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q^{2}+2q+1=0
Divide both sides by 4.
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 q^{2}+aq+bq+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(q^{2}+q\right)+\left(q+1\right)
Rewrite q^{2}+2q+1 as \left(q^{2}+q\right)+\left(q+1\right).
q\left(q+1\right)+q+1
Factor out q in q^{2}+q.
\left(q+1\right)\left(q+1\right)
Factor out common term q+1 by using distributive property.
\left(q+1\right)^{2}
Rewrite as a binomial square.
q=-1
To find equation solution, solve q+1=0.
4q^{2}+8q+4=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.
q=\frac{-8±\sqrt{8^{2}-4\times 4\times 4}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 8 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
q=\frac{-8±\sqrt{64-4\times 4\times 4}}{2\times 4}
Square 8.
q=\frac{-8±\sqrt{64-16\times 4}}{2\times 4}
Multiply -4 times 4.
q=\frac{-8±\sqrt{64-64}}{2\times 4}
Multiply -16 times 4.
q=\frac{-8±\sqrt{0}}{2\times 4}
Add 64 to -64.
q=-\frac{8}{2\times 4}
Take the square root of 0.
q=-\frac{8}{8}
Multiply 2 times 4.
q=-1
Divide -8 by 8.
4q^{2}+8q+4=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.
4q^{2}+8q+4-4=-4
Subtract 4 from both sides of the equation.
4q^{2}+8q=-4
Subtracting 4 from itself leaves 0.
\frac{4q^{2}+8q}{4}=-\frac{4}{4}
Divide both sides by 4.
q^{2}+\frac{8}{4}q=-\frac{4}{4}
Dividing by 4 undoes the multiplication by 4.
q^{2}+2q=-\frac{4}{4}
Divide 8 by 4.
q^{2}+2q=-1
Divide -4 by 4.
q^{2}+2q+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.
q^{2}+2q+1=-1+1
Square 1.
q^{2}+2q+1=0
Add -1 to 1.
\left(q+1\right)^{2}=0
Factor q^{2}+2q+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(q+1\right)^{2}}=\sqrt{0}
Take the square root of both sides of the equation.
q+1=0 q+1=0
Simplify.
q=-1 q=-1
Subtract 1 from both sides of the equation.
q=-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 4
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.