Question 107967
#1 




{{{x^2+4x-1=0}}} Start with the given equation



{{{x^2+4x=1}}} Add 1 to both sides



Take half of the x coefficient 4 to get 2 (ie {{{4/2=2}}})

Now square 2 to get 4 (ie {{{(2)^2=4}}})




{{{x^2+4x+4=1+4}}} Add this result (4) to both sides. Now the expression {{{x^2+4x+4}}} is a perfect square trinomial.





{{{(x+2)^2=1+4}}} Factor {{{x^2+4x+4}}} into {{{(x+2)^2}}}  (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>)




{{{(x+2)^2=5}}} Combine like terms on the right side


{{{x+2=0+-sqrt(5)}}} Take the square root of both sides


{{{x=-2+-sqrt(5)}}} Subtract 2 from both sides to isolate x.


So the expression breaks down to

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



So our answer is approximately

{{{x=0.23606797749979}}} or {{{x=-4.23606797749979}}}


Here is visual proof


{{{ graph( 500, 500, -10, 10, -10, 10, x^2+4x-1) }}} graph of {{{y=x^2+4x-1}}}



When we use the root finder feature on a calculator, we would find that the x-intercepts are {{{x=0.23606797749979}}} and {{{x=-4.23606797749979}}}, so this verifies our answer.




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#2




{{{x^2+6x-4=0}}} Start with the given equation



{{{x^2+6x=4}}} Add 4 to both sides



Take half of the x coefficient 6 to get 3 (ie {{{6/2=3}}})

Now square 3 to get 9 (ie {{{(3)^2=9}}})




{{{x^2+6x+9=4+9}}} Add this result (9) to both sides. Now the expression {{{x^2+6x+9}}} is a perfect square trinomial.





{{{(x+3)^2=4+9}}} Factor {{{x^2+6x+9}}} into {{{(x+3)^2}}}  (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>)




{{{(x+3)^2=13}}} Combine like terms on the right side


{{{x+3=0+-sqrt(13)}}} Take the square root of both sides


{{{x=-3+-sqrt(13)}}} Subtract 3 from both sides to isolate x.


So the expression breaks down to

{{{x=-3+sqrt(13)}}} or {{{x=-3-sqrt(13)}}}



So our answer is approximately

{{{x=0.605551275463989}}} or {{{x=-6.60555127546399}}}


Here is visual proof


{{{ graph( 500, 500, -10, 10, -10, 10, x^2+6x-4) }}} graph of {{{y=x^2+6x-4}}}



When we use the root finder feature on a calculator, we would find that the x-intercepts are {{{x=0.605551275463989}}} and {{{x=-6.60555127546399}}}, so this verifies our answer.




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#3




{{{x^2-2x-5=0}}} Start with the given equation



{{{x^2-2x=5}}} Add 5 to both sides



Take half of the x coefficient -2 to get -1 (ie {{{-2/2=-1}}})

Now square -1 to get 1 (ie {{{(-1)^2=1}}})




{{{x^2-2x+1=5+1}}} Add this result (1) to both sides. Now the expression {{{x^2-2x+1}}} is a perfect square trinomial.





{{{(x-1)^2=5+1}}} Factor {{{x^2-2x+1}}} into {{{(x-1)^2}}}  (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>)




{{{(x-1)^2=6}}} Combine like terms on the right side


{{{x-1=0+-sqrt(6)}}} Take the square root of both sides


{{{x=1+-sqrt(6)}}} Add 1 to both sides to isolate x.


So the expression breaks down to

{{{x=1+sqrt(6)}}} or {{{x=1-sqrt(6)}}}



So our answer is approximately

{{{x=3.44948974278318}}} or {{{x=-1.44948974278318}}}


Here is visual proof


{{{ graph( 500, 500, -10, 10, -10, 10, x^2-2x-5) }}} graph of {{{y=x^2-2x-5}}}



When we use the root finder feature on a calculator, we would find that the x-intercepts are {{{x=3.44948974278318}}} and {{{x=-1.44948974278318}}}, so this verifies our answer.




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