Question 98540
Let's use the quadratic formula to solve for v:



Starting with the general quadratic


{{{av^2+bv+c=0}}}


the general solution using the quadratic equation is:


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


So lets solve {{{3*v^2+4*v-1=0}}} ( notice {{{a=3}}}, {{{b=4}}}, and {{{c=-1}}})


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




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




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




{{{v = (-4 +- sqrt( 28 ))/(2*3)}}} Combine like terms in the radicand (everything under the square root)




{{{v = (-4 +- 2*sqrt(7))/(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>)




{{{v = (-4 +- 2*sqrt(7))/6}}} Multiply 2 and 3 to get 6


So now the expression breaks down into two parts


{{{v = (-4 + 2*sqrt(7))/6}}} or {{{v = (-4 - 2*sqrt(7))/6}}}



Now break up the fraction



{{{v=-4/6+2*sqrt(7)/6}}} or {{{v=-4/6-2*sqrt(7)/6}}}



Simplify



{{{v=-2 / 3+sqrt(7)/3}}} or {{{v=-2 / 3-sqrt(7)/3}}}



So these expressions approximate to


{{{v=0.21525043702153}}} or {{{v=-1.54858377035486}}}



So our solutions are:

{{{v=0.21525043702153}}} or {{{v=-1.54858377035486}}}


Notice when we graph {{{3*x^2+4*x-1}}} (just replace v with x), we get:


{{{ graph( 500, 500, -11.5485837703549, 10.2152504370215, -11.5485837703549, 10.2152504370215,3*x^2+4*x+-1) }}}


when we use the root finder feature on a calculator, we find that {{{x=0.21525043702153}}} and {{{x=-1.54858377035486}}}.So this verifies our answer