Solve for k
k=\sqrt{3}-1\approx 0.732050808
k=-\sqrt{3}-1\approx -2.732050808
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k^{2}+2k-2=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{-2±\sqrt{2^{2}-4\left(-2\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
k=\frac{-2±\sqrt{4-4\left(-2\right)}}{2}
Square 2.
k=\frac{-2±\sqrt{4+8}}{2}
Multiply -4 times -2.
k=\frac{-2±\sqrt{12}}{2}
Add 4 to 8.
k=\frac{-2±2\sqrt{3}}{2}
Take the square root of 12.
k=\frac{2\sqrt{3}-2}{2}
Now solve the equation k=\frac{-2±2\sqrt{3}}{2} when ± is plus. Add -2 to 2\sqrt{3}.
k=\sqrt{3}-1
Divide -2+2\sqrt{3} by 2.
k=\frac{-2\sqrt{3}-2}{2}
Now solve the equation k=\frac{-2±2\sqrt{3}}{2} when ± is minus. Subtract 2\sqrt{3} from -2.
k=-\sqrt{3}-1
Divide -2-2\sqrt{3} by 2.
k=\sqrt{3}-1 k=-\sqrt{3}-1
The equation is now solved.
k^{2}+2k-2=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.
k^{2}+2k-2-\left(-2\right)=-\left(-2\right)
Add 2 to both sides of the equation.
k^{2}+2k=-\left(-2\right)
Subtracting -2 from itself leaves 0.
k^{2}+2k=2
Subtract -2 from 0.
k^{2}+2k+1^{2}=2+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=2+1
Square 1.
k^{2}+2k+1=3
Add 2 to 1.
\left(k+1\right)^{2}=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{3}
Take the square root of both sides of the equation.
k+1=\sqrt{3} k+1=-\sqrt{3}
Simplify.
k=\sqrt{3}-1 k=-\sqrt{3}-1
Subtract 1 from both sides of the equation.
x ^ 2 +2x -2 = 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 = -2 rs = -2
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) = -2
To solve for unknown quantity u, substitute these in the product equation rs = -2
1 - u^2 = -2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -2-1 = -3
Simplify the expression by subtracting 1 on both sides
u^2 = 3 u = \pm\sqrt{3} = \pm \sqrt{3}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-1 - \sqrt{3} = -2.732 s = -1 + \sqrt{3} = 0.732
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}