Solve 2x2 + 8x - 12 = 0 by completing the square. what are the solutions?

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Simplifying 2x2 + 8x + 12 = 0 Reorder the terms: 12 + 8x + 2x2 = 0 Solving 12 + 8x + 2x2 = 0 Solving for variable 'x'. Factor out the Greatest Common Factor (GCF), '2'. 2(6 + 4x + x2) = 0 Ignore the factor 2. Set the factor '(6 + 4x + x2)' equal to zero and attempt to solve: Simplifying 6 + 4x + x2 = 0 Solving 6 + 4x + x2 = 0 Begin completing the square. Move the constant term to the right: Add '-6' to each side of the equation. 6 + 4x + -6 + x2 = 0 + -6 Reorder the terms: 6 + -6 + 4x + x2 = 0 + -6 Combine like terms: 6 + -6 = 0 0 + 4x + x2 = 0 + -6 4x + x2 = 0 + -6 Combine like terms: 0 + -6 = -6 4x + x2 = -6 The x term is 4x. Take half its coefficient (2). Square it (4) and add it to both sides. Add '4' to each side of the equation. 4x + 4 + x2 = -6 + 4 Reorder the terms: 4 + 4x + x2 = -6 + 4 Combine like terms: -6 + 4 = -2 4 + 4x + x2 = -2 Factor a perfect square on the left side: (x + 2)(x + 2) = -2 Can't calculate square root of the right side. The solution to this equation could not be determined. This subproblem is being ignored because a solution could not be determined. The solution to this equation could not be determined.

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2x + 10 = 12 2x = 12 - 10 2x = 2 (divide both sides by 2 to get x) 2x/2 = 2/2 x = 1

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| 3x+2=18 |

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3x = 12 3x = 12 (divide both sides by 3 to get x) 3x/3 = 12/3 x = 4

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| 6x-2=14 | | 2x+10=12 | | 3x+2=18 |

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3x + 2 = 18 3x = 18 - 2 3x = 16 (divide both sides by 3 to get x) 3x/3 = 16/3 x = 5.333

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4x - 2 = 12 4x = 12 + 2 4x = 14 (divide both sides by 4 to get x) 4x/4 = 14/4 x = 3.5

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| 6x-2=14 | | 2x+10=12 | | 3x+2=18 |

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12 + x = 5 x = 5 - 12 x = -7

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| 4x-2=12 | | 6x-2=14 | | 2x+10=12 | | 3x+2=18 |

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2x^{2}-8x-12=0

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\left(-12\right)}}{2\times 2}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -8 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\left(-12\right)}}{2\times 2}

Square -8.

x=\frac{-\left(-8\right)±\sqrt{64-8\left(-12\right)}}{2\times 2}

Multiply -4 times 2.

x=\frac{-\left(-8\right)±\sqrt{64+96}}{2\times 2}

Multiply -8 times -12.

x=\frac{-\left(-8\right)±\sqrt{160}}{2\times 2}

Add 64 to 96.

x=\frac{-\left(-8\right)±4\sqrt{10}}{2\times 2}

Take the square root of 160.

x=\frac{8±4\sqrt{10}}{2\times 2}

The opposite of -8 is 8.

x=\frac{8±4\sqrt{10}}{4}

Multiply 2 times 2.

x=\frac{4\sqrt{10}+8}{4}

Now solve the equation x=\frac{8±4\sqrt{10}}{4} when ± is plus. Add 8 to 4\sqrt{10}.

x=\sqrt{10}+2

Divide 8+4\sqrt{10} by 4.

x=\frac{8-4\sqrt{10}}{4}

Now solve the equation x=\frac{8±4\sqrt{10}}{4} when ± is minus. Subtract 4\sqrt{10} from 8.

x=2-\sqrt{10}

Divide 8-4\sqrt{10} by 4.

x=\sqrt{10}+2 x=2-\sqrt{10}

The equation is now solved.

2x^{2}-8x-12=0

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

2x^{2}-8x-12-\left(-12\right)=-\left(-12\right)

Add 12 to both sides of the equation.

2x^{2}-8x=-\left(-12\right)

Subtracting -12 from itself leaves 0.

2x^{2}-8x=12

Subtract -12 from 0.

\frac{2x^{2}-8x}{2}=\frac{12}{2}

Divide both sides by 2.

x^{2}+\frac{-8}{2}x=\frac{12}{2}

Dividing by 2 undoes the multiplication by 2.

x^{2}-4x=\frac{12}{2}

Divide -8 by 2.

x^{2}-4x=6

Divide 12 by 2.

x^{2}-4x+\left(-2\right)^{2}=6+\left(-2\right)^{2}

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

x^{2}-4x+4=6+4

Square -2.

x^{2}-4x+4=10

Add 6 to 4.

