Solve for x
x = -\frac{33}{2} = -16\frac{1}{2} = -16.5
x=16
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a+b=1 ab=2\left(-528\right)=-1056
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-528. To find a and b, set up a system to be solved.
-1,1056 -2,528 -3,352 -4,264 -6,176 -8,132 -11,96 -12,88 -16,66 -22,48 -24,44 -32,33
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -1056.
-1+1056=1055 -2+528=526 -3+352=349 -4+264=260 -6+176=170 -8+132=124 -11+96=85 -12+88=76 -16+66=50 -22+48=26 -24+44=20 -32+33=1
Calculate the sum for each pair.
a=-32 b=33
The solution is the pair that gives sum 1.
\left(2x^{2}-32x\right)+\left(33x-528\right)
Rewrite 2x^{2}+x-528 as \left(2x^{2}-32x\right)+\left(33x-528\right).
2x\left(x-16\right)+33\left(x-16\right)
Factor out 2x in the first and 33 in the second group.
\left(x-16\right)\left(2x+33\right)
Factor out common term x-16 by using distributive property.
x=16 x=-\frac{33}{2}
To find equation solutions, solve x-16=0 and 2x+33=0.
2x^{2}+x-528=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{-1±\sqrt{1^{2}-4\times 2\left(-528\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 1 for b, and -528 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-1±\sqrt{1-4\times 2\left(-528\right)}}{2\times 2}
Square 1.
x=\frac{-1±\sqrt{1-8\left(-528\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-1±\sqrt{1+4224}}{2\times 2}
Multiply -8 times -528.
x=\frac{-1±\sqrt{4225}}{2\times 2}
Add 1 to 4224.
x=\frac{-1±65}{2\times 2}
Take the square root of 4225.
x=\frac{-1±65}{4}
Multiply 2 times 2.
x=\frac{64}{4}
Now solve the equation x=\frac{-1±65}{4} when ± is plus. Add -1 to 65.
x=16
Divide 64 by 4.
x=-\frac{66}{4}
Now solve the equation x=\frac{-1±65}{4} when ± is minus. Subtract 65 from -1.
x=-\frac{33}{2}
Reduce the fraction \frac{-66}{4} to lowest terms by extracting and canceling out 2.
x=16 x=-\frac{33}{2}
The equation is now solved.
2x^{2}+x-528=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.
2x^{2}+x-528-\left(-528\right)=-\left(-528\right)
Add 528 to both sides of the equation.
2x^{2}+x=-\left(-528\right)
Subtracting -528 from itself leaves 0.
2x^{2}+x=528
Subtract -528 from 0.
\frac{2x^{2}+x}{2}=\frac{528}{2}
Divide both sides by 2.
x^{2}+\frac{1}{2}x=\frac{528}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}+\frac{1}{2}x=264
Divide 528 by 2.
x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=264+\left(\frac{1}{4}\right)^{2}
Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{1}{2}x+\frac{1}{16}=264+\frac{1}{16}
Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{4225}{16}
Add 264 to \frac{1}{16}.
\left(x+\frac{1}{4}\right)^{2}=\frac{4225}{16}
Factor x^{2}+\frac{1}{2}x+\frac{1}{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(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{4225}{16}}
Take the square root of both sides of the equation.
x+\frac{1}{4}=\frac{65}{4} x+\frac{1}{4}=-\frac{65}{4}
Simplify.
x=16 x=-\frac{33}{2}
Subtract \frac{1}{4} from both sides of the equation.
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Limits
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