SOLUTION: Please show me how to Solve the equation x2 + 8x – 2 = 0 using both The quadratic formula and Completing the square.

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Question 389186: Please show me how to Solve the equation x2 + 8x – 2 = 0 using both The quadratic formula and Completing the square.
Found 2 solutions by haileytucki, jim_thompson5910:
Answer by haileytucki(390) About Me  (Show Source):
You can put this solution on YOUR website!
Quadratic First (~=square root)
x^(2)+8x-2=0
Use the quadratic formula to find the solutions. In this case, the values are a=1, b=8, and c=-2.
x=(-b\~(b^(2)-4ac))/(2a) where ax^(2)+bx+c=0
Use the standard form of the equation to find a, b, and c for this quadratic.
a=1, b=8, and c=-2
Substitute in the values of a=1, b=8, and c=-2.
x=(-8\~((8)^(2)-4(1)(-2)))/(2(1))
Simplify the section inside the radical.
x=(-8\6~(2))/(2(1))
Simplify the denominator of the quadratic formula.
x=(-8\6~(2))/(2)
First, solve the + portion of +-.
x=(-8+6~(2))/(2)
Factor out the GCF of 2 from each term in the polynomial.
x=(2(-4)+2(3~(2)))/(2)
Factor out the GCF of 2 from -8+6~(2).
x=(2(-4+3~(2)))/(2)
Reduce the expression (2(-4+3~(2)))/(2) by removing a factor of 2 from the numerator and denominator.
x=(-4+3~(2))
Remove the parentheses around the expression -4+3~(2).
x=-4+3~(2)
Next, solve the - portion of +-.
x=(-8-6~(2))/(2)
Factor out the GCF of 2 from each term in the polynomial.
x=(2(-4)+2(-3~(2)))/(2)
Factor out the GCF of 2 from -8-6~(2).
x=(2(-4-3~(2)))/(2)
Reduce the expression (2(-4-3~(2)))/(2) by removing a factor of 2 from the numerator and denominator.
x=(-4-3~(2))
Remove the parentheses around the expression -4-3~(2).
x=-4-3~(2)
The final answer is the combination of both solutions.
x=-4+3~(2),-4-3~(2)_x=0.2426407,-8.24264

Now, completing the square

x^(2)+8x-2=0
Since -2 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 2 to both sides.
x^(2)+8x=2
To create a trinomial square on the left-hand side of the equation, add a value to both sides of the equation that is equal to the square of half the coefficient of x. In this problem, add (4)^(2) to both sides of the equation.
x^(2)+8x+16=2+16
Add 16 to 2 to get 18.
x^(2)+8x+16=18
Factor the perfect trinomial square into (x+4)^(2).
(x+4)^(2)=18
Take the square root of each side of the equation to setup the solution for x.
~((x+4)^(2))=\~(18)
Remove the perfect root factor (x+4) under the radical to solve for x.
(x+4)=\~(18)
Pull all perfect square roots out from under the radical. In this case, remove the 3 because it is a perfect square.
(x+4)=\3~(2)
First, substitute in the + portion of the \ to find the first solution.
(x+4)=3~(2)
Remove the parentheses around the expression x+4.
x+4=3~(2)
Since 4 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 4 from both sides.
x=-4+3~(2)
Move all terms not containing x to the right-hand side of the equation.
x=3~(2)-4
Next, substitute in the - portion of the \ to find the second solution.
(x+4)=-3~(2)
Remove the parentheses around the expression x+4.
x+4=-3~(2)
Since 4 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 4 from both sides.
x=-4+-3~(2)
Move all terms not containing x to the right-hand side of the equation.
x=-3~(2)-4
The complete solution is the result of both the + and - portions of the solution.
x=3~(2)-4,-3~(2)-4

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Solving by use of the quadratic formula:


x%5E2%2B8x-2=0 Start with the given equation.


Notice that the quadratic x%5E2%2B8x-2 is in the form of Ax%5E2%2BBx%2BC where A=1, B=8, and C=-2


Let's use the quadratic formula to solve for "x":


x+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


x+=+%28-%288%29+%2B-+sqrt%28+%288%29%5E2-4%281%29%28-2%29+%29%29%2F%282%281%29%29 Plug in A=1, B=8, and C=-2


x+=+%28-8+%2B-+sqrt%28+64-4%281%29%28-2%29+%29%29%2F%282%281%29%29 Square 8 to get 64.


x+=+%28-8+%2B-+sqrt%28+64--8+%29%29%2F%282%281%29%29 Multiply 4%281%29%28-2%29 to get -8


x+=+%28-8+%2B-+sqrt%28+64%2B8+%29%29%2F%282%281%29%29 Rewrite sqrt%2864--8%29 as sqrt%2864%2B8%29


x+=+%28-8+%2B-+sqrt%28+72+%29%29%2F%282%281%29%29 Add 64 to 8 to get 72


x+=+%28-8+%2B-+sqrt%28+72+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


x+=+%28-8+%2B-+6%2Asqrt%282%29%29%2F%282%29 Simplify the square root (note: If you need help with simplifying square roots, check out this solver)


x+=+%28-8%29%2F%282%29+%2B-+%286%2Asqrt%282%29%29%2F%282%29 Break up the fraction.


x+=+-4+%2B-+3%2Asqrt%282%29 Reduce.


x+=+-4%2B3%2Asqrt%282%29 or x+=+-4-3%2Asqrt%282%29 Break up the expression.


So the solutions are x+=+-4%2B3%2Asqrt%282%29 or x+=+-4-3%2Asqrt%282%29


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Solving by use of completing the square:


First we need to complete the square for the expression x%5E2%2B8x-2


x%5E2%2B8x-2 Start with the given expression.


Take half of the x coefficient 8 to get 4. In other words, %281%2F2%29%288%29=4.


Now square 4 to get 16. In other words, %284%29%5E2=%284%29%284%29=16


x%5E2%2B8x%2Bhighlight%2816-16%29-2 Now add and subtract 16. Make sure to place this after the "x" term. Notice how 16-16=0. So the expression is not changed.


%28x%5E2%2B8x%2B16%29-16-2 Group the first three terms.


%28x%2B4%29%5E2-16-2 Factor x%5E2%2B8x%2B16 to get %28x%2B4%29%5E2.


%28x%2B4%29%5E2-18 Combine like terms.


So after completing the square, x%5E2%2B8x-2 transforms to %28x%2B4%29%5E2-18. So x%5E2%2B8x-2=%28x%2B4%29%5E2-18.


So x%5E2%2B8x-2=0 is equivalent to %28x%2B4%29%5E2-18=0.


Now let's solve %28x%2B4%29%5E2-18=0


%28x%2B4%29%5E2-18=0 Start with the given equation.


%28x%2B4%29%5E2=0%2B18Add 18 to both sides.


%28x%2B4%29%5E2=18 Combine like terms.


x%2B4=%22%22%2B-sqrt%2818%29 Take the square root of both sides.


x%2B4=sqrt%2818%29 or x%2B4=-sqrt%2818%29 Break up the "plus/minus" to form two equations.


x%2B4=3%2Asqrt%282%29 or x%2B4=-3%2Asqrt%282%29 Simplify the square root.


x=-4%2B3%2Asqrt%282%29 or x=-4-3%2Asqrt%282%29 Subtract 4 from both sides.


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Answer:


So the solutions are x=-4%2B3%2Asqrt%282%29 or x=-4-3%2Asqrt%282%29.


Notice how we get the same answers. So either method works.


If you need more help, email me at jim_thompson5910@hotmail.com

Also, feel free to check out my tutoring website

Jim