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2\left(4q^{2}-8q+5\right)
Factor out 2. Polynomial 4q^{2}-8q+5 is not factored since it does not have any rational roots.
8q^{2}-16q+10=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.
q=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 8\times 10}}{2\times 8}
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{-\left(-16\right)±\sqrt{256-4\times 8\times 10}}{2\times 8}
Square -16.
q=\frac{-\left(-16\right)±\sqrt{256-32\times 10}}{2\times 8}
Multiply -4 times 8.
q=\frac{-\left(-16\right)±\sqrt{256-320}}{2\times 8}
Multiply -32 times 10.
q=\frac{-\left(-16\right)±\sqrt{-64}}{2\times 8}
Add 256 to -320.
8q^{2}-16q+10
Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.
x ^ 2 -2x +\frac{5}{4} = 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 8
r + s = 2 rs = \frac{5}{4}
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) = \frac{5}{4}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{4}
1 - u^2 = \frac{5}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{5}{4}-1 = \frac{1}{4}
Simplify the expression by subtracting 1 on both sides
u^2 = -\frac{1}{4} u = \pm\sqrt{-\frac{1}{4}} = \pm \frac{1}{2}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =1 - \frac{1}{2}i = 1 - 0.500i s = 1 + \frac{1}{2}i = 1 + 0.500i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.