Solve for f
f=-\frac{1}{6}\approx -0.166666667
f=1
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a+b=-5 ab=6\left(-1\right)=-6
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6f^{2}+af+bf-1. To find a and b, set up a system to be solved.
1,-6 2,-3
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 -6.
1-6=-5 2-3=-1
Calculate the sum for each pair.
a=-6 b=1
The solution is the pair that gives sum -5.
\left(6f^{2}-6f\right)+\left(f-1\right)
Rewrite 6f^{2}-5f-1 as \left(6f^{2}-6f\right)+\left(f-1\right).
6f\left(f-1\right)+f-1
Factor out 6f in 6f^{2}-6f.
\left(f-1\right)\left(6f+1\right)
Factor out common term f-1 by using distributive property.
f=1 f=-\frac{1}{6}
To find equation solutions, solve f-1=0 and 6f+1=0.
6f^{2}-5f-1=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.
f=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6\left(-1\right)}}{2\times 6}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -5 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
f=\frac{-\left(-5\right)±\sqrt{25-4\times 6\left(-1\right)}}{2\times 6}
Square -5.
f=\frac{-\left(-5\right)±\sqrt{25-24\left(-1\right)}}{2\times 6}
Multiply -4 times 6.
f=\frac{-\left(-5\right)±\sqrt{25+24}}{2\times 6}
Multiply -24 times -1.
f=\frac{-\left(-5\right)±\sqrt{49}}{2\times 6}
Add 25 to 24.
f=\frac{-\left(-5\right)±7}{2\times 6}
Take the square root of 49.
f=\frac{5±7}{2\times 6}
The opposite of -5 is 5.
f=\frac{5±7}{12}
Multiply 2 times 6.
f=\frac{12}{12}
Now solve the equation f=\frac{5±7}{12} when ± is plus. Add 5 to 7.
f=1
Divide 12 by 12.
f=-\frac{2}{12}
Now solve the equation f=\frac{5±7}{12} when ± is minus. Subtract 7 from 5.
f=-\frac{1}{6}
Reduce the fraction \frac{-2}{12} to lowest terms by extracting and canceling out 2.
f=1 f=-\frac{1}{6}
The equation is now solved.
6f^{2}-5f-1=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.
6f^{2}-5f-1-\left(-1\right)=-\left(-1\right)
Add 1 to both sides of the equation.
6f^{2}-5f=-\left(-1\right)
Subtracting -1 from itself leaves 0.
6f^{2}-5f=1
Subtract -1 from 0.
\frac{6f^{2}-5f}{6}=\frac{1}{6}
Divide both sides by 6.
f^{2}-\frac{5}{6}f=\frac{1}{6}
Dividing by 6 undoes the multiplication by 6.
f^{2}-\frac{5}{6}f+\left(-\frac{5}{12}\right)^{2}=\frac{1}{6}+\left(-\frac{5}{12}\right)^{2}
Divide -\frac{5}{6}, the coefficient of the x term, by 2 to get -\frac{5}{12}. Then add the square of -\frac{5}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
f^{2}-\frac{5}{6}f+\frac{25}{144}=\frac{1}{6}+\frac{25}{144}
Square -\frac{5}{12} by squaring both the numerator and the denominator of the fraction.
f^{2}-\frac{5}{6}f+\frac{25}{144}=\frac{49}{144}
Add \frac{1}{6} to \frac{25}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(f-\frac{5}{12}\right)^{2}=\frac{49}{144}
Factor f^{2}-\frac{5}{6}f+\frac{25}{144}. 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(f-\frac{5}{12}\right)^{2}}=\sqrt{\frac{49}{144}}
Take the square root of both sides of the equation.
f-\frac{5}{12}=\frac{7}{12} f-\frac{5}{12}=-\frac{7}{12}
Simplify.
f=1 f=-\frac{1}{6}
Add \frac{5}{12} to both sides of the equation.
x ^ 2 -\frac{5}{6}x -\frac{1}{6} = 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 6
r + s = \frac{5}{6} rs = -\frac{1}{6}
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{5}{12} - u s = \frac{5}{12} + u
Two numbers r and s sum up to \frac{5}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{6} = \frac{5}{12}. 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{5}{12} - u) (\frac{5}{12} + u) = -\frac{1}{6}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{6}
\frac{25}{144} - u^2 = -\frac{1}{6}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{1}{6}-\frac{25}{144} = -\frac{49}{144}
Simplify the expression by subtracting \frac{25}{144} on both sides
u^2 = \frac{49}{144} u = \pm\sqrt{\frac{49}{144}} = \pm \frac{7}{12}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{5}{12} - \frac{7}{12} = -0.167 s = \frac{5}{12} + \frac{7}{12} = 1
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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