Question 181944
a)



From {{{x^2+3x-3=0}}} we can see that {{{a=1}}}, {{{b=3}}}, and {{{c=-3}}}



{{{D=b^2-4ac}}} Start with the discriminant formula.



{{{D=(3)^2-4(1)(-3)}}} Plug in {{{a=1}}}, {{{b=3}}}, and {{{c=-3}}}



{{{D=9-4(1)(-3)}}} Square {{{3}}} to get {{{9}}}



{{{D=9--12}}} Multiply {{{4(1)(-3)}}} to get {{{(4)(-3)=-12}}}



{{{D=9+12}}} Rewrite {{{D=9--12}}} as {{{D=9+12}}}



{{{D=21}}} Add {{{9}}} to {{{12}}} to get {{{21}}}


So the discriminant is 21.

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b)


From part a), we see that {{{D=21}}}. This means that {{{D>0}}} (ie the discriminant is positive)


Since the discriminant is greater than zero, this means that there are two real solutions.


So we can say that...


Type of Solution(s): Real
Number: 2 (these are distinct)



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c)




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



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=3}}}, and {{{c=-3}}}



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(1)(-3) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=3}}}, and {{{c=-3}}}



{{{x = (-3 +- sqrt( 9-4(1)(-3) ))/(2(1))}}} Square {{{3}}} to get {{{9}}}. 



{{{x = (-3 +- sqrt( 9--12 ))/(2(1))}}} Multiply {{{4(1)(-3)}}} to get {{{-12}}}



{{{x = (-3 +- sqrt( 9+12 ))/(2(1))}}} Rewrite {{{sqrt(9--12)}}} as {{{sqrt(9+12)}}}



{{{x = (-3 +- sqrt( 21 ))/(2(1))}}} Add {{{9}}} to {{{12}}} to get {{{21}}}



{{{x = (-3 +- sqrt( 21 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-3+sqrt(21))/(2)}}} or {{{x = (-3-sqrt(21))/(2)}}} Break up the expression.  



So the answers are {{{x = (-3+sqrt(21))/(2)}}} or {{{x = (-3-sqrt(21))/(2)}}} 



which approximate to {{{x=0.791}}} or {{{x=-3.791}}} 



Notice how there are two real solutions. So this confirms our answer to part b)