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Steps Using the Quadratic Formula
Steps for Completing the Square
Steps Using Direct Factoring Method
Steps Using the Quadratic Formula
All equations of the form can be solved using the quadratic formula: . The quadratic formula gives two solutions, one when is addition and one when it is subtraction.
This equation is in standard form: . Substitute for , for , and for in the quadratic formula, .
Multiply times .
Multiply times .
Add to .
Take the square root of .
The opposite of is .
Multiply times .
Now solve the equation when is plus. Add to .
Divide by .
Now solve the equation when is minus. Subtract from .
Divide by .
The equation is now solved.
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Similar Problems from Web Search

-x^{2}-x-1=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(-1\right)}}{2\left(-1\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -1 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-1\right)±\sqrt{1+4\left(-1\right)}}{2\left(-1\right)}
Multiply -4 times -1.
x=\frac{-\left(-1\right)±\sqrt{1-4}}{2\left(-1\right)}
Multiply 4 times -1.
x=\frac{-\left(-1\right)±\sqrt{-3}}{2\left(-1\right)}
Add 1 to -4.
x=\frac{-\left(-1\right)±\sqrt{3}i}{2\left(-1\right)}
Take the square root of -3.
x=\frac{1±\sqrt{3}i}{2\left(-1\right)}
The opposite of -1 is 1.
x=\frac{1±\sqrt{3}i}{-2}
Multiply 2 times -1.
x=\frac{1+\sqrt{3}i}{-2}
Now solve the equation x=\frac{1±\sqrt{3}i}{-2} when ± is plus. Add 1 to i\sqrt{3}\approx 1.732050808i.
x=\frac{-\sqrt{3}i-1}{2}
Divide 1+i\sqrt{3}\approx 1+1.732050808i by -2.
x=\frac{-\sqrt{3}i+1}{-2}
Now solve the equation x=\frac{1±\sqrt{3}i}{-2} when ± is minus. Subtract i\sqrt{3}\approx 1.732050808i from 1.
x=\frac{-1+\sqrt{3}i}{2}
Divide 1-i\sqrt{3}\approx 1-1.732050808i by -2.
x=\frac{-\sqrt{3}i-1}{2} x=\frac{-1+\sqrt{3}i}{2}
The equation is now solved.
-x^{2}-x-1=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-1-\left(-1\right)=-\left(-1\right)
Add 1 to both sides of the equation.
-x^{2}-x=-\left(-1\right)
Subtracting -1 from itself leaves 0.
-x^{2}-x=1
Subtract -1 from 0.
\frac{-x^{2}-x}{-1}=\frac{1}{-1}
Divide both sides by -1.
x^{2}+\frac{-1}{-1}x=\frac{1}{-1}
Dividing by -1 undoes the multiplication by -1.
x^{2}+x=\frac{1}{-1}
Divide -1 by -1.
x^{2}+x=-1
Divide 1 by -1.
x^{2}+x+\left(\frac{1}{2}\right)^{2}=-1+\left(\frac{1}{2}\right)^{2}
Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}=0.5. Then add the square of \frac{1}{2}=0.5 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}=-1+\frac{1}{4}
Square \frac{1}{2}=0.5 by squaring both the numerator and the denominator of the fraction.
x^{2}+x+\frac{1}{4}=-\frac{3}{4}
Add -1 to \frac{1}{4}=0.25.
\left(x+\frac{1}{2}\right)^{2}=-\frac{3}{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{3}{4}}
Take the square root of both sides of the equation.
x+\frac{1}{2}=\frac{\sqrt{3}i}{2} x+\frac{1}{2}=-\frac{\sqrt{3}i}{2}
Simplify.
x=\frac{-1+\sqrt{3}i}{2} x=\frac{-\sqrt{3}i-1}{2}
Subtract \frac{1}{2}=0.5 from both sides of the equation.
x ^ 2 +1x +1 = 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 = 1
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) = 1
To solve for unknown quantity u, substitute these in the product equation rs = 1
\frac{1}{4} - u^2 = 1
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 1-\frac{1}{4} = \frac{3}{4}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = -\frac{3}{4} u = \pm\sqrt{-\frac{3}{4}} = \pm \frac{\sqrt{3}}{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{3}}{2}i = -0.500 - 0.866i s = -\frac{1}{2} + \frac{\sqrt{3}}{2}i = -0.500 + 0.866i
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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