Skip to main content
Solve for x (complex solution)
Tick mark Image
Graph

Similar Problems from Web Search

Share

54x^{2}-12x+9=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.
x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 54\times 9}}{2\times 54}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 54 for a, -12 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-12\right)±\sqrt{144-4\times 54\times 9}}{2\times 54}
Square -12.
x=\frac{-\left(-12\right)±\sqrt{144-216\times 9}}{2\times 54}
Multiply -4 times 54.
x=\frac{-\left(-12\right)±\sqrt{144-1944}}{2\times 54}
Multiply -216 times 9.
x=\frac{-\left(-12\right)±\sqrt{-1800}}{2\times 54}
Add 144 to -1944.
x=\frac{-\left(-12\right)±30\sqrt{2}i}{2\times 54}
Take the square root of -1800.
x=\frac{12±30\sqrt{2}i}{2\times 54}
The opposite of -12 is 12.
x=\frac{12±30\sqrt{2}i}{108}
Multiply 2 times 54.
x=\frac{12+30\sqrt{2}i}{108}
Now solve the equation x=\frac{12±30\sqrt{2}i}{108} when ± is plus. Add 12 to 30i\sqrt{2}.
x=\frac{5\sqrt{2}i}{18}+\frac{1}{9}
Divide 12+30i\sqrt{2} by 108.
x=\frac{-30\sqrt{2}i+12}{108}
Now solve the equation x=\frac{12±30\sqrt{2}i}{108} when ± is minus. Subtract 30i\sqrt{2} from 12.
x=-\frac{5\sqrt{2}i}{18}+\frac{1}{9}
Divide 12-30i\sqrt{2} by 108.
x=\frac{5\sqrt{2}i}{18}+\frac{1}{9} x=-\frac{5\sqrt{2}i}{18}+\frac{1}{9}
The equation is now solved.
54x^{2}-12x+9=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.
54x^{2}-12x+9-9=-9
Subtract 9 from both sides of the equation.
54x^{2}-12x=-9
Subtracting 9 from itself leaves 0.
\frac{54x^{2}-12x}{54}=-\frac{9}{54}
Divide both sides by 54.
x^{2}+\left(-\frac{12}{54}\right)x=-\frac{9}{54}
Dividing by 54 undoes the multiplication by 54.
x^{2}-\frac{2}{9}x=-\frac{9}{54}
Reduce the fraction \frac{-12}{54} to lowest terms by extracting and canceling out 6.
x^{2}-\frac{2}{9}x=-\frac{1}{6}
Reduce the fraction \frac{-9}{54} to lowest terms by extracting and canceling out 9.
x^{2}-\frac{2}{9}x+\left(-\frac{1}{9}\right)^{2}=-\frac{1}{6}+\left(-\frac{1}{9}\right)^{2}
Divide -\frac{2}{9}, the coefficient of the x term, by 2 to get -\frac{1}{9}. Then add the square of -\frac{1}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{2}{9}x+\frac{1}{81}=-\frac{1}{6}+\frac{1}{81}
Square -\frac{1}{9} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{2}{9}x+\frac{1}{81}=-\frac{25}{162}
Add -\frac{1}{6} to \frac{1}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{9}\right)^{2}=-\frac{25}{162}
Factor x^{2}-\frac{2}{9}x+\frac{1}{81}. 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(x-\frac{1}{9}\right)^{2}}=\sqrt{-\frac{25}{162}}
Take the square root of both sides of the equation.
x-\frac{1}{9}=\frac{5\sqrt{2}i}{18} x-\frac{1}{9}=-\frac{5\sqrt{2}i}{18}
Simplify.
x=\frac{5\sqrt{2}i}{18}+\frac{1}{9} x=-\frac{5\sqrt{2}i}{18}+\frac{1}{9}
Add \frac{1}{9} to both sides of the equation.
x ^ 2 -\frac{2}{9}x +\frac{1}{6} = 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 54
r + s = \frac{2}{9} rs = \frac{1}{6}
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}{9} - u s = \frac{1}{9} + u
Two numbers r and s sum up to \frac{2}{9} exactly when the average of the two numbers is \frac{1}{2}*\frac{2}{9} = \frac{1}{9}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(\frac{1}{9} - u) (\frac{1}{9} + u) = \frac{1}{6}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{6}
\frac{1}{81} - u^2 = \frac{1}{6}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{6}-\frac{1}{81} = \frac{25}{162}
Simplify the expression by subtracting \frac{1}{81} on both sides
u^2 = -\frac{25}{162} u = \pm\sqrt{-\frac{25}{162}} = \pm \frac{5}{\sqrt{162}}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}{9} - \frac{5}{\sqrt{162}}i = 0.111 - 0.393i s = \frac{1}{9} + \frac{5}{\sqrt{162}}i = 0.111 + 0.393i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.