Question 94385
Let's use the quadratic formula to solve for z:



Starting with the general quadratic


{{{az^2+bz+c=0}}}


the general solution using the quadratic equation is:


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


So lets solve {{{3*z^2-8*z+2=0}}} ( notice {{{a=3}}}, {{{b=-8}}}, and {{{c=2}}})


{{{z = (--8 +- sqrt( (-8)^2-4*3*2 ))/(2*3)}}} Plug in a=3, b=-8, and c=2




{{{z = (8 +- sqrt( (-8)^2-4*3*2 ))/(2*3)}}} Negate -8 to get 8




{{{z = (8 +- sqrt( 64-4*3*2 ))/(2*3)}}} Square -8 to get 64  (note: remember when you square -8, you must square the negative as well. This is because {{{(-8)^2=-8*-8=64}}}.)




{{{z = (8 +- sqrt( 64+-24 ))/(2*3)}}} Multiply {{{-4*2*3}}} to get {{{-24}}}




{{{z = (8 +- sqrt( 40 ))/(2*3)}}} Combine like terms in the radicand (everything under the square root)




{{{z = (8 +- 2*sqrt(10))/(2*3)}}} 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>)




{{{z = (8 +- 2*sqrt(10))/6}}} Multiply 2 and 3 to get 6


So now the expression breaks down into two parts


{{{z = (8 + 2*sqrt(10))/6}}} or {{{z = (8 - 2*sqrt(10))/6}}}



Now break up the fraction



{{{z=+8/6+2*sqrt(10)/6}}} or {{{z=+8/6-2*sqrt(10)/6}}}



Simplify



{{{z=4 / 3+sqrt(10)/3}}} or {{{z=4 / 3-sqrt(10)/3}}}



So these expressions approximate to


{{{z=2.38742588672279}}} or {{{z=0.279240779943873}}}



So our solutions are:

{{{z=2.38742588672279}}} or {{{z=0.279240779943873}}}


Notice when we graph {{{3*x^2-8*x+2}}} (just replace z with x), we get:


{{{ graph( 500, 500, -9.72075922005613, 12.3874258867228, -9.72075922005613, 12.3874258867228,3*x^2+-8*x+2) }}}


when we use the root finder feature on a calculator, we find that {{{x=2.38742588672279}}} and {{{x=0.279240779943873}}}.So this verifies our answer