x=5}}} or {{{x=-1}}}


Notice when we graph {{{x^2-4*x-5}}}, we get:


{{{ graph( 500, 500, -11, 15, -11, 15,1*x^2+-4*x+-5) }}}


and we can see that the roots are {{{x=5}}} and {{{x=-1}}}. This verifies our answer