Question 178424
a)


{{{3x^2-5x-11=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=3}}}, {{{b=-5}}}, and {{{c=-11}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(-5) +- sqrt( (-5)^2-4(3)(-11) ))/(2(3))}}} Plug in  {{{a=3}}}, {{{b=-5}}}, and {{{c=-11}}}



{{{x = (5 +- sqrt( (-5)^2-4(3)(-11) ))/(2(3))}}} Negate {{{-5}}} to get {{{5}}}. 



{{{x = (5 +- sqrt( 25-4(3)(-11) ))/(2(3))}}} Square {{{-5}}} to get {{{25}}}. 



{{{x = (5 +- sqrt( 25--132 ))/(2(3))}}} Multiply {{{4(3)(-11)}}} to get {{{-132}}}



{{{x = (5 +- sqrt( 25+132 ))/(2(3))}}} Rewrite {{{sqrt(25--132)}}} as {{{sqrt(25+132)}}}



{{{x = (5 +- sqrt( 157 ))/(2(3))}}} Add {{{25}}} to {{{132}}} to get {{{157}}}



{{{x = (5 +- sqrt( 157 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (5+sqrt(157))/(6)}}} or {{{x = (5-sqrt(157))/(6)}}} Break up the expression.  



So the answers are {{{x = (5+sqrt(157))/(6)}}} or {{{x = (5-sqrt(157))/(6)}}} 



which approximate to {{{x=2.922}}} or {{{x=-1.255}}} 



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b)


Since the solutions are approximately {{{x=2.922}}} or {{{x=-1.255}}}, this means that the x-intercepts are approximately (2.922,0) and (-1.255,0)



Note: the solutions of any quadratic are the x-intercepts of the parabola