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$2 \exponential{x}{2} - x - 36 = 0 $
Solve for x
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a+b=-1 ab=2\left(-36\right)=-72
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-36. To find a and b, set up a system to be solved.
1,-72 2,-36 3,-24 4,-18 6,-12 8,-9
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 -72.
1-72=-71 2-36=-34 3-24=-21 4-18=-14 6-12=-6 8-9=-1
Calculate the sum for each pair.
a=-9 b=8
The solution is the pair that gives sum -1.
\left(2x^{2}-9x\right)+\left(8x-36\right)
Rewrite 2x^{2}-x-36 as \left(2x^{2}-9x\right)+\left(8x-36\right).
x\left(2x-9\right)+4\left(2x-9\right)
Factor out x in the first and 4 in the second group.
\left(2x-9\right)\left(x+4\right)
Factor out common term 2x-9 by using distributive property.
x=\frac{9}{2} x=-4
To find equation solutions, solve 2x-9=0 and x+4=0.
2x^{2}-x-36=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 2\left(-36\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -1 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-1\right)±\sqrt{1-8\left(-36\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-1\right)±\sqrt{1+288}}{2\times 2}
Multiply -8 times -36.
x=\frac{-\left(-1\right)±\sqrt{289}}{2\times 2}
Add 1 to 288.
x=\frac{-\left(-1\right)±17}{2\times 2}
Take the square root of 289.
x=\frac{1±17}{2\times 2}
The opposite of -1 is 1.
x=\frac{1±17}{4}
Multiply 2 times 2.
x=\frac{18}{4}
Now solve the equation x=\frac{1±17}{4} when ± is plus. Add 1 to 17.
x=\frac{9}{2}
Reduce the fraction \frac{18}{4} to lowest terms by extracting and canceling out 2.
x=-\frac{16}{4}
Now solve the equation x=\frac{1±17}{4} when ± is minus. Subtract 17 from 1.
x=-4
Divide -16 by 4.
x=\frac{9}{2} x=-4
The equation is now solved.
2x^{2}-x-36=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-36-\left(-36\right)=-\left(-36\right)
Add 36 to both sides of the equation.
2x^{2}-x=-\left(-36\right)
Subtracting -36 from itself leaves 0.
2x^{2}-x=36
Subtract -36 from 0.
\frac{2x^{2}-x}{2}=\frac{36}{2}
Divide both sides by 2.
x^{2}-\frac{1}{2}x=\frac{36}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-\frac{1}{2}x=18
Divide 36 by 2.
x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=18+\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}=18+\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{289}{16}
Add 18 to \frac{1}{16}.
\left(x-\frac{1}{4}\right)^{2}=\frac{289}{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{289}{16}}
Take the square root of both sides of the equation.
x-\frac{1}{4}=\frac{17}{4} x-\frac{1}{4}=-\frac{17}{4}
Simplify.
x=\frac{9}{2} x=-4
Add \frac{1}{4} to both sides of the equation.
x ^ 2 -\frac{1}{2}x -18 = 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{1}{2} rs = -18
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}{4} - u s = \frac{1}{4} + u
Two numbers r and s sum up to \frac{1}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{2} = \frac{1}{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{1}{4} - u) (\frac{1}{4} + u) = -18
To solve for unknown quantity u, substitute these in the product equation rs = -18
\frac{1}{16} - u^2 = -18
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -18-\frac{1}{16} = -\frac{289}{16}
Simplify the expression by subtracting \frac{1}{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{1}{4} - \frac{17}{4} = -4 s = \frac{1}{4} + \frac{17}{4} = 4.500
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.