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a+b=-25 ab=12\times 12=144
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 12z^{2}+az+bz+12. 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=-16 b=-9
The solution is the pair that gives sum -25.
\left(12z^{2}-16z\right)+\left(-9z+12\right)
Rewrite 12z^{2}-25z+12 as \left(12z^{2}-16z\right)+\left(-9z+12\right).
4z\left(3z-4\right)-3\left(3z-4\right)
Factor out 4z in the first and -3 in the second group.
\left(3z-4\right)\left(4z-3\right)
Factor out common term 3z-4 by using distributive property.
z=\frac{4}{3} z=\frac{3}{4}
To find equation solutions, solve 3z-4=0 and 4z-3=0.
12z^{2}-25z+12=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.
z=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 12\times 12}}{2\times 12}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -25 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
z=\frac{-\left(-25\right)±\sqrt{625-4\times 12\times 12}}{2\times 12}
Square -25.
z=\frac{-\left(-25\right)±\sqrt{625-48\times 12}}{2\times 12}
Multiply -4 times 12.
z=\frac{-\left(-25\right)±\sqrt{625-576}}{2\times 12}
Multiply -48 times 12.
z=\frac{-\left(-25\right)±\sqrt{49}}{2\times 12}
Add 625 to -576.
z=\frac{-\left(-25\right)±7}{2\times 12}
Take the square root of 49.
z=\frac{25±7}{2\times 12}
The opposite of -25 is 25.
z=\frac{25±7}{24}
Multiply 2 times 12.
z=\frac{32}{24}
Now solve the equation z=\frac{25±7}{24} when ± is plus. Add 25 to 7.
z=\frac{4}{3}
Reduce the fraction \frac{32}{24} to lowest terms by extracting and canceling out 8.
z=\frac{18}{24}
Now solve the equation z=\frac{25±7}{24} when ± is minus. Subtract 7 from 25.
z=\frac{3}{4}
Reduce the fraction \frac{18}{24} to lowest terms by extracting and canceling out 6.
z=\frac{4}{3} z=\frac{3}{4}
The equation is now solved.
12z^{2}-25z+12=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.
12z^{2}-25z+12-12=-12
Subtract 12 from both sides of the equation.
12z^{2}-25z=-12
Subtracting 12 from itself leaves 0.
\frac{12z^{2}-25z}{12}=-\frac{12}{12}
Divide both sides by 12.
z^{2}-\frac{25}{12}z=-\frac{12}{12}
Dividing by 12 undoes the multiplication by 12.
z^{2}-\frac{25}{12}z=-1
Divide -12 by 12.
z^{2}-\frac{25}{12}z+\left(-\frac{25}{24}\right)^{2}=-1+\left(-\frac{25}{24}\right)^{2}
Divide -\frac{25}{12}, the coefficient of the x term, by 2 to get -\frac{25}{24}. Then add the square of -\frac{25}{24} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
z^{2}-\frac{25}{12}z+\frac{625}{576}=-1+\frac{625}{576}
Square -\frac{25}{24} by squaring both the numerator and the denominator of the fraction.
z^{2}-\frac{25}{12}z+\frac{625}{576}=\frac{49}{576}
Add -1 to \frac{625}{576}.
\left(z-\frac{25}{24}\right)^{2}=\frac{49}{576}
Factor z^{2}-\frac{25}{12}z+\frac{625}{576}. 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(z-\frac{25}{24}\right)^{2}}=\sqrt{\frac{49}{576}}
Take the square root of both sides of the equation.
z-\frac{25}{24}=\frac{7}{24} z-\frac{25}{24}=-\frac{7}{24}
Simplify.
z=\frac{4}{3} z=\frac{3}{4}
Add \frac{25}{24} to both sides of the equation.
x ^ 2 -\frac{25}{12}x +1 = 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 12
r + s = \frac{25}{12} rs = 1
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{25}{24} - u s = \frac{25}{24} + u
Two numbers r and s sum up to \frac{25}{12} exactly when the average of the two numbers is \frac{1}{2}*\frac{25}{12} = \frac{25}{24}. 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{25}{24} - u) (\frac{25}{24} + u) = 1
To solve for unknown quantity u, substitute these in the product equation rs = 1
\frac{625}{576} - u^2 = 1
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 1-\frac{625}{576} = -\frac{49}{576}
Simplify the expression by subtracting \frac{625}{576} on both sides
u^2 = \frac{49}{576} u = \pm\sqrt{\frac{49}{576}} = \pm \frac{7}{24}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{25}{24} - \frac{7}{24} = 0.750 s = \frac{25}{24} + \frac{7}{24} = 1.333
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.