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v^{2}+2v+1=0
Divide both sides by 2.
a+b=2 ab=1\times 1=1
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as v^{2}+av+bv+1. To find a and b, set up a system to be solved.
a=1 b=1
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.
\left(v^{2}+v\right)+\left(v+1\right)
Rewrite v^{2}+2v+1 as \left(v^{2}+v\right)+\left(v+1\right).
v\left(v+1\right)+v+1
Factor out v in v^{2}+v.
\left(v+1\right)\left(v+1\right)
Factor out common term v+1 by using distributive property.
\left(v+1\right)^{2}
Rewrite as a binomial square.
v=-1
To find equation solution, solve v+1=0.
2v^{2}+4v+2=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.
v=\frac{-4±\sqrt{4^{2}-4\times 2\times 2}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 4 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
v=\frac{-4±\sqrt{16-4\times 2\times 2}}{2\times 2}
Square 4.
v=\frac{-4±\sqrt{16-8\times 2}}{2\times 2}
Multiply -4 times 2.
v=\frac{-4±\sqrt{16-16}}{2\times 2}
Multiply -8 times 2.
v=\frac{-4±\sqrt{0}}{2\times 2}
Add 16 to -16.
v=-\frac{4}{2\times 2}
Take the square root of 0.
v=-\frac{4}{4}
Multiply 2 times 2.
v=-1
Divide -4 by 4.
2v^{2}+4v+2=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.
2v^{2}+4v+2-2=-2
Subtract 2 from both sides of the equation.
2v^{2}+4v=-2
Subtracting 2 from itself leaves 0.
\frac{2v^{2}+4v}{2}=-\frac{2}{2}
Divide both sides by 2.
v^{2}+\frac{4}{2}v=-\frac{2}{2}
Dividing by 2 undoes the multiplication by 2.
v^{2}+2v=-\frac{2}{2}
Divide 4 by 2.
v^{2}+2v=-1
Divide -2 by 2.
v^{2}+2v+1^{2}=-1+1^{2}
Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
v^{2}+2v+1=-1+1
Square 1.
v^{2}+2v+1=0
Add -1 to 1.
\left(v+1\right)^{2}=0
Factor v^{2}+2v+1. 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(v+1\right)^{2}}=\sqrt{0}
Take the square root of both sides of the equation.
v+1=0 v+1=0
Simplify.
v=-1 v=-1
Subtract 1 from both sides of the equation.
v=-1
The equation is now solved. Solutions are the same.
x ^ 2 +2x +1 = 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 = -2 rs = 1
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 = -1 - u s = -1 + u
Two numbers r and s sum up to -2 exactly when the average of the two numbers is \frac{1}{2}*-2 = -1. 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>
(-1 - u) (-1 + u) = 1
To solve for unknown quantity u, substitute these in the product equation rs = 1
1 - u^2 = 1
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 1-1 = 0
Simplify the expression by subtracting 1 on both sides
u^2 = 0 u = 0
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r = s = -1
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.