Question 89113
{{{x^2 - 4x = 3}}}


{{{x^2 - 4x -3=0}}} Subtract 3 from both sides


Now 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-3=0}}} ( notice {{{a=1}}}, {{{b=-4}}}, and {{{c=-3}}})


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




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




{{{x = (4 +- sqrt( 16-4*1*-3 ))/(2*1)}}} Square -4 to get 16




{{{x = (4 +- sqrt( 16+12 ))/(2*1)}}} Multiply {{{-4*-3*1}}} to get {{{12}}}




{{{x = (4 +- sqrt( 28 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (4 +- 2*sqrt(7))/(2*1)}}} Simplify the square root




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


So now the expression breaks down into two parts


{{{x = (4 + 2*sqrt(7))/2}}} or {{{x = (4 - 2*sqrt(7))/2}}}



which split up to



{{{x=+4/2+2*sqrt(7)/2}}} or {{{x=+4/2-2*sqrt(7)/2}}}



and simplify to



{{{x=2+1*sqrt(7)}}} or {{{x=2-1*sqrt(7)}}}



Which approximate to


{{{x=4.64575131106459}}} or {{{x=-0.645751311064591}}}



So our solutions are:

{{{x=4.64575131106459}}} or {{{x=-0.645751311064591}}}


Notice when we graph {{{x^2-4*x-3}}}, we get:


{{{ graph( 500, 500, -10.6457513110646, 14.6457513110646, -10.6457513110646, 14.6457513110646,1*x^2+-4*x+-3) }}}


when we use the root finder feature on a calculator, we find that {{{x=4.64575131106459}}} and {{{x=-0.645751311064591}}}.So this verifies our answer