Solve for k
k=-2
k=1
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a+b=1 ab=-2
To solve the equation, factor k^{2}+k-2 using formula k^{2}+\left(a+b\right)k+ab=\left(k+a\right)\left(k+b\right). To find a and b, set up a system to be solved.
a=-1 b=2
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.
\left(k-1\right)\left(k+2\right)
Rewrite factored expression \left(k+a\right)\left(k+b\right) using the obtained values.
k=1 k=-2
To find equation solutions, solve k-1=0 and k+2=0.
a+b=1 ab=1\left(-2\right)=-2
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as k^{2}+ak+bk-2. To find a and b, set up a system to be solved.
a=-1 b=2
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.
\left(k^{2}-k\right)+\left(2k-2\right)
Rewrite k^{2}+k-2 as \left(k^{2}-k\right)+\left(2k-2\right).
k\left(k-1\right)+2\left(k-1\right)
Factor out k in the first and 2 in the second group.
\left(k-1\right)\left(k+2\right)
Factor out common term k-1 by using distributive property.
k=1 k=-2
To find equation solutions, solve k-1=0 and k+2=0.
k^{2}+k-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{-1±\sqrt{1^{2}-4\left(-2\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
k=\frac{-1±\sqrt{1-4\left(-2\right)}}{2}
Square 1.
k=\frac{-1±\sqrt{1+8}}{2}
Multiply -4 times -2.
k=\frac{-1±\sqrt{9}}{2}
Add 1 to 8.
k=\frac{-1±3}{2}
Take the square root of 9.
k=\frac{2}{2}
Now solve the equation k=\frac{-1±3}{2} when ± is plus. Add -1 to 3.
k=1
Divide 2 by 2.
k=-\frac{4}{2}
Now solve the equation k=\frac{-1±3}{2} when ± is minus. Subtract 3 from -1.
k=-2
Divide -4 by 2.
k=1 k=-2
The equation is now solved.
k^{2}+k-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}+k-2-\left(-2\right)=-\left(-2\right)
Add 2 to both sides of the equation.
k^{2}+k=-\left(-2\right)
Subtracting -2 from itself leaves 0.
k^{2}+k=2
Subtract -2 from 0.
k^{2}+k+\left(\frac{1}{2}\right)^{2}=2+\left(\frac{1}{2}\right)^{2}
Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
k^{2}+k+\frac{1}{4}=2+\frac{1}{4}
Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.
k^{2}+k+\frac{1}{4}=\frac{9}{4}
Add 2 to \frac{1}{4}.
\left(k+\frac{1}{2}\right)^{2}=\frac{9}{4}
Factor k^{2}+k+\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(k+\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}
Take the square root of both sides of the equation.
k+\frac{1}{2}=\frac{3}{2} k+\frac{1}{2}=-\frac{3}{2}
Simplify.
k=1 k=-2
Subtract \frac{1}{2} from both sides of the equation.
x ^ 2 +1x -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 = -1 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 = -\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) = -2
To solve for unknown quantity u, substitute these in the product equation rs = -2
\frac{1}{4} - u^2 = -2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -2-\frac{1}{4} = -\frac{9}{4}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}
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{3}{2} = -2 s = -\frac{1}{2} + \frac{3}{2} = 1
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
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{ 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}