Solve for z
z=\frac{5}{3}+\frac{4}{3}i\approx 1.666666667+1.333333333i
z=\frac{5}{3}-\frac{4}{3}i\approx 1.666666667-1.333333333i
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9z^{2}-30z+41=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.
z=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 9\times 41}}{2\times 9}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -30 for b, and 41 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
z=\frac{-\left(-30\right)±\sqrt{900-4\times 9\times 41}}{2\times 9}
Square -30.
z=\frac{-\left(-30\right)±\sqrt{900-36\times 41}}{2\times 9}
Multiply -4 times 9.
z=\frac{-\left(-30\right)±\sqrt{900-1476}}{2\times 9}
Multiply -36 times 41.
z=\frac{-\left(-30\right)±\sqrt{-576}}{2\times 9}
Add 900 to -1476.
z=\frac{-\left(-30\right)±24i}{2\times 9}
Take the square root of -576.
z=\frac{30±24i}{2\times 9}
The opposite of -30 is 30.
z=\frac{30±24i}{18}
Multiply 2 times 9.
z=\frac{30+24i}{18}
Now solve the equation z=\frac{30±24i}{18} when ± is plus. Add 30 to 24i.
z=\frac{5}{3}+\frac{4}{3}i
Divide 30+24i by 18.
z=\frac{30-24i}{18}
Now solve the equation z=\frac{30±24i}{18} when ± is minus. Subtract 24i from 30.
z=\frac{5}{3}-\frac{4}{3}i
Divide 30-24i by 18.
z=\frac{5}{3}+\frac{4}{3}i z=\frac{5}{3}-\frac{4}{3}i
The equation is now solved.
9z^{2}-30z+41=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.
9z^{2}-30z+41-41=-41
Subtract 41 from both sides of the equation.
9z^{2}-30z=-41
Subtracting 41 from itself leaves 0.
\frac{9z^{2}-30z}{9}=-\frac{41}{9}
Divide both sides by 9.
z^{2}+\left(-\frac{30}{9}\right)z=-\frac{41}{9}
Dividing by 9 undoes the multiplication by 9.
z^{2}-\frac{10}{3}z=-\frac{41}{9}
Reduce the fraction \frac{-30}{9} to lowest terms by extracting and canceling out 3.
z^{2}-\frac{10}{3}z+\left(-\frac{5}{3}\right)^{2}=-\frac{41}{9}+\left(-\frac{5}{3}\right)^{2}
Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
z^{2}-\frac{10}{3}z+\frac{25}{9}=\frac{-41+25}{9}
Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.
z^{2}-\frac{10}{3}z+\frac{25}{9}=-\frac{16}{9}
Add -\frac{41}{9} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(z-\frac{5}{3}\right)^{2}=-\frac{16}{9}
Factor z^{2}-\frac{10}{3}z+\frac{25}{9}. 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(z-\frac{5}{3}\right)^{2}}=\sqrt{-\frac{16}{9}}
Take the square root of both sides of the equation.
z-\frac{5}{3}=\frac{4}{3}i z-\frac{5}{3}=-\frac{4}{3}i
Simplify.
z=\frac{5}{3}+\frac{4}{3}i z=\frac{5}{3}-\frac{4}{3}i
Add \frac{5}{3} to both sides of the equation.
x ^ 2 -\frac{10}{3}x +\frac{41}{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{10}{3} rs = \frac{41}{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{5}{3} - u s = \frac{5}{3} + u
Two numbers r and s sum up to \frac{10}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{10}{3} = \frac{5}{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{5}{3} - u) (\frac{5}{3} + u) = \frac{41}{9}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{41}{9}
\frac{25}{9} - u^2 = \frac{41}{9}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{41}{9}-\frac{25}{9} = \frac{16}{9}
Simplify the expression by subtracting \frac{25}{9} on both sides
u^2 = -\frac{16}{9} u = \pm\sqrt{-\frac{16}{9}} = \pm \frac{4}{3}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{5}{3} - \frac{4}{3}i = 1.667 - 1.333i s = \frac{5}{3} + \frac{4}{3}i = 1.667 + 1.333i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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