Question 99819
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^2-2*x-3=0}}} ( notice {{{a=1}}}, {{{b=-2}}}, and {{{c=-3}}})





{{{x = (--2 +- sqrt( (-2)^2-4*1*-3 ))/(2*1)}}} Plug in a=1, b=-2, and c=-3




{{{x = (2 +- sqrt( (-2)^2-4*1*-3 ))/(2*1)}}} Negate -2 to get 2




{{{x = (2 +- sqrt( 4-4*1*-3 ))/(2*1)}}} Square -2 to get 4  (note: remember when you square -2, you must square the negative as well. This is because {{{(-2)^2=-2*-2=4}}}.)




{{{x = (2 +- sqrt( 4+12 ))/(2*1)}}} Multiply {{{-4*-3*1}}} to get {{{12}}}




{{{x = (2 +- sqrt( 16 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




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


So now the expression breaks down into two parts


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


Lets look at the first part:


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


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

{{{x=-1}}} Divide


So another answer is

{{{x=-1}}}


So our solutions are:

{{{x=3}}} or {{{x=-1}}}


Notice when we graph {{{x^2-2*x-3}}}, we get:


{{{ graph( 500, 500, -11, 13, -11, 13,1*x^2+-2*x+-3) }}}


and we can see that the roots are {{{x=3}}} and {{{x=-1}}}. This verifies our answer