Question 838873
how do I solve k^2-4k-5=0


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1.) You can either factor or use the quadratic formula.


For this step I'll show you factoring since this is a nice trinomial.


k^2-4k-5=0


We need to find two numbers that will give us -5 when multiplied together and -4 when added together.



I like to set it up like this when I'm solving...



[_]+[_]=-4
[_]x[_]=-5



I know the only way to get -5 when multiplying is by having 1 & -5(or -1 and 5) but looking at what it needs to add up to it has to be 1 and -5.


[1]+[-5]=-4
[1]x[-5]=-5



Now use those numbers to factor.


k^2-4k-5=0


k^2+1k-5k-5=0


(k^2+1k)+(-5k-5)=0


k(k+1)-5(k+1)=0



k-5=0___&___k+1=0


k-5+5=0+5_&_k+1-1=0-1


k=5_____&___k=-1


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2.) If you wanted to use the **Quadratic Formula**, this is the step to look at:



k^2-4k-5=0


a=1, b=-4, c=-5

Quadratic Formula:


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


{{{x = (-(-4) +- sqrt( (-4)^2-4*(1)*(-5) ))/(2*(1)) }}}


{{{x = (4 +- sqrt( (-4)^2-4*(1)*(-5) ))/(2*(1)) }}}


{{{x = (4 +- sqrt( (-4)^2-4*(1)*(-5) ))/(2) }}}


{{{x = (4 +- sqrt( 16-4*(1)*(-5) ))/(2) }}}


{{{x = (4 +- sqrt( 16-4*(-5) ))/(2) }}}


{{{x = (4 +- sqrt( 16-(-20) ))/(2) }}}


{{{x = (4 +- sqrt( 36 ))/(2) }}}


{{{x = (4 +- 6 )/(2) }}}


{{{x = (4 +- 6 )/(2) }}}



Solve for {{{x = (4 + 6 )/(2) }}}


and


Solve for {{{x = (4 - 6 )/(2) }}}



{{{x = (4 + 6 )/(2) }}}


{{{x = (10 )/(2) }}}


{{{x = 5 }}}


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{{{x = (4 - 6 )/(2) }}}


{{{x = (- 2 )/(2) }}}


{{{x = -1 }}}