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a+b=9 ab=2\left(-26\right)=-52
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2a^{2}+aa+ba-26. To find a and b, set up a system to be solved.
-1,52 -2,26 -4,13
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 -52.
-1+52=51 -2+26=24 -4+13=9
Calculate the sum for each pair.
a=-4 b=13
The solution is the pair that gives sum 9.
\left(2a^{2}-4a\right)+\left(13a-26\right)
Rewrite 2a^{2}+9a-26 as \left(2a^{2}-4a\right)+\left(13a-26\right).
2a\left(a-2\right)+13\left(a-2\right)
Factor out 2a in the first and 13 in the second group.
\left(a-2\right)\left(2a+13\right)
Factor out common term a-2 by using distributive property.
a=2 a=-\frac{13}{2}
To find equation solutions, solve a-2=0 and 2a+13=0.
2a^{2}+9a-26=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{-9±\sqrt{9^{2}-4\times 2\left(-26\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 9 for b, and -26 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-9±\sqrt{81-4\times 2\left(-26\right)}}{2\times 2}
Square 9.
a=\frac{-9±\sqrt{81-8\left(-26\right)}}{2\times 2}
Multiply -4 times 2.
a=\frac{-9±\sqrt{81+208}}{2\times 2}
Multiply -8 times -26.
a=\frac{-9±\sqrt{289}}{2\times 2}
Add 81 to 208.
a=\frac{-9±17}{2\times 2}
Take the square root of 289.
a=\frac{-9±17}{4}
Multiply 2 times 2.
a=\frac{8}{4}
Now solve the equation a=\frac{-9±17}{4} when ± is plus. Add -9 to 17.
a=2
Divide 8 by 4.
a=-\frac{26}{4}
Now solve the equation a=\frac{-9±17}{4} when ± is minus. Subtract 17 from -9.
a=-\frac{13}{2}
Reduce the fraction \frac{-26}{4} to lowest terms by extracting and canceling out 2.
a=2 a=-\frac{13}{2}
The equation is now solved.
2a^{2}+9a-26=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}+9a-26-\left(-26\right)=-\left(-26\right)
Add 26 to both sides of the equation.
2a^{2}+9a=-\left(-26\right)
Subtracting -26 from itself leaves 0.
2a^{2}+9a=26
Subtract -26 from 0.
\frac{2a^{2}+9a}{2}=\frac{26}{2}
Divide both sides by 2.
a^{2}+\frac{9}{2}a=\frac{26}{2}
Dividing by 2 undoes the multiplication by 2.
a^{2}+\frac{9}{2}a=13
Divide 26 by 2.
a^{2}+\frac{9}{2}a+\left(\frac{9}{4}\right)^{2}=13+\left(\frac{9}{4}\right)^{2}
Divide \frac{9}{2}, the coefficient of the x term, by 2 to get \frac{9}{4}. Then add the square of \frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
a^{2}+\frac{9}{2}a+\frac{81}{16}=13+\frac{81}{16}
Square \frac{9}{4} by squaring both the numerator and the denominator of the fraction.
a^{2}+\frac{9}{2}a+\frac{81}{16}=\frac{289}{16}
Add 13 to \frac{81}{16}.
\left(a+\frac{9}{4}\right)^{2}=\frac{289}{16}
Factor a^{2}+\frac{9}{2}a+\frac{81}{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{9}{4}\right)^{2}}=\sqrt{\frac{289}{16}}
Take the square root of both sides of the equation.
a+\frac{9}{4}=\frac{17}{4} a+\frac{9}{4}=-\frac{17}{4}
Simplify.
a=2 a=-\frac{13}{2}
Subtract \frac{9}{4} from both sides of the equation.
x ^ 2 +\frac{9}{2}x -13 = 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{9}{2} rs = -13
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{9}{4} - u s = -\frac{9}{4} + u
Two numbers r and s sum up to -\frac{9}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{9}{2} = -\frac{9}{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{9}{4} - u) (-\frac{9}{4} + u) = -13
To solve for unknown quantity u, substitute these in the product equation rs = -13
\frac{81}{16} - u^2 = -13
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -13-\frac{81}{16} = -\frac{289}{16}
Simplify the expression by subtracting \frac{81}{16} on both sides
u^2 = \frac{289}{16} u = \pm\sqrt{\frac{289}{16}} = \pm \frac{17}{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{9}{4} - \frac{17}{4} = -6.500 s = -\frac{9}{4} + \frac{17}{4} = 2
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.