Solve for x (complex solution)
x=\frac{-\sqrt{15}i-1}{2}\approx -0.5-1.936491673i
x=\frac{-1+\sqrt{15}i}{2}\approx -0.5+1.936491673i
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-x^{2}-x-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.
x=\frac{-\left(-1\right)±\sqrt{1-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -1 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-1\right)±\sqrt{1+4\left(-4\right)}}{2\left(-1\right)}
Multiply -4 times -1.
x=\frac{-\left(-1\right)±\sqrt{1-16}}{2\left(-1\right)}
Multiply 4 times -4.
x=\frac{-\left(-1\right)±\sqrt{-15}}{2\left(-1\right)}
Add 1 to -16.
x=\frac{-\left(-1\right)±\sqrt{15}i}{2\left(-1\right)}
Take the square root of -15.
x=\frac{1±\sqrt{15}i}{2\left(-1\right)}
The opposite of -1 is 1.
x=\frac{1±\sqrt{15}i}{-2}
Multiply 2 times -1.
x=\frac{1+\sqrt{15}i}{-2}
Now solve the equation x=\frac{1±\sqrt{15}i}{-2} when ± is plus. Add 1 to i\sqrt{15}.
x=\frac{-\sqrt{15}i-1}{2}
Divide 1+i\sqrt{15} by -2.
x=\frac{-\sqrt{15}i+1}{-2}
Now solve the equation x=\frac{1±\sqrt{15}i}{-2} when ± is minus. Subtract i\sqrt{15} from 1.
x=\frac{-1+\sqrt{15}i}{2}
Divide 1-i\sqrt{15} by -2.
x=\frac{-\sqrt{15}i-1}{2} x=\frac{-1+\sqrt{15}i}{2}
The equation is now solved.
-x^{2}-x-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.
-x^{2}-x-4-\left(-4\right)=-\left(-4\right)
Add 4 to both sides of the equation.
-x^{2}-x=-\left(-4\right)
Subtracting -4 from itself leaves 0.
-x^{2}-x=4
Subtract -4 from 0.
\frac{-x^{2}-x}{-1}=\frac{4}{-1}
Divide both sides by -1.
x^{2}+\left(-\frac{1}{-1}\right)x=\frac{4}{-1}
Dividing by -1 undoes the multiplication by -1.
x^{2}+x=\frac{4}{-1}
Divide -1 by -1.
x^{2}+x=-4
Divide 4 by -1.
x^{2}+x+\left(\frac{1}{2}\right)^{2}=-4+\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.
x^{2}+x+\frac{1}{4}=-4+\frac{1}{4}
Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+x+\frac{1}{4}=-\frac{15}{4}
Add -4 to \frac{1}{4}.
\left(x+\frac{1}{2}\right)^{2}=-\frac{15}{4}
Factor x^{2}+x+\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(x+\frac{1}{2}\right)^{2}}=\sqrt{-\frac{15}{4}}
Take the square root of both sides of the equation.
x+\frac{1}{2}=\frac{\sqrt{15}i}{2} x+\frac{1}{2}=-\frac{\sqrt{15}i}{2}
Simplify.
x=\frac{-1+\sqrt{15}i}{2} x=\frac{-\sqrt{15}i-1}{2}
Subtract \frac{1}{2} from both sides of the equation.
x ^ 2 +1x +4 = 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.
r + s = -1 rs = 4
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) = 4
To solve for unknown quantity u, substitute these in the product equation rs = 4
\frac{1}{4} - u^2 = 4
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 4-\frac{1}{4} = \frac{15}{4}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = -\frac{15}{4} u = \pm\sqrt{-\frac{15}{4}} = \pm \frac{\sqrt{15}}{2}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}{2} - \frac{\sqrt{15}}{2}i = -0.500 - 1.936i s = -\frac{1}{2} + \frac{\sqrt{15}}{2}i = -0.500 + 1.936i
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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