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3\left(8x^{2}-4x+3\right)
Factor out 3. Polynomial 8x^{2}-4x+3 is not factored since it does not have any rational roots.
24x^{2}-12x+9=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{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 24\times 9}}{2\times 24}
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(-12\right)±\sqrt{144-4\times 24\times 9}}{2\times 24}
Square -12.
x=\frac{-\left(-12\right)±\sqrt{144-96\times 9}}{2\times 24}
Multiply -4 times 24.
x=\frac{-\left(-12\right)±\sqrt{144-864}}{2\times 24}
Multiply -96 times 9.
x=\frac{-\left(-12\right)±\sqrt{-720}}{2\times 24}
Add 144 to -864.
24x^{2}-12x+9
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 -\frac{1}{2}x +\frac{3}{8} = 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 24
r + s = \frac{1}{2} rs = \frac{3}{8}
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{1}{4} - u s = \frac{1}{4} + u
Two numbers r and s sum up to \frac{1}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{2} = \frac{1}{4}. 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{1}{4} - u) (\frac{1}{4} + u) = \frac{3}{8}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{8}
\frac{1}{16} - u^2 = \frac{3}{8}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{3}{8}-\frac{1}{16} = \frac{5}{16}
Simplify the expression by subtracting \frac{1}{16} on both sides
u^2 = -\frac{5}{16} u = \pm\sqrt{-\frac{5}{16}} = \pm \frac{\sqrt{5}}{4}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{1}{4} - \frac{\sqrt{5}}{4}i = 0.250 - 0.559i s = \frac{1}{4} + \frac{\sqrt{5}}{4}i = 0.250 + 0.559i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.