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7x^{2}+6x+1=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-6±\sqrt{6^{2}-4\times 7}}{2\times 7}
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{-6±\sqrt{36-4\times 7}}{2\times 7}
Square 6.
x=\frac{-6±\sqrt{36-28}}{2\times 7}
Multiply -4 times 7.
x=\frac{-6±\sqrt{8}}{2\times 7}
Add 36 to -28.
x=\frac{-6±2\sqrt{2}}{2\times 7}
Take the square root of 8.
x=\frac{-6±2\sqrt{2}}{14}
Multiply 2 times 7.
x=\frac{2\sqrt{2}-6}{14}
Now solve the equation x=\frac{-6±2\sqrt{2}}{14} when ± is plus. Add -6 to 2\sqrt{2}.
x=\frac{\sqrt{2}-3}{7}
Divide -6+2\sqrt{2} by 14.
x=\frac{-2\sqrt{2}-6}{14}
Now solve the equation x=\frac{-6±2\sqrt{2}}{14} when ± is minus. Subtract 2\sqrt{2} from -6.
x=\frac{-\sqrt{2}-3}{7}
Divide -6-2\sqrt{2} by 14.
7x^{2}+6x+1=7\left(x-\frac{\sqrt{2}-3}{7}\right)\left(x-\frac{-\sqrt{2}-3}{7}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3+\sqrt{2}}{7} for x_{1} and \frac{-3-\sqrt{2}}{7} for x_{2}.
x ^ 2 +\frac{6}{7}x +\frac{1}{7} = 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 7
r + s = -\frac{6}{7} rs = \frac{1}{7}
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 = -\frac{3}{7} - u s = -\frac{3}{7} + u
Two numbers r and s sum up to -\frac{6}{7} exactly when the average of the two numbers is \frac{1}{2}*-\frac{6}{7} = -\frac{3}{7}. 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>
(-\frac{3}{7} - u) (-\frac{3}{7} + u) = \frac{1}{7}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{7}
\frac{9}{49} - u^2 = \frac{1}{7}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{7}-\frac{9}{49} = -\frac{2}{49}
Simplify the expression by subtracting \frac{9}{49} on both sides
u^2 = \frac{2}{49} u = \pm\sqrt{\frac{2}{49}} = \pm \frac{\sqrt{2}}{7}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{3}{7} - \frac{\sqrt{2}}{7} = -0.631 s = -\frac{3}{7} + \frac{\sqrt{2}}{7} = -0.227
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.