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9x^{2}+24x+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{-24±\sqrt{24^{2}-4\times 9}}{2\times 9}
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{-24±\sqrt{576-4\times 9}}{2\times 9}
Square 24.
x=\frac{-24±\sqrt{576-36}}{2\times 9}
Multiply -4 times 9.
x=\frac{-24±\sqrt{540}}{2\times 9}
Add 576 to -36.
x=\frac{-24±6\sqrt{15}}{2\times 9}
Take the square root of 540.
x=\frac{-24±6\sqrt{15}}{18}
Multiply 2 times 9.
x=\frac{6\sqrt{15}-24}{18}
Now solve the equation x=\frac{-24±6\sqrt{15}}{18} when ± is plus. Add -24 to 6\sqrt{15}.
x=\frac{\sqrt{15}-4}{3}
Divide -24+6\sqrt{15} by 18.
x=\frac{-6\sqrt{15}-24}{18}
Now solve the equation x=\frac{-24±6\sqrt{15}}{18} when ± is minus. Subtract 6\sqrt{15} from -24.
x=\frac{-\sqrt{15}-4}{3}
Divide -24-6\sqrt{15} by 18.
9x^{2}+24x+1=9\left(x-\frac{\sqrt{15}-4}{3}\right)\left(x-\frac{-\sqrt{15}-4}{3}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-4+\sqrt{15}}{3} for x_{1} and \frac{-4-\sqrt{15}}{3} for x_{2}.
x ^ 2 +\frac{8}{3}x +\frac{1}{9} = 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 9
r + s = -\frac{8}{3} rs = \frac{1}{9}
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{4}{3} - u s = -\frac{4}{3} + u
Two numbers r and s sum up to -\frac{8}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{8}{3} = -\frac{4}{3}. 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{4}{3} - u) (-\frac{4}{3} + u) = \frac{1}{9}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{9}
\frac{16}{9} - u^2 = \frac{1}{9}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{9}-\frac{16}{9} = -\frac{5}{3}
Simplify the expression by subtracting \frac{16}{9} on both sides
u^2 = \frac{5}{3} u = \pm\sqrt{\frac{5}{3}} = \pm \frac{\sqrt{5}}{\sqrt{3}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{4}{3} - \frac{\sqrt{5}}{\sqrt{3}} = -2.624 s = -\frac{4}{3} + \frac{\sqrt{5}}{\sqrt{3}} = -0.042
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.