Question 88435
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 {{{2*x^2-1=0}}} (note: since the polynomial does not have an "x" term, the 2nd coefficient is zero. In other words, b=0. So that means the polynomial really looks like {{{2*x^2+0*x-1=0}}} notice {{{a=2}}}, {{{b=0}}}, and {{{c=-1}}})


{{{x = (-0 +- sqrt( (0)^2-4*2*-1 ))/(2*2)}}} Plug in a=2, b=0, and c=-1




{{{x = (-0 +- sqrt( 0-4*2*-1 ))/(2*2)}}} Square 0 to get 0




{{{x = (-0 +- sqrt( 0+8 ))/(2*2)}}} Multiply {{{-4*-1*2}}} to get {{{8}}}




{{{x = (-0 +- sqrt( 8 ))/(2*2)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-0 +- 2*sqrt(2))/(2*2)}}} Simplify the square root




{{{x = (-0 +- 2*sqrt(2))/4}}} Multiply 2 and 2 to get 4


So now the expression breaks down into two parts


{{{x = (-0 + 2*sqrt(2))/4}}} or {{{x = (-0 - 2*sqrt(2))/4}}}



Which simplifies to 


{{{x = sqrt(2)/2}}} or {{{x = - sqrt(2)/2}}}



So the roots approximate to


{{{x=0.707106781186548}}} or {{{x=-0.707106781186548}}}



So our solutions are:

{{{x=0.707106781186548}}} or {{{x=-0.707106781186548}}}


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


{{{ graph( 500, 500, -10.7071067811865, 10.7071067811865, -10.7071067811865, 10.7071067811865,2*x^2+0*x+-1) }}}


when we use the root finder feature on a calculator, we find that {{{x=0.707106781186548}}} and {{{x=-0.707106781186548}}}.So this verifies our answer



note:

Since {{{0+-sqrt(2)/2=0+-sqrt(1/2)}}} this means you were really close. You just need to make sure that the negative is placed on the outside of the square root.