Question 130760
*[Tex \LARGE 4x^2=12] Start with the given equation



*[Tex \LARGE x^2=3] Divide both sides by 4



*[Tex \LARGE x=\pm sqrt{3}] Take the square root of both sides






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Answer:


So the solution to *[Tex \LARGE 4x^2=12] is:


*[Tex \LARGE x=\pm sqrt{3}]



Which breaks down to:


*[Tex \LARGE x=sqrt{3}] or *[Tex \LARGE x=-sqrt{3}]






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{{{4x^2-12=0}}} Start with the given equation


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 {{{4*x^2-12=0}}} (notice {{{a=4}}}, {{{b=0}}}, and {{{c=-12}}}. This is due to {{{4*x^2-12=0}}} really looking like {{{4*x^2+0*x-12=0}}})





{{{x = (0 +- sqrt( (0)^2-4*4*-12 ))/(2*4)}}} Plug in a=4, b=0, and c=-12




{{{x = (0 +- sqrt( 0-4*4*-12 ))/(2*4)}}} Square 0 to get 0  




{{{x = (0 +- sqrt( 0+192 ))/(2*4)}}} Multiply {{{-4*-12*4}}} to get {{{192}}}




{{{x = (0 +- sqrt( 192 ))/(2*4)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (0 +- 8*sqrt(3))/(2*4)}}} 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))/8}}} Multiply 2 and 4 to get 8


So now the expression breaks down into two parts


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



Now break up the fraction



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



Simplify



{{{x=sqrt(3)}}} or {{{x=-sqrt(3)}}} (notice how these solutions are identical to the previous set of solutions)



So these expressions approximate to


{{{x=1.73205080756888}}} or {{{x=-1.73205080756888}}}



So the x-intercepts of {{{y=4*x^2-12}}} are


{{{x=1.73205080756888}}} and {{{x=-1.73205080756888}}}