SOLUTION: solve the equation by completing the square {{{x^2+4x-1=0}}} {{{x^2+6x-4=0}}} {{{x^2-2x-5=0}}}

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: solve the equation by completing the square {{{x^2+4x-1=0}}} {{{x^2+6x-4=0}}} {{{x^2-2x-5=0}}}      Log On


   

Question 107967: solve the equation by completing the square
x%5E2%2B4x-1=0
x%5E2%2B6x-4=0
x%5E2-2x-5=0
Found 2 solutions by jim_thompson5910, MathLover1:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
#1



x%5E2%2B4x-1=0 Start with the given equation


x%5E2%2B4x=1 Add 1 to both sides


Take half of the x coefficient 4 to get 2 (ie 4%2F2=2)
Now square 2 to get 4 (ie %282%29%5E2=4)



x%5E2%2B4x%2B4=1%2B4 Add this result (4) to both sides. Now the expression x%5E2%2B4x%2B4 is a perfect square trinomial.




%28x%2B2%29%5E2=1%2B4 Factor x%5E2%2B4x%2B4 into %28x%2B2%29%5E2 (note: if you need help with factoring, check out this solver)



%28x%2B2%29%5E2=5 Combine like terms on the right side

x%2B2=0%2B-sqrt%285%29 Take the square root of both sides

x=-2%2B-sqrt%285%29 Subtract 2 from both sides to isolate x.

So the expression breaks down to
x=-2%2Bsqrt%285%29 or x=-2-sqrt%285%29


So our answer is approximately
x=0.23606797749979 or x=-4.23606797749979

Here is visual proof

+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E2%2B4x-1%29+ graph of y=x%5E2%2B4x-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.





#2



x%5E2%2B6x-4=0 Start with the given equation


x%5E2%2B6x=4 Add 4 to both sides


Take half of the x coefficient 6 to get 3 (ie 6%2F2=3)
Now square 3 to get 9 (ie %283%29%5E2=9)



x%5E2%2B6x%2B9=4%2B9 Add this result (9) to both sides. Now the expression x%5E2%2B6x%2B9 is a perfect square trinomial.




%28x%2B3%29%5E2=4%2B9 Factor x%5E2%2B6x%2B9 into %28x%2B3%29%5E2 (note: if you need help with factoring, check out this solver)



%28x%2B3%29%5E2=13 Combine like terms on the right side

x%2B3=0%2B-sqrt%2813%29 Take the square root of both sides

x=-3%2B-sqrt%2813%29 Subtract 3 from both sides to isolate x.

So the expression breaks down to
x=-3%2Bsqrt%2813%29 or x=-3-sqrt%2813%29


So our answer is approximately
x=0.605551275463989 or x=-6.60555127546399

Here is visual proof

+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E2%2B6x-4%29+ graph of y=x%5E2%2B6x-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.






#3



x%5E2-2x-5=0 Start with the given equation


x%5E2-2x=5 Add 5 to both sides


Take half of the x coefficient -2 to get -1 (ie -2%2F2=-1)
Now square -1 to get 1 (ie %28-1%29%5E2=1)



x%5E2-2x%2B1=5%2B1 Add this result (1) to both sides. Now the expression x%5E2-2x%2B1 is a perfect square trinomial.




%28x-1%29%5E2=5%2B1 Factor x%5E2-2x%2B1 into %28x-1%29%5E2 (note: if you need help with factoring, check out this solver)



%28x-1%29%5E2=6 Combine like terms on the right side

x-1=0%2B-sqrt%286%29 Take the square root of both sides

x=1%2B-sqrt%286%29 Add 1 to both sides to isolate x.

So the expression breaks down to
x=1%2Bsqrt%286%29 or x=1-sqrt%286%29


So our answer is approximately
x=3.44948974278318 or x=-1.44948974278318

Here is visual proof

+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E2-2x-5%29+ graph of y=x%5E2-2x-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.




Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

x%5E2%2B4x-1=0


Solved by pluggable solver: COMPLETING THE SQUARE solver for quadratics
Read this lesson on completing the square by prince_abubu, if you do not know how to complete the square.
Let's convert 1x%5E2%2B4x%2B-1=0 to standard form by dividing both sides by 1:
We have: 1x%5E2%2B4x%2B-1=0. What we want to do now is to change this equation to a complete square %28x%2Bsomenumber%29%5E2+%2B+othernumber. How can we find out values of somenumber and othernumber that would make it work?
Look at %28x%2Bsomenumber%29%5E2: %28x%2Bsomenumber%29%5E2+=+x%5E2%2B2%2Asomenumber%2Ax+%2B+somenumber%5E2. Since the coefficient in our equation 1x%5E2%2Bhighlight_red%28+4%29+%2A+x%2B-1=0 that goes in front of x is 4, we know that 4=2*somenumber, or somenumber+=+4%2F2. So, we know that our equation can be rewritten as %28x%2B4%2F2%29%5E2+%2B+othernumber, and we do not yet know the other number.
We are almost there. Finding the other number is simply a matter of not making too many mistakes. We need to find 'other number' such that %28x%2B4%2F2%29%5E2+%2B+othernumber is equivalent to our original equation 1x%5E2%2B4x%2Bhighlight_green%28+-1+%29=0.


The highlighted red part must be equal to -1 (highlighted green part).

4%5E2%2F4+%2B+othernumber+=+-1, or othernumber+=+-1-4%5E2%2F4+=+-5.
So, the equation converts to %28x%2B4%2F2%29%5E2+%2B+-5+=+0, or %28x%2B4%2F2%29%5E2+=+5.

Our equation converted to a square %28x%2B4%2F2%29%5E2, equated to a number (5).

Since the right part 5 is greater than zero, there are two solutions:

system%28+%28x%2B4%2F2%29+=+%2Bsqrt%28+5+%29%2C+%28x%2B4%2F2%29+=+-sqrt%28+5+%29+%29
, or

system%28+%28x%2B4%2F2%29+=+2.23606797749979%2C+%28x%2B4%2F2%29+=+-2.23606797749979+%29
system%28+x%2B4%2F2+=+2.23606797749979%2C+x%2B4%2F2+=+-2.23606797749979+%29
system%28+x+=+2.23606797749979-4%2F2%2C+x+=+-2.23606797749979-4%2F2+%29

system%28+x+=+0.23606797749979%2C+x+=+-4.23606797749979+%29
Answer: x=0.23606797749979, -4.23606797749979.




x%5E2%2B6x-4=0

Solved by pluggable solver: COMPLETING THE SQUARE solver for quadratics
Read this lesson on completing the square by prince_abubu, if you do not know how to complete the square.
Let's convert 1x%5E2%2B6x%2B-4=0 to standard form by dividing both sides by 1:
We have: 1x%5E2%2B6x%2B-4=0. What we want to do now is to change this equation to a complete square %28x%2Bsomenumber%29%5E2+%2B+othernumber. How can we find out values of somenumber and othernumber that would make it work?
Look at %28x%2Bsomenumber%29%5E2: %28x%2Bsomenumber%29%5E2+=+x%5E2%2B2%2Asomenumber%2Ax+%2B+somenumber%5E2. Since the coefficient in our equation 1x%5E2%2Bhighlight_red%28+6%29+%2A+x%2B-4=0 that goes in front of x is 6, we know that 6=2*somenumber, or somenumber+=+6%2F2. So, we know that our equation can be rewritten as %28x%2B6%2F2%29%5E2+%2B+othernumber, and we do not yet know the other number.
We are almost there. Finding the other number is simply a matter of not making too many mistakes. We need to find 'other number' such that %28x%2B6%2F2%29%5E2+%2B+othernumber is equivalent to our original equation 1x%5E2%2B6x%2Bhighlight_green%28+-4+%29=0.


