Question 128742
{{{x^2-3x=4x-1}}} Start with the given equation



{{{x^2-3x-4x+1=0}}}  Subtract 4x from both sides.  Add 1 to both sides. 



{{{x^2-7x+1=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^2-7*x+1=0}}} ( notice {{{a=1}}}, {{{b=-7}}}, and {{{c=1}}})





{{{x = (--7 +- sqrt( (-7)^2-4*1*1 ))/(2*1)}}} Plug in a=1, b=-7, and c=1




{{{x = (7 +- sqrt( (-7)^2-4*1*1 ))/(2*1)}}} Negate -7 to get 7




{{{x = (7 +- sqrt( 49-4*1*1 ))/(2*1)}}} Square -7 to get 49  (note: remember when you square -7, you must square the negative as well. This is because {{{(-7)^2=-7*-7=49}}}.)




{{{x = (7 +- sqrt( 49+-4 ))/(2*1)}}} Multiply {{{-4*1*1}}} to get {{{-4}}}




{{{x = (7 +- sqrt( 45 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (7 +- 3*sqrt(5))/(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 = (7 +- 3*sqrt(5))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (7 + 3*sqrt(5))/2}}} or {{{x = (7 - 3*sqrt(5))/2}}}



Now break up the fraction



{{{x=+7/2+3*sqrt(5)/2}}} or {{{x=+7/2-3*sqrt(5)/2}}}



Simplify



{{{x=7 / 2+3*sqrt(5)/2}}} or {{{x=7 / 2-3*sqrt(5)/2}}}



So these expressions approximate to


{{{x=6.85410196624968}}} or {{{x=0.145898033750315}}}



So our solutions are:

{{{x=6.85410196624968}}} or {{{x=0.145898033750315}}}


Notice when we graph {{{x^2-7*x+1}}}, we get:


{{{ graph( 500, 500, -9.85410196624968, 16.8541019662497, -9.85410196624968, 16.8541019662497,1*x^2+-7*x+1) }}}


when we use the root finder feature on a calculator, we find that {{{x=6.85410196624968}}} and {{{x=0.145898033750315}}}.So this verifies our answer