Question 132984

{{{x^2-5x-1=-7}}} Start with the given equation



{{{x^2-5x-1+7=0}}}  Add 7 to both sides. 



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


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^2-5*x+6=0}}} ( notice {{{a=1}}}, {{{b=-5}}}, and {{{c=6}}})





{{{x = (--5 +- sqrt( (-5)^2-4*1*6 ))/(2*1)}}} Plug in a=1, b=-5, and c=6




{{{x = (5 +- sqrt( (-5)^2-4*1*6 ))/(2*1)}}} Negate -5 to get 5




{{{x = (5 +- sqrt( 25-4*1*6 ))/(2*1)}}} Square -5 to get 25  (note: remember when you square -5, you must square the negative as well. This is because {{{(-5)^2=-5*-5=25}}}.)




{{{x = (5 +- sqrt( 25+-24 ))/(2*1)}}} Multiply {{{-4*6*1}}} to get {{{-24}}}




{{{x = (5 +- sqrt( 1 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




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


So now the expression breaks down into two parts


{{{x = (5 + 1)/2}}} or {{{x = (5 - 1)/2}}}


Lets look at the first part:


{{{x=(5 + 1)/2}}}


{{{x=6/2}}} Add the terms in the numerator

{{{x=3}}} Divide


So one answer is

{{{x=3}}}




Now lets look at the second part:


{{{x=(5 - 1)/2}}}


{{{x=4/2}}} Subtract the terms in the numerator

{{{x=2}}} Divide


So another answer is

{{{x=2}}}


So our solutions are:

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


Notice when we graph {{{x^2-5*x+6}}}, we get:


{{{ graph( 500, 500, -8, 13, -8, 13,1*x^2+-5*x+6) }}}


and we can see that the roots are {{{x=3}}} and {{{x=2}}}. This verifies our answer