a+b=-9 ab=2\times 9=18
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2j^{2}+aj+bj+9. To find a and b, set up a system to be solved.
-1,-18 -2,-9 -3,-6
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 18.
-1-18=-19 -2-9=-11 -3-6=-9
Calculate the sum for each pair.
a=-6 b=-3
The solution is the pair that gives sum -9.
\left(2j^{2}-6j\right)+\left(-3j+9\right)
Rewrite 2j^{2}-9j+9 as \left(2j^{2}-6j\right)+\left(-3j+9\right).
2j\left(j-3\right)-3\left(j-3\right)
Factor out 2j in the first and -3 in the second group.
\left(j-3\right)\left(2j-3\right)
Factor out common term j-3 by using distributive property.
j=3 j=\frac{3}{2}
To find equation solutions, solve j-3=0 and 2j-3=0.
2j^{2}-9j+9=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.
j=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 2\times 9}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -9 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
j=\frac{-\left(-9\right)±\sqrt{81-4\times 2\times 9}}{2\times 2}
Square -9.
j=\frac{-\left(-9\right)±\sqrt{81-8\times 9}}{2\times 2}
Multiply -4 times 2.
j=\frac{-\left(-9\right)±\sqrt{81-72}}{2\times 2}
Multiply -8 times 9.
j=\frac{-\left(-9\right)±\sqrt{9}}{2\times 2}
Add 81 to -72.
j=\frac{-\left(-9\right)±3}{2\times 2}
Take the square root of 9.
j=\frac{9±3}{2\times 2}
The opposite of -9 is 9.
j=\frac{9±3}{4}
Multiply 2 times 2.
j=\frac{12}{4}
Now solve the equation j=\frac{9±3}{4} when ± is plus. Add 9 to 3.
j=3
Divide 12 by 4.
j=\frac{6}{4}
Now solve the equation j=\frac{9±3}{4} when ± is minus. Subtract 3 from 9.
j=\frac{3}{2}
Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.
j=3 j=\frac{3}{2}
The equation is now solved.
2j^{2}-9j+9=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.
2j^{2}-9j+9-9=-9
Subtract 9 from both sides of the equation.
2j^{2}-9j=-9
Subtracting 9 from itself leaves 0.
\frac{2j^{2}-9j}{2}=-\frac{9}{2}
Divide both sides by 2.
j^{2}-\frac{9}{2}j=-\frac{9}{2}
Dividing by 2 undoes the multiplication by 2.
j^{2}-\frac{9}{2}j+\left(-\frac{9}{4}\right)^{2}=-\frac{9}{2}+\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.
j^{2}-\frac{9}{2}j+\frac{81}{16}=-\frac{9}{2}+\frac{81}{16}
Square -\frac{9}{4} by squaring both the numerator and the denominator of the fraction.
j^{2}-\frac{9}{2}j+\frac{81}{16}=\frac{9}{16}
Add -\frac{9}{2} to \frac{81}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(j-\frac{9}{4}\right)^{2}=\frac{9}{16}
Factor j^{2}-\frac{9}{2}j+\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(j-\frac{9}{4}\right)^{2}}=\sqrt{\frac{9}{16}}
Take the square root of both sides of the equation.
j-\frac{9}{4}=\frac{3}{4} j-\frac{9}{4}=-\frac{3}{4}
Simplify.
j=3 j=\frac{3}{2}
Add \frac{9}{4} to both sides of the equation.
x ^ 2 -\frac{9}{2}x +\frac{9}{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{9}{2} rs = \frac{9}{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{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) = \frac{9}{2}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{9}{2}
\frac{81}{16} - u^2 = \frac{9}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{9}{2}-\frac{81}{16} = -\frac{9}{16}
Simplify the expression by subtracting \frac{81}{16} on both sides
u^2 = \frac{9}{16} u = \pm\sqrt{\frac{9}{16}} = \pm \frac{3}{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{3}{4} = 1.500 s = \frac{9}{4} + \frac{3}{4} = 3
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
