Question 155616
a)


{{{2x^2=6}}} Start with the given equation



{{{x^2=3}}} Divide both sides by 2 



{{{x=sqrt(3)}}} or {{{x=-sqrt(3)}}} Take the square root of both sides. 



So the solutions are {{{x=sqrt(3)}}} or {{{x=-sqrt(3)}}} 



which approximate to {{{x=1.732}}} or {{{x=-1.732}}} 



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b)


{{{f(x)=2x^2-6}}} Start with the given function



{{{0=2x^2-6}}} Plug in {{{f(x)=0}}}



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=2}}}, {{{b=0}}}, and {{{c=-6}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(0) +- sqrt( (0)^2-4(2)(-6) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=0}}}, and {{{c=-6}}}



{{{x = (-0 +- sqrt( 0-4(2)(-6) ))/(2(2))}}} Square {{{0}}} to get {{{0}}}. 



{{{x = (-0 +- sqrt( 0--48 ))/(2(2))}}} Multiply {{{4(2)(-6)}}} to get {{{-48}}}



{{{x = (-0 +- sqrt( 0+48 ))/(2(2))}}} Rewrite {{{sqrt(0--48)}}} as {{{sqrt(0+48)}}}



{{{x = (-0 +- sqrt( 48 ))/(2(2))}}} Add {{{0}}} to {{{48}}} to get {{{48}}}



{{{x = (-0 +- sqrt( 48 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-0 +- 4*sqrt(3))/(4)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (-0)/(4) +- (4*sqrt(3))/(4)}}} Break up the fraction.  



{{{x = 0 +- sqrt(3)}}} Reduce.  



{{{x = sqrt(3)}}} or {{{x = -sqrt(3)}}} Break up the "plus/minus" to form two equations.



So the answers are {{{x = sqrt(3)}}} or {{{x = -sqrt(3)}}} 



which approximate to {{{x=1.732}}} or {{{x=-1.732}}} 



So this means that the x-intercepts are *[Tex \LARGE \left(\sqrt{3},0\right)] and *[Tex \LARGE \left(-\sqrt{3},0\right)]


which in decimal form are (1.732,0) and (-1.732,0)



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Notice how the solutions of the first equation are simply the x-coordinates of the x-intercepts.