Solve for x
x=\frac{\sqrt{3}+6}{11}\approx 0.70291371
x=\frac{6-\sqrt{3}}{11}\approx 0.387995381
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11x^{2}-12x+3=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(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 11\times 3}}{2\times 11}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 11 for a, -12 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-12\right)±\sqrt{144-4\times 11\times 3}}{2\times 11}
Square -12.
x=\frac{-\left(-12\right)±\sqrt{144-44\times 3}}{2\times 11}
Multiply -4 times 11.
x=\frac{-\left(-12\right)±\sqrt{144-132}}{2\times 11}
Multiply -44 times 3.
x=\frac{-\left(-12\right)±\sqrt{12}}{2\times 11}
Add 144 to -132.
x=\frac{-\left(-12\right)±2\sqrt{3}}{2\times 11}
Take the square root of 12.
x=\frac{12±2\sqrt{3}}{2\times 11}
The opposite of -12 is 12.
x=\frac{12±2\sqrt{3}}{22}
Multiply 2 times 11.
x=\frac{2\sqrt{3}+12}{22}
Now solve the equation x=\frac{12±2\sqrt{3}}{22} when ± is plus. Add 12 to 2\sqrt{3}.
x=\frac{\sqrt{3}+6}{11}
Divide 12+2\sqrt{3} by 22.
x=\frac{12-2\sqrt{3}}{22}
Now solve the equation x=\frac{12±2\sqrt{3}}{22} when ± is minus. Subtract 2\sqrt{3} from 12.
x=\frac{6-\sqrt{3}}{11}
Divide 12-2\sqrt{3} by 22.
x=\frac{\sqrt{3}+6}{11} x=\frac{6-\sqrt{3}}{11}
The equation is now solved.
11x^{2}-12x+3=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.
11x^{2}-12x+3-3=-3
Subtract 3 from both sides of the equation.
11x^{2}-12x=-3
Subtracting 3 from itself leaves 0.
\frac{11x^{2}-12x}{11}=-\frac{3}{11}
Divide both sides by 11.
x^{2}-\frac{12}{11}x=-\frac{3}{11}
Dividing by 11 undoes the multiplication by 11.
x^{2}-\frac{12}{11}x+\left(-\frac{6}{11}\right)^{2}=-\frac{3}{11}+\left(-\frac{6}{11}\right)^{2}
Divide -\frac{12}{11}, the coefficient of the x term, by 2 to get -\frac{6}{11}. Then add the square of -\frac{6}{11} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{12}{11}x+\frac{36}{121}=-\frac{3}{11}+\frac{36}{121}
Square -\frac{6}{11} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{12}{11}x+\frac{36}{121}=\frac{3}{121}
Add -\frac{3}{11} to \frac{36}{121} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{6}{11}\right)^{2}=\frac{3}{121}
Factor x^{2}-\frac{12}{11}x+\frac{36}{121}. 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{6}{11}\right)^{2}}=\sqrt{\frac{3}{121}}
Take the square root of both sides of the equation.
x-\frac{6}{11}=\frac{\sqrt{3}}{11} x-\frac{6}{11}=-\frac{\sqrt{3}}{11}
Simplify.
x=\frac{\sqrt{3}+6}{11} x=\frac{6-\sqrt{3}}{11}
Add \frac{6}{11} to both sides of the equation.
x ^ 2 -\frac{12}{11}x +\frac{3}{11} = 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 11
r + s = \frac{12}{11} rs = \frac{3}{11}
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{6}{11} - u s = \frac{6}{11} + u
Two numbers r and s sum up to \frac{12}{11} exactly when the average of the two numbers is \frac{1}{2}*\frac{12}{11} = \frac{6}{11}. 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{6}{11} - u) (\frac{6}{11} + u) = \frac{3}{11}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{11}
\frac{36}{121} - u^2 = \frac{3}{11}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{3}{11}-\frac{36}{121} = -\frac{3}{121}
Simplify the expression by subtracting \frac{36}{121} on both sides
u^2 = \frac{3}{121} u = \pm\sqrt{\frac{3}{121}} = \pm \frac{\sqrt{3}}{11}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{6}{11} - \frac{\sqrt{3}}{11} = 0.388 s = \frac{6}{11} + \frac{\sqrt{3}}{11} = 0.703
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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