Question 96856
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^2-48=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 {{{x^2+0*x-48=0}}}  notice {{{a=1}}}, {{{b=0}}}, and {{{c=-48}}})


{{{x = (0 +- sqrt( (0)^2-4*1*-48 ))/(2*1)}}} Plug in a=1, b=0, and c=-48




{{{x = (0 +- sqrt( 0-4*1*-48 ))/(2*1)}}} Square 0 to get 0  




{{{x = (0 +- sqrt( 0+192 ))/(2*1)}}} Multiply {{{-4*-48*1}}} to get {{{192}}}




{{{x = (0 +- sqrt( 192 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (0 +- 8*sqrt(3))/(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 = (0 +- 8*sqrt(3))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (0 + 8*sqrt(3))/2}}} or {{{x = (0 - 8*sqrt(3))/2}}}



Now break up the fraction



{{{x=0/2+8*sqrt(3)/2}}} or {{{x=0/2-8*sqrt(3)/2}}}



Simplify



{{{x=4*sqrt(3)}}} or {{{x=-4*sqrt(3)}}}



So these expressions approximate to


{{{x=6.92820323027551}}} or {{{x=-6.92820323027551}}}



So our solutions are:

{{{x=6.92820323027551}}} or {{{x=-6.92820323027551}}}


Notice when we graph {{{x^2-48}}}, we get:


{{{ graph( 500, 500, -16.9282032302755, 16.9282032302755, -16.9282032302755, 16.9282032302755,1*x^2+0*x+-48) }}}


when we use the root finder feature on a calculator, we find that {{{x=6.92820323027551}}} and {{{x=-6.92820323027551}}}.So this verifies our answer