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

Question 107967: solve the equation by completing the square

Found 2 solutions by jim_thompson5910, MathLover1:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
#1

Start with the given equation

Add 1 to both sides

Take half of the x coefficient 4 to get 2 (ie )
Now square 2 to get 4 (ie )

Add this result (4) to both sides. Now the expression is a perfect square trinomial.

Factor into (note: if you need help with factoring, check out this solver)

Combine like terms on the right side

Take the square root of both sides

Subtract 2 from both sides to isolate x.

So the expression breaks down to
or

So our answer is approximately
or

Here is visual proof

graph of

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

#2

Start with the given equation

Add 4 to both sides

Take half of the x coefficient 6 to get 3 (ie )
Now square 3 to get 9 (ie )

Add this result (9) to both sides. Now the expression is a perfect square trinomial.

Factor into (note: if you need help with factoring, check out this solver)

Combine like terms on the right side

Take the square root of both sides

Subtract 3 from both sides to isolate x.

So the expression breaks down to
or

So our answer is approximately
or

Here is visual proof

graph of

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

#3

Start with the given equation

Add 5 to both sides

Take half of the x coefficient -2 to get -1 (ie )
Now square -1 to get 1 (ie )

Add this result (1) to both sides. Now the expression is a perfect square trinomial.

Factor into (note: if you need help with factoring, check out this solver)

Combine like terms on the right side

Take the square root of both sides

Add 1 to both sides to isolate x.

So the expression breaks down to
or

So our answer is approximately
or

Here is visual proof

graph of

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

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

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 to standard form by dividing both sides by 1:
We have: . What we want to do now is to change this equation to a complete square . How can we find out values of somenumber and othernumber that would make it work?
Look at : . Since the coefficient in our equation that goes in front of x is 4, we know that 4=2*somenumber, or . So, we know that our equation can be rewritten as , 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 is equivalent to our original equation .

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

, or .
So, the equation converts to , or .

Our equation converted to a square , equated to a number (5).

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

, or

Answer: x=0.23606797749979, -4.23606797749979.

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 to standard form by dividing both sides by 1:
We have: . What we want to do now is to change this equation to a complete square . How can we find out values of somenumber and othernumber that would make it work?
Look at : . Since the coefficient in our equation that goes in front of x is 6, we know that 6=2*somenumber, or . So, we know that our equation can be rewritten as , 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 is equivalent to our original equation .

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

, or .
So, the equation converts to , or .

Our equation converted to a square , equated to a number (13).

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

, or

Answer: x=0.605551275463989, -6.60555127546399.

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 to standard form by dividing both sides by 1:
We have: . What we want to do now is to change this equation to a complete square . How can we find out values of somenumber and othernumber that would make it work?
Look at : . Since the coefficient in our equation that goes in front of x is -2, we know that -2=2*somenumber, or . So, we know that our equation can be rewritten as , 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 is equivalent to our original equation .

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

, or .
So, the equation converts to , or .

Our equation converted to a square , equated to a number (6).

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

, or

Answer: x=3.44948974278318, -1.44948974278318.

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