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a+b=32 ab=12\times 5=60
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 12x^{2}+ax+bx+5. To find a and b, set up a system to be solved.
1,60 2,30 3,20 4,15 5,12 6,10
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 60.
1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16
Calculate the sum for each pair.
a=2 b=30
The solution is the pair that gives sum 32.
\left(12x^{2}+2x\right)+\left(30x+5\right)
Rewrite 12x^{2}+32x+5 as \left(12x^{2}+2x\right)+\left(30x+5\right).
2x\left(6x+1\right)+5\left(6x+1\right)
Factor out 2x in the first and 5 in the second group.
\left(6x+1\right)\left(2x+5\right)
Factor out common term 6x+1 by using distributive property.
x=-\frac{1}{6} x=-\frac{5}{2}
To find equation solutions, solve 6x+1=0 and 2x+5=0.
12x^{2}+32x+5=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{-32±\sqrt{32^{2}-4\times 12\times 5}}{2\times 12}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, 32 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-32±\sqrt{1024-4\times 12\times 5}}{2\times 12}
Square 32.
x=\frac{-32±\sqrt{1024-48\times 5}}{2\times 12}
Multiply -4 times 12.
x=\frac{-32±\sqrt{1024-240}}{2\times 12}
Multiply -48 times 5.
x=\frac{-32±\sqrt{784}}{2\times 12}
Add 1024 to -240.
x=\frac{-32±28}{2\times 12}
Take the square root of 784.
x=\frac{-32±28}{24}
Multiply 2 times 12.
x=\frac{-4}{24}
Now solve the equation x=\frac{-32±28}{24} when ± is plus. Add -32 to 28.
x=-\frac{1}{6}
Reduce the fraction \frac{-4}{24} to lowest terms by extracting and canceling out 4.
x=\frac{-60}{24}
Now solve the equation x=\frac{-32±28}{24} when ± is minus. Subtract 28 from -32.
x=-\frac{5}{2}
Reduce the fraction \frac{-60}{24} to lowest terms by extracting and canceling out 12.
x=-\frac{1}{6} x=-\frac{5}{2}
The equation is now solved.
12x^{2}+32x+5=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.
12x^{2}+32x+5-5=-5
Subtract 5 from both sides of the equation.
12x^{2}+32x=-5
Subtracting 5 from itself leaves 0.
\frac{12x^{2}+32x}{12}=\frac{-5}{12}
Divide both sides by 12.
x^{2}+\frac{32}{12}x=\frac{-5}{12}
Dividing by 12 undoes the multiplication by 12.
x^{2}+\frac{8}{3}x=\frac{-5}{12}
Reduce the fraction \frac{32}{12} to lowest terms by extracting and canceling out 4.
x^{2}+\frac{8}{3}x=-\frac{5}{12}
Divide -5 by 12.
x^{2}+\frac{8}{3}x+\left(\frac{4}{3}\right)^{2}=-\frac{5}{12}+\left(\frac{4}{3}\right)^{2}
Divide \frac{8}{3}, the coefficient of the x term, by 2 to get \frac{4}{3}. Then add the square of \frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{8}{3}x+\frac{16}{9}=-\frac{5}{12}+\frac{16}{9}
Square \frac{4}{3} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{8}{3}x+\frac{16}{9}=\frac{49}{36}
Add -\frac{5}{12} to \frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{4}{3}\right)^{2}=\frac{49}{36}
Factor x^{2}+\frac{8}{3}x+\frac{16}{9}. 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{4}{3}\right)^{2}}=\sqrt{\frac{49}{36}}
Take the square root of both sides of the equation.
x+\frac{4}{3}=\frac{7}{6} x+\frac{4}{3}=-\frac{7}{6}
Simplify.
x=-\frac{1}{6} x=-\frac{5}{2}
Subtract \frac{4}{3} from both sides of the equation.
x ^ 2 +\frac{8}{3}x +\frac{5}{12} = 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{8}{3} rs = \frac{5}{12}
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{4}{3} - u s = -\frac{4}{3} + u
Two numbers r and s sum up to -\frac{8}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{8}{3} = -\frac{4}{3}. 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{4}{3} - u) (-\frac{4}{3} + u) = \frac{5}{12}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{12}
\frac{16}{9} - u^2 = \frac{5}{12}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{5}{12}-\frac{16}{9} = -\frac{49}{36}
Simplify the expression by subtracting \frac{16}{9} on both sides
u^2 = \frac{49}{36} u = \pm\sqrt{\frac{49}{36}} = \pm \frac{7}{6}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{4}{3} - \frac{7}{6} = -2.500 s = -\frac{4}{3} + \frac{7}{6} = -0.167
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.