Question 187810
Remember, the quadratic formula is 


{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}}



which really breaks down to 



{{{x[1] = (-b + sqrt( b^2-4ac ))/(2a)}}} or {{{x[2] = (-b - sqrt( b^2-4ac ))/(2a)}}}



where {{{x[1]}}} and {{{x[2]}}} are the two roots



a)


find a formula for the sum of the roots



{{{x[1]+x[2]}}} Add the roots {{{x[1]}}} and {{{x[2]}}}



{{{(-b + sqrt( b^2-4ac ))/(2a)+(-b - sqrt( b^2-4ac ))/(2a)}}} Plug in the formulas from above.



{{{(-b + sqrt( b^2-4ac ) - b - sqrt( b^2-4ac ))/(2a)}}} Combine the fractions



{{{(-2b)/(2a)}}} Combine like terms.



{{{-b/a}}} Reduce



So the sum of the roots {{{x[1]}}} and {{{x[2]}}} is {{{-b/a}}}



In other words, {{{x[1]+x[2]=-b/a}}}




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b) 


find a formula for the product of the roots



{{{x[1]*x[2]}}} Multiply the two roots.



{{{((-b + sqrt( b^2-4ac ))/(2a))((-b - sqrt( b^2-4ac ))/(2a))}}} Plug in the formulas above



{{{((-b + sqrt( b^2-4ac ))(-b - sqrt( b^2-4ac )))/((2a)(2a))}}} Combine the fractions.



{{{((-b)^2 + (sqrt( b^2-4ac ))^2)/((2a)(2a))}}} FOIL by using the difference of squares formula



{{{(b^2-(b^2-4ac))/(4a^2)}}} Square each term.



{{{(b^2-b^2+4ac)/(4a^2)}}} Distribute



{{{(4ac)/(4a^2)}}} Combine like terms.



{{{c/a}}} Reduce



So the formula for the product of the roots {{{x[1]}}} and {{{x[2]}}} is {{{c/a}}}



In other words, {{{x[1]*x[2]=c/a}}}



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c)


without solving tell the sum and product of the solutions of 2x^2+5x-3=0



From {{{2x^2+5x-3=0}}}, we see that {{{a=2}}}, {{{b=5}}}, and {{{c=-3}}}



Sum of the roots: {{{x[1]+x[2]=-b/a=-5/2}}}



Product of the roots: {{{x[1]*x[2]=-3/2}}}