Question 192184
# 1


{{{4x^3-8x^2=0}}} Start with the given equation



{{{4x^2(x-2)=0}}} Factor out the GCF {{{4x^2}}}



{{{4x^2=0}}} or {{{x-2=0}}} Set each factor equal to zero



{{{x=0}}} or {{{x=2}}} Solve for "x" in each case



So the solutions are {{{x=0}}} or {{{x=2}}}




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# 2


{{{5x^2-3x+6=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=5}}}, {{{b=-3}}}, 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 = (-(-3) +- sqrt( (-3)^2-4(5)(6) ))/(2(5))}}} Plug in  {{{a=5}}}, {{{b=-3}}}, and {{{c=6}}}



{{{x = (3 +- sqrt( (-3)^2-4(5)(6) ))/(2(5))}}} Negate {{{-3}}} to get {{{3}}}. 



{{{x = (3 +- sqrt( 9-4(5)(6) ))/(2(5))}}} Square {{{-3}}} to get {{{9}}}. 



{{{x = (3 +- sqrt( 9-120 ))/(2(5))}}} Multiply {{{4(5)(6)}}} to get {{{120}}}



{{{x = (3 +- sqrt( -111 ))/(2(5))}}} Subtract {{{120}}} from {{{9}}} to get {{{-111}}}



{{{x = (3 +- sqrt( -111 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{x = (3 +- i*sqrt(111))/(10)}}} Simplify the square root. Note: if you have no clue what the term "i" is (ie you haven't learned about it yet), then this means that the answer is simply "no solutions"



If you know what "i" is, then...



{{{x = (3+i*sqrt(111))/(10)}}} or {{{x = (3-i*sqrt(111))/(10)}}} Break up the expression.  



So the answers are {{{x = (3+i*sqrt(111))/(10)}}} or {{{x = (3-i*sqrt(111))/(10)}}} 



which approximate to {{{x=1.354*i}}} or {{{x=-0.754*i}}} 



Note: Once again, if you have no clue what the term "i" is (ie you haven't learned about it yet), then this means that the answer is simply "no solutions"