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a+b=-25 ab=6\times 24=144
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx+24. 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(6x^{2}-16x\right)+\left(-9x+24\right)
Rewrite 6x^{2}-25x+24 as \left(6x^{2}-16x\right)+\left(-9x+24\right).
2x\left(3x-8\right)-3\left(3x-8\right)
Factor out 2x in the first and -3 in the second group.
\left(3x-8\right)\left(2x-3\right)
Factor out common term 3x-8 by using distributive property.
x=\frac{8}{3} x=\frac{3}{2}
To find equation solutions, solve 3x-8=0 and 2x-3=0.
6x^{2}-25x+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(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 6\times 24}}{2\times 6}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -25 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-25\right)±\sqrt{625-4\times 6\times 24}}{2\times 6}
Square -25.
x=\frac{-\left(-25\right)±\sqrt{625-24\times 24}}{2\times 6}
Multiply -4 times 6.
x=\frac{-\left(-25\right)±\sqrt{625-576}}{2\times 6}
Multiply -24 times 24.
x=\frac{-\left(-25\right)±\sqrt{49}}{2\times 6}
Add 625 to -576.
x=\frac{-\left(-25\right)±7}{2\times 6}
Take the square root of 49.
x=\frac{25±7}{2\times 6}
The opposite of -25 is 25.
x=\frac{25±7}{12}
Multiply 2 times 6.
x=\frac{32}{12}
Now solve the equation x=\frac{25±7}{12} when ± is plus. Add 25 to 7.
x=\frac{8}{3}
Reduce the fraction \frac{32}{12} to lowest terms by extracting and canceling out 4.
x=\frac{18}{12}
Now solve the equation x=\frac{25±7}{12} when ± is minus. Subtract 7 from 25.
x=\frac{3}{2}
Reduce the fraction \frac{18}{12} to lowest terms by extracting and canceling out 6.
x=\frac{8}{3} x=\frac{3}{2}
The equation is now solved.
6x^{2}-25x+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.
6x^{2}-25x+24-24=-24
Subtract 24 from both sides of the equation.
6x^{2}-25x=-24
Subtracting 24 from itself leaves 0.
\frac{6x^{2}-25x}{6}=-\frac{24}{6}
Divide both sides by 6.
x^{2}-\frac{25}{6}x=-\frac{24}{6}
Dividing by 6 undoes the multiplication by 6.
x^{2}-\frac{25}{6}x=-4
Divide -24 by 6.
x^{2}-\frac{25}{6}x+\left(-\frac{25}{12}\right)^{2}=-4+\left(-\frac{25}{12}\right)^{2}
Divide -\frac{25}{6}, the coefficient of the x term, by 2 to get -\frac{25}{12}. Then add the square of -\frac{25}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{25}{6}x+\frac{625}{144}=-4+\frac{625}{144}
Square -\frac{25}{12} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{25}{6}x+\frac{625}{144}=\frac{49}{144}
Add -4 to \frac{625}{144}.
\left(x-\frac{25}{12}\right)^{2}=\frac{49}{144}
Factor x^{2}-\frac{25}{6}x+\frac{625}{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{25}{12}\right)^{2}}=\sqrt{\frac{49}{144}}
Take the square root of both sides of the equation.
x-\frac{25}{12}=\frac{7}{12} x-\frac{25}{12}=-\frac{7}{12}
Simplify.
x=\frac{8}{3} x=\frac{3}{2}
Add \frac{25}{12} to both sides of the equation.
x ^ 2 -\frac{25}{6}x +4 = 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{25}{6} rs = 4
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}{12} - u s = \frac{25}{12} + u
Two numbers r and s sum up to \frac{25}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{25}{6} = \frac{25}{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{25}{12} - u) (\frac{25}{12} + u) = 4
To solve for unknown quantity u, substitute these in the product equation rs = 4
\frac{625}{144} - u^2 = 4
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 4-\frac{625}{144} = -\frac{49}{144}
Simplify the expression by subtracting \frac{625}{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{25}{12} - \frac{7}{12} = 1.500 s = \frac{25}{12} + \frac{7}{12} = 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.