\left(x-2\right)^{2}=10

Factor x^{2}-4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

\sqrt{\left(x-2\right)^{2}}=\sqrt{10}

Take the square root of both sides of the equation.

x-2=\sqrt{10} x-2=-\sqrt{10}

Simplify.

x=\sqrt{10}+2 x=2-\sqrt{10}

Add 2 to both sides of the equation.

x ^ 2 -4x -6 = 0

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 2

r + s = 4 rs = -6

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

r = 2 - u s = 2 + u

Two numbers r and s sum up to 4 exactly when the average of the two numbers is \frac{1}{2}*4 = 2. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

(2 - u) (2 + u) = -6

To solve for unknown quantity u, substitute these in the product equation rs = -6

4 - u^2 = -6

Simplify by expanding (a -b) (a + b) = a^2 – b^2

-u^2 = -6-4 = -10

Simplify the expression by subtracting 4 on both sides

u^2 = 10 u = \pm\sqrt{10} = \pm \sqrt{10}

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

r =2 - \sqrt{10} = -1.162 s = 2 + \sqrt{10} = 5.162

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

Changes made to your input should not affect the solution:

 (1): "x2"   was replaced by   "x^2". 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

Step  2  :

Step  3  :

Pulling out like terms :

 3.1     Pull out like factors :

   -2x2 - 8x - 12  =   -2 • (x2 + 4x + 6) 

Trying to factor by splitting the middle term

 3.2     Factoring  x2 + 4x + 6 

The first term is,  x2  its coefficient is  1 .

Step-1 : Multiply the coefficient of the first term by the constant   1 • 6 = 6 

Step-2 : Find two factors of  6  whose sum equals the coefficient of the middle term, which is   4 .

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Equation at the end of step  3  :

Step  4  :

Equations which are never true :

 4.1      Solve :    -2   =  0

This equation has no solution.
A a non-zero constant never equals zero.

Parabola, Finding the Vertex :

 4.2      Find the Vertex of   Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  "y"  because the coefficient of the first term, 1 , is positive (greater than zero).Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.For any parabola,Ax2+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is  -2.0000 Plugging into the parabola formula  -2.0000  for  x  we can calculate the  y -coordinate : 
  y = 1.0 * -2.00 * -2.00 + 4.0 * -2.00 + 6.0
or   y = 2.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = x2+4x+6
Axis of Symmetry (dashed)  {x}={-2.00} 
Vertex at  {x,y} = {-2.00, 2.00}  Function has no real roots

Solve Quadratic Equation by Completing The Square

 4.3     Solving   x2+4x+6 = 0 by Completing The SquareSubtract  6  from both side of the equation :
   x2+4x = -6

Now the clever bit: Take the coefficient of  x , which is  4 , divide by two, giving  2 , and finally square it giving  4 

Add  4  to both sides of the equation :

   -6  +  4    or,  (-6/1)+(4/1) 

   x2+4x+4 = -2

Adding  4  has completed the left hand side into a perfect square :

   x2+4x+4 = -2 and

   (x+2)2 = -2

We'll refer to this Equation as  Eq. #4.3.1  

The Square Root Principle says that When two things are equal, their square roots are equal.

   (x+2)2   is

Now, applying the Square Root Principle to  Eq. #4.3.1  we get:

Subtract  2  from both sides to obtain:

   x2 + 4x + 6 = 0

  x = -2 + √ 2 •  i 

  x = -2 - √ 2 •  i 

Solve Quadratic Equation using the Quadratic Formula

 4.4     Solving    x2+4x+6 = 0 by the Quadratic FormulaAccording to the Quadratic Formula,  x  , the solution for   Ax2+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :

                                     
            - B  ±  √ B2-4AC  x =   ————————                      2A

               In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written  (a+b*i) 

Both   i   and   -i   are the square roots of minus 1

Accordingly,√ -8  = 
                    √ 8 • (-1)  =
                    √ 8  • √ -1   =
                    ±  √ 8  • i

Can  √ 8 be simplified ?

Yes!   The prime factorization of  8   is

√ 8   =  √ 2•2•2   =
                ±  2 • √ 2

  √ 2   , rounded to 4 decimal digits, is   1.4142
 So now we are looking at:

Two imaginary solutions :

Two solutions were found :

1.  x =(-4-√-8)/2=-2-i√ 2 = -2.0000-1.4142i
2.  x =(-4+√-8)/2=-2+i√ 2 = -2.0000+1.4142i