Question 124652
# 1



{{{m^2-6m=16}}} Start with the given equation



{{{m^2-6m-16=0}}}  Subtract 16 from both sides.


{{{(m-8)(m+2)=0}}} Factor the left side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:

{{{m-8=0}}} or  {{{m+2=0}}} 


{{{m=8}}} or  {{{m=-2}}}    Now solve for m in each case



So our answer is 

 {{{m=8}}} or  {{{m=-2}}} 



Notice if we graph {{{y=x^2-6x-16}}}  (just replace m with x)  we can see that the roots are {{{x=8}}} and  {{{x=-2}}} . So this visually verifies our answer.



{{{ graph(500,500,-10,10,-10,10,0, x^2-6x-16) }}}




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# 2



{{{x^2+5x=3}}} Start with the given equation



{{{x^2+5x-3=0}}}  Subtract 3 from both sides. 


Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


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




So lets solve {{{x^2+5*x-3=0}}} ( notice {{{a=1}}}, {{{b=5}}}, and {{{c=-3}}})





{{{x = (-5 +- sqrt( (5)^2-4*1*-3 ))/(2*1)}}} Plug in a=1, b=5, and c=-3




{{{x = (-5 +- sqrt( 25-4*1*-3 ))/(2*1)}}} Square 5 to get 25  




{{{x = (-5 +- sqrt( 25+12 ))/(2*1)}}} Multiply {{{-4*-3*1}}} to get {{{12}}}




{{{x = (-5 +- sqrt( 37 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-5 +- sqrt(37))/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (-5 +- sqrt(37))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (-5 + sqrt(37))/2}}} or {{{x = (-5 - sqrt(37))/2}}}



Now break up the fraction



{{{x=-5/2+sqrt(37)/2}}} or {{{x=-5/2-sqrt(37)/2}}}





So these expressions approximate to


{{{x=0.54138126514911}}} or {{{x=-5.54138126514911}}}



So our solutions are:

{{{x=0.54138126514911}}} or {{{x=-5.54138126514911}}}


Notice when we graph {{{x^2+5*x-3}}}, we get:


{{{ graph( 500, 500, -15.5413812651491, 10.5413812651491, -15.5413812651491, 10.5413812651491,1*x^2+5*x+-3) }}}


when we use the root finder feature on a calculator, we find that {{{x=0.54138126514911}}} and {{{x=-5.54138126514911}}}.So this verifies our answer