The highlighted red part must be equal to -4 (highlighted green part).

6%5E2%2F4+%2B+othernumber+=+-4, or othernumber+=+-4-6%5E2%2F4+=+-13.
So, the equation converts to %28x%2B6%2F2%29%5E2+%2B+-13+=+0, or %28x%2B6%2F2%29%5E2+=+13.

Our equation converted to a square %28x%2B6%2F2%29%5E2, equated to a number (13).

Since the right part 13 is greater than zero, there are two solutions:

system%28+%28x%2B6%2F2%29+=+%2Bsqrt%28+13+%29%2C+%28x%2B6%2F2%29+=+-sqrt%28+13+%29+%29
, or

system%28+%28x%2B6%2F2%29+=+3.60555127546399%2C+%28x%2B6%2F2%29+=+-3.60555127546399+%29
system%28+x%2B6%2F2+=+3.60555127546399%2C+x%2B6%2F2+=+-3.60555127546399+%29
system%28+x+=+3.60555127546399-6%2F2%2C+x+=+-3.60555127546399-6%2F2+%29

system%28+x+=+0.605551275463989%2C+x+=+-6.60555127546399+%29
Answer: x=0.605551275463989, -6.60555127546399.



x%5E2-2x-5=0

Solved by pluggable solver: COMPLETING THE SQUARE solver for quadratics
Read this lesson on completing the square by prince_abubu, if you do not know how to complete the square.
Let's convert 1x%5E2%2B-2x%2B-5=0 to standard form by dividing both sides by 1:
We have: 1x%5E2%2B-2x%2B-5=0. What we want to do now is to change this equation to a complete square %28x%2Bsomenumber%29%5E2+%2B+othernumber. How can we find out values of somenumber and othernumber that would make it work?
Look at %28x%2Bsomenumber%29%5E2: %28x%2Bsomenumber%29%5E2+=+x%5E2%2B2%2Asomenumber%2Ax+%2B+somenumber%5E2. Since the coefficient in our equation 1x%5E2%2Bhighlight_red%28+-2%29+%2A+x%2B-5=0 that goes in front of x is -2, we know that -2=2*somenumber, or somenumber+=+-2%2F2. So, we know that our equation can be rewritten as %28x%2B-2%2F2%29%5E2+%2B+othernumber, and we do not yet know the other number.
We are almost there. Finding the other number is simply a matter of not making too many mistakes. We need to find 'other number' such that %28x%2B-2%2F2%29%5E2+%2B+othernumber is equivalent to our original equation 1x%5E2%2B-2x%2Bhighlight_green%28+-5+%29=0.


The highlighted red part must be equal to -5 (highlighted green part).

-2%5E2%2F4+%2B+othernumber+=+-5, or othernumber+=+-5--2%5E2%2F4+=+-6.
So, the equation converts to %28x%2B-2%2F2%29%5E2+%2B+-6+=+0, or %28x%2B-2%2F2%29%5E2+=+6.

Our equation converted to a square %28x%2B-2%2F2%29%5E2, equated to a number (6).

Since the right part 6 is greater than zero, there are two solutions:

system%28+%28x%2B-2%2F2%29+=+%2Bsqrt%28+6+%29%2C+%28x%2B-2%2F2%29+=+-sqrt%28+6+%29+%29
, or

system%28+%28x%2B-2%2F2%29+=+2.44948974278318%2C+%28x%2B-2%2F2%29+=+-2.44948974278318+%29
system%28+x%2B-2%2F2+=+2.44948974278318%2C+x%2B-2%2F2+=+-2.44948974278318+%29
system%28+x+=+2.44948974278318--2%2F2%2C+x+=+-2.44948974278318--2%2F2+%29

system%28+x+=+3.44948974278318%2C+x+=+-1.44948974278318+%29
Answer: x=3.44948974278318, -1.44948974278318.