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r^{2}+3r+2=0
Divide both sides by 3.
a+b=3 ab=1\times 2=2
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as r^{2}+ar+br+2. To find a and b, set up a system to be solved.
a=1 b=2
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.
\left(r^{2}+r\right)+\left(2r+2\right)
Rewrite r^{2}+3r+2 as \left(r^{2}+r\right)+\left(2r+2\right).
r\left(r+1\right)+2\left(r+1\right)
Factor out r in the first and 2 in the second group.
\left(r+1\right)\left(r+2\right)
Factor out common term r+1 by using distributive property.
r=-1 r=-2
To find equation solutions, solve r+1=0 and r+2=0.
3r^{2}+9r+6=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.
r=\frac{-9±\sqrt{9^{2}-4\times 3\times 6}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 9 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
r=\frac{-9±\sqrt{81-4\times 3\times 6}}{2\times 3}
Square 9.
r=\frac{-9±\sqrt{81-12\times 6}}{2\times 3}
Multiply -4 times 3.
r=\frac{-9±\sqrt{81-72}}{2\times 3}
Multiply -12 times 6.
r=\frac{-9±\sqrt{9}}{2\times 3}
Add 81 to -72.
r=\frac{-9±3}{2\times 3}
Take the square root of 9.
r=\frac{-9±3}{6}
Multiply 2 times 3.
r=-\frac{6}{6}
Now solve the equation r=\frac{-9±3}{6} when ± is plus. Add -9 to 3.
r=-1
Divide -6 by 6.
r=-\frac{12}{6}
Now solve the equation r=\frac{-9±3}{6} when ± is minus. Subtract 3 from -9.
r=-2
Divide -12 by 6.
r=-1 r=-2
The equation is now solved.
3r^{2}+9r+6=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.
3r^{2}+9r+6-6=-6
Subtract 6 from both sides of the equation.
3r^{2}+9r=-6
Subtracting 6 from itself leaves 0.
\frac{3r^{2}+9r}{3}=-\frac{6}{3}
Divide both sides by 3.
r^{2}+\frac{9}{3}r=-\frac{6}{3}
Dividing by 3 undoes the multiplication by 3.
r^{2}+3r=-\frac{6}{3}
Divide 9 by 3.
r^{2}+3r=-2
Divide -6 by 3.
r^{2}+3r+\left(\frac{3}{2}\right)^{2}=-2+\left(\frac{3}{2}\right)^{2}
Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
r^{2}+3r+\frac{9}{4}=-2+\frac{9}{4}
Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.
r^{2}+3r+\frac{9}{4}=\frac{1}{4}
Add -2 to \frac{9}{4}.
\left(r+\frac{3}{2}\right)^{2}=\frac{1}{4}
Factor r^{2}+3r+\frac{9}{4}. 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(r+\frac{3}{2}\right)^{2}}=\sqrt{\frac{1}{4}}
Take the square root of both sides of the equation.
r+\frac{3}{2}=\frac{1}{2} r+\frac{3}{2}=-\frac{1}{2}
Simplify.
r=-1 r=-2
Subtract \frac{3}{2} from both sides of the equation.
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.This is achieved by dividing both sides of the equation by 3
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} = -2 s = -\frac{3}{2} + \frac{1}{2} = -1
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.