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z^{2}+z-2=0
Divide both sides by 2.
a+b=1 ab=1\left(-2\right)=-2
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as z^{2}+az+bz-2. To find a and b, set up a system to be solved.
a=-1 b=2
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.
\left(z^{2}-z\right)+\left(2z-2\right)
Rewrite z^{2}+z-2 as \left(z^{2}-z\right)+\left(2z-2\right).
z\left(z-1\right)+2\left(z-1\right)
Factor out z in the first and 2 in the second group.
\left(z-1\right)\left(z+2\right)
Factor out common term z-1 by using distributive property.
z=1 z=-2
To find equation solutions, solve z-1=0 and z+2=0.
2z^{2}+2z-4=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{-2±\sqrt{2^{2}-4\times 2\left(-4\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 2 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
z=\frac{-2±\sqrt{4-4\times 2\left(-4\right)}}{2\times 2}
Square 2.
z=\frac{-2±\sqrt{4-8\left(-4\right)}}{2\times 2}
Multiply -4 times 2.
z=\frac{-2±\sqrt{4+32}}{2\times 2}
Multiply -8 times -4.
z=\frac{-2±\sqrt{36}}{2\times 2}
Add 4 to 32.
z=\frac{-2±6}{2\times 2}
Take the square root of 36.
z=\frac{-2±6}{4}
Multiply 2 times 2.
z=\frac{4}{4}
Now solve the equation z=\frac{-2±6}{4} when ± is plus. Add -2 to 6.
z=1
Divide 4 by 4.
z=-\frac{8}{4}
Now solve the equation z=\frac{-2±6}{4} when ± is minus. Subtract 6 from -2.
z=-2
Divide -8 by 4.
z=1 z=-2
The equation is now solved.
2z^{2}+2z-4=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.
2z^{2}+2z-4-\left(-4\right)=-\left(-4\right)
Add 4 to both sides of the equation.
2z^{2}+2z=-\left(-4\right)
Subtracting -4 from itself leaves 0.
2z^{2}+2z=4
Subtract -4 from 0.
\frac{2z^{2}+2z}{2}=\frac{4}{2}
Divide both sides by 2.
z^{2}+\frac{2}{2}z=\frac{4}{2}
Dividing by 2 undoes the multiplication by 2.
z^{2}+z=\frac{4}{2}
Divide 2 by 2.
z^{2}+z=2
Divide 4 by 2.
z^{2}+z+\left(\frac{1}{2}\right)^{2}=2+\left(\frac{1}{2}\right)^{2}
Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
z^{2}+z+\frac{1}{4}=2+\frac{1}{4}
Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.
z^{2}+z+\frac{1}{4}=\frac{9}{4}
Add 2 to \frac{1}{4}.
\left(z+\frac{1}{2}\right)^{2}=\frac{9}{4}
Factor z^{2}+z+\frac{1}{4}. 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{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}
Take the square root of both sides of the equation.
z+\frac{1}{2}=\frac{3}{2} z+\frac{1}{2}=-\frac{3}{2}
Simplify.
z=1 z=-2
Subtract \frac{1}{2} from both sides of the equation.
x ^ 2 +1x -2 = 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 2
r + s = -1 rs = -2
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}{2} - u s = -\frac{1}{2} + u
Two numbers r and s sum up to -1 exactly when the average of the two numbers is \frac{1}{2}*-1 = -\frac{1}{2}. 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}{2} - u) (-\frac{1}{2} + u) = -2
To solve for unknown quantity u, substitute these in the product equation rs = -2
\frac{1}{4} - u^2 = -2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -2-\frac{1}{4} = -\frac{9}{4}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}
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}{2} - \frac{3}{2} = -2 s = -\frac{1}{2} + \frac{3}{2} = 1
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.