Skip to main content
Solve for z
Tick mark Image

Similar Problems from Web Search

Share

3z^{2}+6z+8=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{-6±\sqrt{6^{2}-4\times 3\times 8}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 6 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
z=\frac{-6±\sqrt{36-4\times 3\times 8}}{2\times 3}
Square 6.
z=\frac{-6±\sqrt{36-12\times 8}}{2\times 3}
Multiply -4 times 3.
z=\frac{-6±\sqrt{36-96}}{2\times 3}
Multiply -12 times 8.
z=\frac{-6±\sqrt{-60}}{2\times 3}
Add 36 to -96.
z=\frac{-6±2\sqrt{15}i}{2\times 3}
Take the square root of -60.
z=\frac{-6±2\sqrt{15}i}{6}
Multiply 2 times 3.
z=\frac{-6+2\sqrt{15}i}{6}
Now solve the equation z=\frac{-6±2\sqrt{15}i}{6} when ± is plus. Add -6 to 2i\sqrt{15}.
z=\frac{\sqrt{15}i}{3}-1
Divide -6+2i\sqrt{15} by 6.
z=\frac{-2\sqrt{15}i-6}{6}
Now solve the equation z=\frac{-6±2\sqrt{15}i}{6} when ± is minus. Subtract 2i\sqrt{15} from -6.
z=-\frac{\sqrt{15}i}{3}-1
Divide -6-2i\sqrt{15} by 6.
z=\frac{\sqrt{15}i}{3}-1 z=-\frac{\sqrt{15}i}{3}-1
The equation is now solved.
3z^{2}+6z+8=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.
3z^{2}+6z+8-8=-8
Subtract 8 from both sides of the equation.
3z^{2}+6z=-8
Subtracting 8 from itself leaves 0.
\frac{3z^{2}+6z}{3}=-\frac{8}{3}
Divide both sides by 3.
z^{2}+\frac{6}{3}z=-\frac{8}{3}
Dividing by 3 undoes the multiplication by 3.
z^{2}+2z=-\frac{8}{3}
Divide 6 by 3.
z^{2}+2z+1^{2}=-\frac{8}{3}+1^{2}
Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
z^{2}+2z+1=-\frac{8}{3}+1
Square 1.
z^{2}+2z+1=-\frac{5}{3}
Add -\frac{8}{3} to 1.
\left(z+1\right)^{2}=-\frac{5}{3}
Factor z^{2}+2z+1. 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+1\right)^{2}}=\sqrt{-\frac{5}{3}}
Take the square root of both sides of the equation.
z+1=\frac{\sqrt{15}i}{3} z+1=-\frac{\sqrt{15}i}{3}
Simplify.
z=\frac{\sqrt{15}i}{3}-1 z=-\frac{\sqrt{15}i}{3}-1
Subtract 1 from both sides of the equation.
x ^ 2 +2x +\frac{8}{3} = 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 3
r + s = -2 rs = \frac{8}{3}
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{8}{3}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{8}{3}
1 - u^2 = \frac{8}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{8}{3}-1 = \frac{5}{3}
Simplify the expression by subtracting 1 on both sides
u^2 = -\frac{5}{3} u = \pm\sqrt{-\frac{5}{3}} = \pm \frac{\sqrt{5}}{\sqrt{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 =-1 - \frac{\sqrt{5}}{\sqrt{3}}i = -1 - 1.291i s = -1 + \frac{\sqrt{5}}{\sqrt{3}}i = -1 + 1.291i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.