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a+b=1 ab=28\left(-2\right)=-56
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 28k^{2}+ak+bk-2. To find a and b, set up a system to be solved.
-1,56 -2,28 -4,14 -7,8
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. List all such integer pairs that give product -56.
-1+56=55 -2+28=26 -4+14=10 -7+8=1
Calculate the sum for each pair.
a=-7 b=8
The solution is the pair that gives sum 1.
\left(28k^{2}-7k\right)+\left(8k-2\right)
Rewrite 28k^{2}+k-2 as \left(28k^{2}-7k\right)+\left(8k-2\right).
7k\left(4k-1\right)+2\left(4k-1\right)
Factor out 7k in the first and 2 in the second group.
\left(4k-1\right)\left(7k+2\right)
Factor out common term 4k-1 by using distributive property.
k=\frac{1}{4} k=-\frac{2}{7}
To find equation solutions, solve 4k-1=0 and 7k+2=0.
28k^{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\times 28\left(-2\right)}}{2\times 28}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 28 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\times 28\left(-2\right)}}{2\times 28}
Square 1.
k=\frac{-1±\sqrt{1-112\left(-2\right)}}{2\times 28}
Multiply -4 times 28.
k=\frac{-1±\sqrt{1+224}}{2\times 28}
Multiply -112 times -2.
k=\frac{-1±\sqrt{225}}{2\times 28}
Add 1 to 224.
k=\frac{-1±15}{2\times 28}
Take the square root of 225.
k=\frac{-1±15}{56}
Multiply 2 times 28.
k=\frac{14}{56}
Now solve the equation k=\frac{-1±15}{56} when ± is plus. Add -1 to 15.
k=\frac{1}{4}
Reduce the fraction \frac{14}{56} to lowest terms by extracting and canceling out 14.
k=-\frac{16}{56}
Now solve the equation k=\frac{-1±15}{56} when ± is minus. Subtract 15 from -1.
k=-\frac{2}{7}
Reduce the fraction \frac{-16}{56} to lowest terms by extracting and canceling out 8.
k=\frac{1}{4} k=-\frac{2}{7}
The equation is now solved.
28k^{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.
28k^{2}+k-2-\left(-2\right)=-\left(-2\right)
Add 2 to both sides of the equation.
28k^{2}+k=-\left(-2\right)
Subtracting -2 from itself leaves 0.
28k^{2}+k=2
Subtract -2 from 0.
\frac{28k^{2}+k}{28}=\frac{2}{28}
Divide both sides by 28.
k^{2}+\frac{1}{28}k=\frac{2}{28}
Dividing by 28 undoes the multiplication by 28.
k^{2}+\frac{1}{28}k=\frac{1}{14}
Reduce the fraction \frac{2}{28} to lowest terms by extracting and canceling out 2.
k^{2}+\frac{1}{28}k+\left(\frac{1}{56}\right)^{2}=\frac{1}{14}+\left(\frac{1}{56}\right)^{2}
Divide \frac{1}{28}, the coefficient of the x term, by 2 to get \frac{1}{56}. Then add the square of \frac{1}{56} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
k^{2}+\frac{1}{28}k+\frac{1}{3136}=\frac{1}{14}+\frac{1}{3136}
Square \frac{1}{56} by squaring both the numerator and the denominator of the fraction.
k^{2}+\frac{1}{28}k+\frac{1}{3136}=\frac{225}{3136}
Add \frac{1}{14} to \frac{1}{3136} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(k+\frac{1}{56}\right)^{2}=\frac{225}{3136}
Factor k^{2}+\frac{1}{28}k+\frac{1}{3136}. 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}{56}\right)^{2}}=\sqrt{\frac{225}{3136}}
Take the square root of both sides of the equation.
k+\frac{1}{56}=\frac{15}{56} k+\frac{1}{56}=-\frac{15}{56}
Simplify.
k=\frac{1}{4} k=-\frac{2}{7}
Subtract \frac{1}{56} from both sides of the equation.
x ^ 2 +\frac{1}{28}x -\frac{1}{14} = 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 28
r + s = -\frac{1}{28} rs = -\frac{1}{14}
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}{56} - u s = -\frac{1}{56} + u
Two numbers r and s sum up to -\frac{1}{28} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{28} = -\frac{1}{56}. 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}{56} - u) (-\frac{1}{56} + u) = -\frac{1}{14}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{14}
\frac{1}{3136} - u^2 = -\frac{1}{14}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{1}{14}-\frac{1}{3136} = -\frac{225}{3136}
Simplify the expression by subtracting \frac{1}{3136} on both sides
u^2 = \frac{225}{3136} u = \pm\sqrt{\frac{225}{3136}} = \pm \frac{15}{56}
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}{56} - \frac{15}{56} = -0.286 s = -\frac{1}{56} + \frac{15}{56} = 0.250
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.