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a+b=-3 ab=9\left(-2\right)=-18
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9m^{2}+am+bm-2. To find a and b, set up a system to be solved.
1,-18 2,-9 3,-6
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 -18.
1-18=-17 2-9=-7 3-6=-3
Calculate the sum for each pair.
a=-6 b=3
The solution is the pair that gives sum -3.
\left(9m^{2}-6m\right)+\left(3m-2\right)
Rewrite 9m^{2}-3m-2 as \left(9m^{2}-6m\right)+\left(3m-2\right).
3m\left(3m-2\right)+3m-2
Factor out 3m in 9m^{2}-6m.
\left(3m-2\right)\left(3m+1\right)
Factor out common term 3m-2 by using distributive property.
m=\frac{2}{3} m=-\frac{1}{3}
To find equation solutions, solve 3m-2=0 and 3m+1=0.
9m^{2}-3m-2=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.
m=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 9\left(-2\right)}}{2\times 9}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -3 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
m=\frac{-\left(-3\right)±\sqrt{9-4\times 9\left(-2\right)}}{2\times 9}
Square -3.
m=\frac{-\left(-3\right)±\sqrt{9-36\left(-2\right)}}{2\times 9}
Multiply -4 times 9.
m=\frac{-\left(-3\right)±\sqrt{9+72}}{2\times 9}
Multiply -36 times -2.
m=\frac{-\left(-3\right)±\sqrt{81}}{2\times 9}
Add 9 to 72.
m=\frac{-\left(-3\right)±9}{2\times 9}
Take the square root of 81.
m=\frac{3±9}{2\times 9}
The opposite of -3 is 3.
m=\frac{3±9}{18}
Multiply 2 times 9.
m=\frac{12}{18}
Now solve the equation m=\frac{3±9}{18} when ± is plus. Add 3 to 9.
m=\frac{2}{3}
Reduce the fraction \frac{12}{18} to lowest terms by extracting and canceling out 6.
m=-\frac{6}{18}
Now solve the equation m=\frac{3±9}{18} when ± is minus. Subtract 9 from 3.
m=-\frac{1}{3}
Reduce the fraction \frac{-6}{18} to lowest terms by extracting and canceling out 6.
m=\frac{2}{3} m=-\frac{1}{3}
The equation is now solved.
9m^{2}-3m-2=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.
9m^{2}-3m-2-\left(-2\right)=-\left(-2\right)
Add 2 to both sides of the equation.
9m^{2}-3m=-\left(-2\right)
Subtracting -2 from itself leaves 0.
9m^{2}-3m=2
Subtract -2 from 0.
\frac{9m^{2}-3m}{9}=\frac{2}{9}
Divide both sides by 9.
m^{2}+\left(-\frac{3}{9}\right)m=\frac{2}{9}
Dividing by 9 undoes the multiplication by 9.
m^{2}-\frac{1}{3}m=\frac{2}{9}
Reduce the fraction \frac{-3}{9} to lowest terms by extracting and canceling out 3.
m^{2}-\frac{1}{3}m+\left(-\frac{1}{6}\right)^{2}=\frac{2}{9}+\left(-\frac{1}{6}\right)^{2}
Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
m^{2}-\frac{1}{3}m+\frac{1}{36}=\frac{2}{9}+\frac{1}{36}
Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.
m^{2}-\frac{1}{3}m+\frac{1}{36}=\frac{1}{4}
Add \frac{2}{9} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(m-\frac{1}{6}\right)^{2}=\frac{1}{4}
Factor m^{2}-\frac{1}{3}m+\frac{1}{36}. 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(m-\frac{1}{6}\right)^{2}}=\sqrt{\frac{1}{4}}
Take the square root of both sides of the equation.
m-\frac{1}{6}=\frac{1}{2} m-\frac{1}{6}=-\frac{1}{2}
Simplify.
m=\frac{2}{3} m=-\frac{1}{3}
Add \frac{1}{6} to both sides of the equation.
x ^ 2 -\frac{1}{3}x -\frac{2}{9} = 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 9
r + s = \frac{1}{3} rs = -\frac{2}{9}
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}{6} - u s = \frac{1}{6} + u
Two numbers r and s sum up to \frac{1}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{3} = \frac{1}{6}. 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}{6} - u) (\frac{1}{6} + u) = -\frac{2}{9}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{9}
\frac{1}{36} - u^2 = -\frac{2}{9}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{2}{9}-\frac{1}{36} = -\frac{1}{4}
Simplify the expression by subtracting \frac{1}{36} 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{1}{6} - \frac{1}{2} = -0.333 s = \frac{1}{6} + \frac{1}{2} = 0.667
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.