Question 118006
I'll do the first two help you get started


#1


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+6*x-5=0}}} ( notice {{{a=1}}}, {{{b=6}}}, and {{{c=-5}}})





{{{x = (-6 +- sqrt( (6)^2-4*1*-5 ))/(2*1)}}} Plug in a=1, b=6, and c=-5




{{{x = (-6 +- sqrt( 36-4*1*-5 ))/(2*1)}}} Square 6 to get 36  




{{{x = (-6 +- sqrt( 36+20 ))/(2*1)}}} Multiply {{{-4*-5*1}}} to get {{{20}}}




{{{x = (-6 +- sqrt( 56 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-6 +- 2*sqrt(14))/(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 = (-6 +- 2*sqrt(14))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (-6 + 2*sqrt(14))/2}}} or {{{x = (-6 - 2*sqrt(14))/2}}}



Now break up the fraction



{{{x=-6/2+2*sqrt(14)/2}}} or {{{x=-6/2-2*sqrt(14)/2}}}



Simplify



{{{x=-3+sqrt(14)}}} or {{{x=-3-sqrt(14)}}}



So these expressions approximate to


{{{x=0.741657386773941}}} or {{{x=-6.74165738677394}}}



So our solutions are:

{{{x=0.741657386773941}}} or {{{x=-6.74165738677394}}}


Notice when we graph {{{x^2+6*x-5}}}, we get:


{{{ graph( 500, 500, -16.7416573867739, 10.7416573867739, -16.7416573867739, 10.7416573867739,1*x^2+6*x+-5) }}}


when we use the root finder feature on a calculator, we find that {{{x=0.741657386773941}}} and {{{x=-6.74165738677394}}}.So this verifies our answer


<hr>


#2



{{{4x^2+20x=0}}} Start with the given equation


{{{4x(x+5)=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:

{{{4x=0}}} or  {{{x+5=0}}} 


{{{x=0}}} or  {{{x=-5}}}    Now solve for x in each case



So our answer is 

 {{{x=0}}} or  {{{x=-5}}} 



Notice if we graph {{{y=4x^2+20x}}} we can see that the roots are {{{x=0}}} and  {{{x=-5}}} . So this visually verifies our answer.



{{{ graph(500,500,-10,10,-10,10,0, 4x^2+20x) }}}