Solve for x
Tick mark Image
Graph

Similar Problems from Web Search

Share

3x^{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\times 3\left(-1\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 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{1-12\left(-1\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-1\right)±\sqrt{1+12}}{2\times 3}
Multiply -12 times -1.
x=\frac{-\left(-1\right)±\sqrt{13}}{2\times 3}
Add 1 to 12.
x=\frac{1±\sqrt{13}}{2\times 3}
The opposite of -1 is 1.
x=\frac{1±\sqrt{13}}{6}
Multiply 2 times 3.
x=\frac{\sqrt{13}+1}{6}
Now solve the equation x=\frac{1±\sqrt{13}}{6} when ± is plus. Add 1 to \sqrt{13}.
x=\frac{1-\sqrt{13}}{6}
Now solve the equation x=\frac{1±\sqrt{13}}{6} when ± is minus. Subtract \sqrt{13} from 1.
x=\frac{\sqrt{13}+1}{6} x=\frac{1-\sqrt{13}}{6}
The equation is now solved.
3x^{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.
3x^{2}-x-1-\left(-1\right)=-\left(-1\right)
Add 1 to both sides of the equation.
3x^{2}-x=-\left(-1\right)
Subtracting -1 from itself leaves 0.
3x^{2}-x=1
Subtract -1 from 0.
\frac{3x^{2}-x}{3}=\frac{1}{3}
Divide both sides by 3.
x^{2}+\frac{-1}{3}x=\frac{1}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-\frac{1}{3}x=\frac{1}{3}
Divide -1 by 3.
x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{1}{3}+\left(-\frac{1}{6}\right)^{2}
Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{1}{3}+\frac{1}{36}
Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{13}{36}
Add \frac{1}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{6}\right)^{2}=\frac{13}{36}
Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. 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}{6}\right)^{2}}=\sqrt{\frac{13}{36}}
Take the square root of both sides of the equation.
x-\frac{1}{6}=\frac{\sqrt{13}}{6} x-\frac{1}{6}=-\frac{\sqrt{13}}{6}
Simplify.
x=\frac{\sqrt{13}+1}{6} x=\frac{1-\sqrt{13}}{6}
Add \frac{1}{6} to both sides of the equation.
x ^ 2 -\frac{1}{3}x -\frac{1}{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 = \frac{1}{3} rs = -\frac{1}{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 = \frac{1}{6} - u s = \frac{1}{6} + u
Two numbers r and s sum up to \frac{1}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{3} = \frac{1}{6}. 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}{6} - u) (\frac{1}{6} + u) = -\frac{1}{3}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{3}
\frac{1}{36} - u^2 = -\frac{1}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{1}{3}-\frac{1}{36} = -\frac{13}{36}
Simplify the expression by subtracting \frac{1}{36} on both sides
u^2 = \frac{13}{36} u = \pm\sqrt{\frac{13}{36}} = \pm \frac{\sqrt{13}}{6}
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}{6} - \frac{\sqrt{13}}{6} = -0.434 s = \frac{1}{6} + \frac{\sqrt{13}}{6} = 0.768
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.