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a+b=-1 ab=3\left(-52\right)=-156
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-52. To find a and b, set up a system to be solved.
1,-156 2,-78 3,-52 4,-39 6,-26 12,-13
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -156.
1-156=-155 2-78=-76 3-52=-49 4-39=-35 6-26=-20 12-13=-1
Calculate the sum for each pair.
a=-13 b=12
The solution is the pair that gives sum -1.
\left(3x^{2}-13x\right)+\left(12x-52\right)
Rewrite 3x^{2}-x-52 as \left(3x^{2}-13x\right)+\left(12x-52\right).
x\left(3x-13\right)+4\left(3x-13\right)
Factor out x in the first and 4 in the second group.
\left(3x-13\right)\left(x+4\right)
Factor out common term 3x-13 by using distributive property.
x=\frac{13}{3} x=-4
To find equation solutions, solve 3x-13=0 and x+4=0.
3x^{2}-x-52=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(-1\right)±\sqrt{1-4\times 3\left(-52\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -1 for b, and -52 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-1\right)±\sqrt{1-12\left(-52\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-1\right)±\sqrt{1+624}}{2\times 3}
Multiply -12 times -52.
x=\frac{-\left(-1\right)±\sqrt{625}}{2\times 3}
Add 1 to 624.
x=\frac{-\left(-1\right)±25}{2\times 3}
Take the square root of 625.
x=\frac{1±25}{2\times 3}
The opposite of -1 is 1.
x=\frac{1±25}{6}
Multiply 2 times 3.
x=\frac{26}{6}
Now solve the equation x=\frac{1±25}{6} when ± is plus. Add 1 to 25.
x=\frac{13}{3}
Reduce the fraction \frac{26}{6} to lowest terms by extracting and canceling out 2.
x=-\frac{24}{6}
Now solve the equation x=\frac{1±25}{6} when ± is minus. Subtract 25 from 1.
x=-4
Divide -24 by 6.
x=\frac{13}{3} x=-4
The equation is now solved.
3x^{2}-x-52=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.
3x^{2}-x-52-\left(-52\right)=-\left(-52\right)
Add 52 to both sides of the equation.
3x^{2}-x=-\left(-52\right)
Subtracting -52 from itself leaves 0.
3x^{2}-x=52
Subtract -52 from 0.
\frac{3x^{2}-x}{3}=\frac{52}{3}
Divide both sides by 3.
x^{2}-\frac{1}{3}x=\frac{52}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{52}{3}+\left(-\frac{1}{6}\right)^{2}
Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{52}{3}+\frac{1}{36}
Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{625}{36}
Add \frac{52}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{6}\right)^{2}=\frac{625}{36}
Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. 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}{6}\right)^{2}}=\sqrt{\frac{625}{36}}
Take the square root of both sides of the equation.
x-\frac{1}{6}=\frac{25}{6} x-\frac{1}{6}=-\frac{25}{6}
Simplify.
x=\frac{13}{3} x=-4
Add \frac{1}{6} to both sides of the equation.
x ^ 2 -\frac{1}{3}x -\frac{52}{3} = 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 3
r + s = \frac{1}{3} rs = -\frac{52}{3}
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}{6} - u s = \frac{1}{6} + u
Two numbers r and s sum up to \frac{1}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{3} = \frac{1}{6}. 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}{6} - u) (\frac{1}{6} + u) = -\frac{52}{3}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{52}{3}
\frac{1}{36} - u^2 = -\frac{52}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{52}{3}-\frac{1}{36} = -\frac{625}{36}
Simplify the expression by subtracting \frac{1}{36} on both sides
u^2 = \frac{625}{36} u = \pm\sqrt{\frac{625}{36}} = \pm \frac{25}{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{1}{6} - \frac{25}{6} = -4 s = \frac{1}{6} + \frac{25}{6} = 4.333
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.