Solve for x
x=\frac{1}{12}\approx 0.083333333
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a+b=-24 ab=144\times 1=144
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 144x^{2}+ax+bx+1. To find a and b, set up a system to be solved.
-1,-144 -2,-72 -3,-48 -4,-36 -6,-24 -8,-18 -9,-16 -12,-12
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 144.
-1-144=-145 -2-72=-74 -3-48=-51 -4-36=-40 -6-24=-30 -8-18=-26 -9-16=-25 -12-12=-24
Calculate the sum for each pair.
a=-12 b=-12
The solution is the pair that gives sum -24.
\left(144x^{2}-12x\right)+\left(-12x+1\right)
Rewrite 144x^{2}-24x+1 as \left(144x^{2}-12x\right)+\left(-12x+1\right).
12x\left(12x-1\right)-\left(12x-1\right)
Factor out 12x in the first and -1 in the second group.
\left(12x-1\right)\left(12x-1\right)
Factor out common term 12x-1 by using distributive property.
\left(12x-1\right)^{2}
Rewrite as a binomial square.
x=\frac{1}{12}
To find equation solution, solve 12x-1=0.
144x^{2}-24x+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.
x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 144}}{2\times 144}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 144 for a, -24 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-24\right)±\sqrt{576-4\times 144}}{2\times 144}
Square -24.
x=\frac{-\left(-24\right)±\sqrt{576-576}}{2\times 144}
Multiply -4 times 144.
x=\frac{-\left(-24\right)±\sqrt{0}}{2\times 144}
Add 576 to -576.
x=-\frac{-24}{2\times 144}
Take the square root of 0.
x=\frac{24}{2\times 144}
The opposite of -24 is 24.
x=\frac{24}{288}
Multiply 2 times 144.
x=\frac{1}{12}
Reduce the fraction \frac{24}{288} to lowest terms by extracting and canceling out 24.
144x^{2}-24x+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.
144x^{2}-24x+1-1=-1
Subtract 1 from both sides of the equation.
144x^{2}-24x=-1
Subtracting 1 from itself leaves 0.
\frac{144x^{2}-24x}{144}=-\frac{1}{144}
Divide both sides by 144.
x^{2}+\left(-\frac{24}{144}\right)x=-\frac{1}{144}
Dividing by 144 undoes the multiplication by 144.
x^{2}-\frac{1}{6}x=-\frac{1}{144}
Reduce the fraction \frac{-24}{144} to lowest terms by extracting and canceling out 24.
x^{2}-\frac{1}{6}x+\left(-\frac{1}{12}\right)^{2}=-\frac{1}{144}+\left(-\frac{1}{12}\right)^{2}
Divide -\frac{1}{6}, the coefficient of the x term, by 2 to get -\frac{1}{12}. Then add the square of -\frac{1}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{1}{6}x+\frac{1}{144}=\frac{-1+1}{144}
Square -\frac{1}{12} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{1}{6}x+\frac{1}{144}=0
Add -\frac{1}{144} to \frac{1}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{12}\right)^{2}=0
Factor x^{2}-\frac{1}{6}x+\frac{1}{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(x-\frac{1}{12}\right)^{2}}=\sqrt{0}
Take the square root of both sides of the equation.
x-\frac{1}{12}=0 x-\frac{1}{12}=0
Simplify.
x=\frac{1}{12} x=\frac{1}{12}
Add \frac{1}{12} to both sides of the equation.
x=\frac{1}{12}
The equation is now solved. Solutions are the same.
x ^ 2 -\frac{1}{6}x +\frac{1}{144} = 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 144
r + s = \frac{1}{6} rs = \frac{1}{144}
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}{12} - u s = \frac{1}{12} + u
Two numbers r and s sum up to \frac{1}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{6} = \frac{1}{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{1}{12} - u) (\frac{1}{12} + u) = \frac{1}{144}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{144}
\frac{1}{144} - u^2 = \frac{1}{144}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{144}-\frac{1}{144} = 0
Simplify the expression by subtracting \frac{1}{144} on both sides
u^2 = 0 u = 0
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r = s = \frac{1}{12} = 0.083
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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