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a+b=-57 ab=2\times 81=162
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2a^{2}+aa+ba+81. To find a and b, set up a system to be solved.
-1,-162 -2,-81 -3,-54 -6,-27 -9,-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 162.
-1-162=-163 -2-81=-83 -3-54=-57 -6-27=-33 -9-18=-27
Calculate the sum for each pair.
a=-54 b=-3
The solution is the pair that gives sum -57.
\left(2a^{2}-54a\right)+\left(-3a+81\right)
Rewrite 2a^{2}-57a+81 as \left(2a^{2}-54a\right)+\left(-3a+81\right).
2a\left(a-27\right)-3\left(a-27\right)
Factor out 2a in the first and -3 in the second group.
\left(a-27\right)\left(2a-3\right)
Factor out common term a-27 by using distributive property.
a=27 a=\frac{3}{2}
To find equation solutions, solve a-27=0 and 2a-3=0.
2a^{2}-57a+81=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.
a=\frac{-\left(-57\right)±\sqrt{\left(-57\right)^{2}-4\times 2\times 81}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -57 for b, and 81 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-\left(-57\right)±\sqrt{3249-4\times 2\times 81}}{2\times 2}
Square -57.
a=\frac{-\left(-57\right)±\sqrt{3249-8\times 81}}{2\times 2}
Multiply -4 times 2.
a=\frac{-\left(-57\right)±\sqrt{3249-648}}{2\times 2}
Multiply -8 times 81.
a=\frac{-\left(-57\right)±\sqrt{2601}}{2\times 2}
Add 3249 to -648.
a=\frac{-\left(-57\right)±51}{2\times 2}
Take the square root of 2601.
a=\frac{57±51}{2\times 2}
The opposite of -57 is 57.
a=\frac{57±51}{4}
Multiply 2 times 2.
a=\frac{108}{4}
Now solve the equation a=\frac{57±51}{4} when ± is plus. Add 57 to 51.
a=27
Divide 108 by 4.
a=\frac{6}{4}
Now solve the equation a=\frac{57±51}{4} when ± is minus. Subtract 51 from 57.
a=\frac{3}{2}
Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.
a=27 a=\frac{3}{2}
The equation is now solved.
2a^{2}-57a+81=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.
2a^{2}-57a+81-81=-81
Subtract 81 from both sides of the equation.
2a^{2}-57a=-81
Subtracting 81 from itself leaves 0.
\frac{2a^{2}-57a}{2}=-\frac{81}{2}
Divide both sides by 2.
a^{2}-\frac{57}{2}a=-\frac{81}{2}
Dividing by 2 undoes the multiplication by 2.
a^{2}-\frac{57}{2}a+\left(-\frac{57}{4}\right)^{2}=-\frac{81}{2}+\left(-\frac{57}{4}\right)^{2}
Divide -\frac{57}{2}, the coefficient of the x term, by 2 to get -\frac{57}{4}. Then add the square of -\frac{57}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
a^{2}-\frac{57}{2}a+\frac{3249}{16}=-\frac{81}{2}+\frac{3249}{16}
Square -\frac{57}{4} by squaring both the numerator and the denominator of the fraction.
a^{2}-\frac{57}{2}a+\frac{3249}{16}=\frac{2601}{16}
Add -\frac{81}{2} to \frac{3249}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(a-\frac{57}{4}\right)^{2}=\frac{2601}{16}
Factor a^{2}-\frac{57}{2}a+\frac{3249}{16}. 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(a-\frac{57}{4}\right)^{2}}=\sqrt{\frac{2601}{16}}
Take the square root of both sides of the equation.
a-\frac{57}{4}=\frac{51}{4} a-\frac{57}{4}=-\frac{51}{4}
Simplify.
a=27 a=\frac{3}{2}
Add \frac{57}{4} to both sides of the equation.
x ^ 2 -\frac{57}{2}x +\frac{81}{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 2
r + s = \frac{57}{2} rs = \frac{81}{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{57}{4} - u s = \frac{57}{4} + u
Two numbers r and s sum up to \frac{57}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{57}{2} = \frac{57}{4}. 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{57}{4} - u) (\frac{57}{4} + u) = \frac{81}{2}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{81}{2}
\frac{3249}{16} - u^2 = \frac{81}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{81}{2}-\frac{3249}{16} = -\frac{2601}{16}
Simplify the expression by subtracting \frac{3249}{16} on both sides
u^2 = \frac{2601}{16} u = \pm\sqrt{\frac{2601}{16}} = \pm \frac{51}{4}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{57}{4} - \frac{51}{4} = 1.500 s = \frac{57}{4} + \frac{51}{4} = 27
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.