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3k^{2}+6k+11=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.
k=\frac{-6±\sqrt{6^{2}-4\times 3\times 11}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 6 for b, and 11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
k=\frac{-6±\sqrt{36-4\times 3\times 11}}{2\times 3}
Square 6.
k=\frac{-6±\sqrt{36-12\times 11}}{2\times 3}
Multiply -4 times 3.
k=\frac{-6±\sqrt{36-132}}{2\times 3}
Multiply -12 times 11.
k=\frac{-6±\sqrt{-96}}{2\times 3}
Add 36 to -132.
k=\frac{-6±4\sqrt{6}i}{2\times 3}
Take the square root of -96.
k=\frac{-6±4\sqrt{6}i}{6}
Multiply 2 times 3.
k=\frac{-6+4\sqrt{6}i}{6}
Now solve the equation k=\frac{-6±4\sqrt{6}i}{6} when ± is plus. Add -6 to 4i\sqrt{6}.
k=\frac{2\sqrt{6}i}{3}-1
Divide -6+4i\sqrt{6} by 6.
k=\frac{-4\sqrt{6}i-6}{6}
Now solve the equation k=\frac{-6±4\sqrt{6}i}{6} when ± is minus. Subtract 4i\sqrt{6} from -6.
k=-\frac{2\sqrt{6}i}{3}-1
Divide -6-4i\sqrt{6} by 6.
k=\frac{2\sqrt{6}i}{3}-1 k=-\frac{2\sqrt{6}i}{3}-1
The equation is now solved.
3k^{2}+6k+11=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.
3k^{2}+6k+11-11=-11
Subtract 11 from both sides of the equation.
3k^{2}+6k=-11
Subtracting 11 from itself leaves 0.
\frac{3k^{2}+6k}{3}=-\frac{11}{3}
Divide both sides by 3.
k^{2}+\frac{6}{3}k=-\frac{11}{3}
Dividing by 3 undoes the multiplication by 3.
k^{2}+2k=-\frac{11}{3}
Divide 6 by 3.
k^{2}+2k+1^{2}=-\frac{11}{3}+1^{2}
Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
k^{2}+2k+1=-\frac{11}{3}+1
Square 1.
k^{2}+2k+1=-\frac{8}{3}
Add -\frac{11}{3} to 1.
\left(k+1\right)^{2}=-\frac{8}{3}
Factor k^{2}+2k+1. 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(k+1\right)^{2}}=\sqrt{-\frac{8}{3}}
Take the square root of both sides of the equation.
k+1=\frac{2\sqrt{6}i}{3} k+1=-\frac{2\sqrt{6}i}{3}
Simplify.
k=\frac{2\sqrt{6}i}{3}-1 k=-\frac{2\sqrt{6}i}{3}-1
Subtract 1 from both sides of the equation.
x ^ 2 +2x +\frac{11}{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 = -2 rs = \frac{11}{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 = -1 - u s = -1 + u
Two numbers r and s sum up to -2 exactly when the average of the two numbers is \frac{1}{2}*-2 = -1. 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>
(-1 - u) (-1 + u) = \frac{11}{3}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{11}{3}
1 - u^2 = \frac{11}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{11}{3}-1 = \frac{8}{3}
Simplify the expression by subtracting 1 on both sides
u^2 = -\frac{8}{3} u = \pm\sqrt{-\frac{8}{3}} = \pm \frac{\sqrt{8}}{\sqrt{3}}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-1 - \frac{\sqrt{8}}{\sqrt{3}}i = -1 - 1.633i s = -1 + \frac{\sqrt{8}}{\sqrt{3}}i = -1 + 1.633i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.