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a+b=-7 ab=2\left(-4\right)=-8
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2k^{2}+ak+bk-4. To find a and b, set up a system to be solved.
1,-8 2,-4
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -8.
1-8=-7 2-4=-2
Calculate the sum for each pair.
a=-8 b=1
The solution is the pair that gives sum -7.
\left(2k^{2}-8k\right)+\left(k-4\right)
Rewrite 2k^{2}-7k-4 as \left(2k^{2}-8k\right)+\left(k-4\right).
2k\left(k-4\right)+k-4
Factor out 2k in 2k^{2}-8k.
\left(k-4\right)\left(2k+1\right)
Factor out common term k-4 by using distributive property.
k=4 k=-\frac{1}{2}
To find equation solutions, solve k-4=0 and 2k+1=0.
2k^{2}-7k-4=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{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\left(-4\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -7 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
k=\frac{-\left(-7\right)±\sqrt{49-4\times 2\left(-4\right)}}{2\times 2}
Square -7.
k=\frac{-\left(-7\right)±\sqrt{49-8\left(-4\right)}}{2\times 2}
Multiply -4 times 2.
k=\frac{-\left(-7\right)±\sqrt{49+32}}{2\times 2}
Multiply -8 times -4.
k=\frac{-\left(-7\right)±\sqrt{81}}{2\times 2}
Add 49 to 32.
k=\frac{-\left(-7\right)±9}{2\times 2}
Take the square root of 81.
k=\frac{7±9}{2\times 2}
The opposite of -7 is 7.
k=\frac{7±9}{4}
Multiply 2 times 2.
k=\frac{16}{4}
Now solve the equation k=\frac{7±9}{4} when ± is plus. Add 7 to 9.
k=4
Divide 16 by 4.
k=-\frac{2}{4}
Now solve the equation k=\frac{7±9}{4} when ± is minus. Subtract 9 from 7.
k=-\frac{1}{2}
Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.
k=4 k=-\frac{1}{2}
The equation is now solved.
2k^{2}-7k-4=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.
2k^{2}-7k-4-\left(-4\right)=-\left(-4\right)
Add 4 to both sides of the equation.
2k^{2}-7k=-\left(-4\right)
Subtracting -4 from itself leaves 0.
2k^{2}-7k=4
Subtract -4 from 0.
\frac{2k^{2}-7k}{2}=\frac{4}{2}
Divide both sides by 2.
k^{2}-\frac{7}{2}k=\frac{4}{2}
Dividing by 2 undoes the multiplication by 2.
k^{2}-\frac{7}{2}k=2
Divide 4 by 2.
k^{2}-\frac{7}{2}k+\left(-\frac{7}{4}\right)^{2}=2+\left(-\frac{7}{4}\right)^{2}
Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
k^{2}-\frac{7}{2}k+\frac{49}{16}=2+\frac{49}{16}
Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.
k^{2}-\frac{7}{2}k+\frac{49}{16}=\frac{81}{16}
Add 2 to \frac{49}{16}.
\left(k-\frac{7}{4}\right)^{2}=\frac{81}{16}
Factor k^{2}-\frac{7}{2}k+\frac{49}{16}. 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{7}{4}\right)^{2}}=\sqrt{\frac{81}{16}}
Take the square root of both sides of the equation.
k-\frac{7}{4}=\frac{9}{4} k-\frac{7}{4}=-\frac{9}{4}
Simplify.
k=4 k=-\frac{1}{2}
Add \frac{7}{4} to both sides of the equation.
x ^ 2 -\frac{7}{2}x -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.This is achieved by dividing both sides of the equation by 2
r + s = \frac{7}{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 = \frac{7}{4} - u s = \frac{7}{4} + u
Two numbers r and s sum up to \frac{7}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{2} = \frac{7}{4}. 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{7}{4} - u) (\frac{7}{4} + u) = -2
To solve for unknown quantity u, substitute these in the product equation rs = -2
\frac{49}{16} - u^2 = -2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -2-\frac{49}{16} = -\frac{81}{16}
Simplify the expression by subtracting \frac{49}{16} on both sides
u^2 = \frac{81}{16} u = \pm\sqrt{\frac{81}{16}} = \pm \frac{9}{4}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{7}{4} - \frac{9}{4} = -0.500 s = \frac{7}{4} + \frac{9}{4} = 4
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.