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a+b=-41 ab=12\times 24=288
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 12x^{2}+ax+bx+24. To find a and b, set up a system to be solved.
-1,-288 -2,-144 -3,-96 -4,-72 -6,-48 -8,-36 -9,-32 -12,-24 -16,-18
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 288.
-1-288=-289 -2-144=-146 -3-96=-99 -4-72=-76 -6-48=-54 -8-36=-44 -9-32=-41 -12-24=-36 -16-18=-34
Calculate the sum for each pair.
a=-32 b=-9
The solution is the pair that gives sum -41.
\left(12x^{2}-32x\right)+\left(-9x+24\right)
Rewrite 12x^{2}-41x+24 as \left(12x^{2}-32x\right)+\left(-9x+24\right).
4x\left(3x-8\right)-3\left(3x-8\right)
Factor out 4x in the first and -3 in the second group.
\left(3x-8\right)\left(4x-3\right)
Factor out common term 3x-8 by using distributive property.
x=\frac{8}{3} x=\frac{3}{4}
To find equation solutions, solve 3x-8=0 and 4x-3=0.
12x^{2}-41x+24=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(-41\right)±\sqrt{\left(-41\right)^{2}-4\times 12\times 24}}{2\times 12}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -41 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-41\right)±\sqrt{1681-4\times 12\times 24}}{2\times 12}
Square -41.
x=\frac{-\left(-41\right)±\sqrt{1681-48\times 24}}{2\times 12}
Multiply -4 times 12.
x=\frac{-\left(-41\right)±\sqrt{1681-1152}}{2\times 12}
Multiply -48 times 24.
x=\frac{-\left(-41\right)±\sqrt{529}}{2\times 12}
Add 1681 to -1152.
x=\frac{-\left(-41\right)±23}{2\times 12}
Take the square root of 529.
x=\frac{41±23}{2\times 12}
The opposite of -41 is 41.
x=\frac{41±23}{24}
Multiply 2 times 12.
x=\frac{64}{24}
Now solve the equation x=\frac{41±23}{24} when ± is plus. Add 41 to 23.
x=\frac{8}{3}
Reduce the fraction \frac{64}{24} to lowest terms by extracting and canceling out 8.
x=\frac{18}{24}
Now solve the equation x=\frac{41±23}{24} when ± is minus. Subtract 23 from 41.
x=\frac{3}{4}
Reduce the fraction \frac{18}{24} to lowest terms by extracting and canceling out 6.
x=\frac{8}{3} x=\frac{3}{4}
The equation is now solved.
12x^{2}-41x+24=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}-41x+24-24=-24
Subtract 24 from both sides of the equation.
12x^{2}-41x=-24
Subtracting 24 from itself leaves 0.
\frac{12x^{2}-41x}{12}=-\frac{24}{12}
Divide both sides by 12.
x^{2}-\frac{41}{12}x=-\frac{24}{12}
Dividing by 12 undoes the multiplication by 12.
x^{2}-\frac{41}{12}x=-2
Divide -24 by 12.
x^{2}-\frac{41}{12}x+\left(-\frac{41}{24}\right)^{2}=-2+\left(-\frac{41}{24}\right)^{2}
Divide -\frac{41}{12}, the coefficient of the x term, by 2 to get -\frac{41}{24}. Then add the square of -\frac{41}{24} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{41}{12}x+\frac{1681}{576}=-2+\frac{1681}{576}
Square -\frac{41}{24} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{41}{12}x+\frac{1681}{576}=\frac{529}{576}
Add -2 to \frac{1681}{576}.
\left(x-\frac{41}{24}\right)^{2}=\frac{529}{576}
Factor x^{2}-\frac{41}{12}x+\frac{1681}{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(x-\frac{41}{24}\right)^{2}}=\sqrt{\frac{529}{576}}
Take the square root of both sides of the equation.
x-\frac{41}{24}=\frac{23}{24} x-\frac{41}{24}=-\frac{23}{24}
Simplify.
x=\frac{8}{3} x=\frac{3}{4}
Add \frac{41}{24} to both sides of the equation.
x ^ 2 -\frac{41}{12}x +2 = 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{41}{12} rs = 2
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{41}{24} - u s = \frac{41}{24} + u
Two numbers r and s sum up to \frac{41}{12} exactly when the average of the two numbers is \frac{1}{2}*\frac{41}{12} = \frac{41}{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{41}{24} - u) (\frac{41}{24} + u) = 2
To solve for unknown quantity u, substitute these in the product equation rs = 2
\frac{1681}{576} - u^2 = 2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 2-\frac{1681}{576} = -\frac{529}{576}
Simplify the expression by subtracting \frac{1681}{576} on both sides
u^2 = \frac{529}{576} u = \pm\sqrt{\frac{529}{576}} = \pm \frac{23}{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{41}{24} - \frac{23}{24} = 0.750 s = \frac{41}{24} + \frac{23}{24} = 2.667
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.