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a+b=-3 ab=1\times 2=2
Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+2. To find a and b, set up a system to be solved.
a=-2 b=-1
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.
\left(x^{2}-2x\right)+\left(-x+2\right)
Rewrite x^{2}-3x+2 as \left(x^{2}-2x\right)+\left(-x+2\right).
x\left(x-2\right)-\left(x-2\right)
Factor out x in the first and -1 in the second group.
\left(x-2\right)\left(x-1\right)
Factor out common term x-2 by using distributive property.
x^{2}-3x+2=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2}}{2}
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(-3\right)±\sqrt{9-4\times 2}}{2}
Square -3.
x=\frac{-\left(-3\right)±\sqrt{9-8}}{2}
Multiply -4 times 2.
x=\frac{-\left(-3\right)±\sqrt{1}}{2}
Add 9 to -8.
x=\frac{-\left(-3\right)±1}{2}
Take the square root of 1.
x=\frac{3±1}{2}
The opposite of -3 is 3.
x=\frac{4}{2}
Now solve the equation x=\frac{3±1}{2} when ± is plus. Add 3 to 1.
x=2
Divide 4 by 2.
x=\frac{2}{2}
Now solve the equation x=\frac{3±1}{2} when ± is minus. Subtract 1 from 3.
x=1
Divide 2 by 2.
x^{2}-3x+2=\left(x-2\right)\left(x-1\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and 1 for x_{2}.
x ^ 2 -3x +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.
r + s = 3 rs = 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{3}{2} - u s = \frac{3}{2} + u
Two numbers r and s sum up to 3 exactly when the average of the two numbers is \frac{1}{2}*3 = \frac{3}{2}. 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{3}{2} - u) (\frac{3}{2} + u) = 2
To solve for unknown quantity u, substitute these in the product equation rs = 2
\frac{9}{4} - u^2 = 2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 2-\frac{9}{4} = -\frac{1}{4}
Simplify the expression by subtracting \frac{9}{4} on both sides
u^2 = \frac{1}{4} u = \pm\sqrt{\frac{1}{4}} = \pm \frac{1}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{3}{2} - \frac{1}{2} = 1 s = \frac{3}{2} + \frac{1}{2} = 2
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.