Solve for a
a=-\frac{2}{3}\approx -0.666666667
a=3
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a+b=-7 ab=3\left(-6\right)=-18
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3a^{2}+aa+ba-6. 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=-9 b=2
The solution is the pair that gives sum -7.
\left(3a^{2}-9a\right)+\left(2a-6\right)
Rewrite 3a^{2}-7a-6 as \left(3a^{2}-9a\right)+\left(2a-6\right).
3a\left(a-3\right)+2\left(a-3\right)
Factor out 3a in the first and 2 in the second group.
\left(a-3\right)\left(3a+2\right)
Factor out common term a-3 by using distributive property.
a=3 a=-\frac{2}{3}
To find equation solutions, solve a-3=0 and 3a+2=0.
3a^{2}-7a-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.
a=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 3\left(-6\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -7 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-\left(-7\right)±\sqrt{49-4\times 3\left(-6\right)}}{2\times 3}
Square -7.
a=\frac{-\left(-7\right)±\sqrt{49-12\left(-6\right)}}{2\times 3}
Multiply -4 times 3.
a=\frac{-\left(-7\right)±\sqrt{49+72}}{2\times 3}
Multiply -12 times -6.
a=\frac{-\left(-7\right)±\sqrt{121}}{2\times 3}
Add 49 to 72.
a=\frac{-\left(-7\right)±11}{2\times 3}
Take the square root of 121.
a=\frac{7±11}{2\times 3}
The opposite of -7 is 7.
a=\frac{7±11}{6}
Multiply 2 times 3.
a=\frac{18}{6}
Now solve the equation a=\frac{7±11}{6} when ± is plus. Add 7 to 11.
a=3
Divide 18 by 6.
a=-\frac{4}{6}
Now solve the equation a=\frac{7±11}{6} when ± is minus. Subtract 11 from 7.
a=-\frac{2}{3}
Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.
a=3 a=-\frac{2}{3}
The equation is now solved.
3a^{2}-7a-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.
3a^{2}-7a-6-\left(-6\right)=-\left(-6\right)
Add 6 to both sides of the equation.
3a^{2}-7a=-\left(-6\right)
Subtracting -6 from itself leaves 0.
3a^{2}-7a=6
Subtract -6 from 0.
\frac{3a^{2}-7a}{3}=\frac{6}{3}
Divide both sides by 3.
a^{2}-\frac{7}{3}a=\frac{6}{3}
Dividing by 3 undoes the multiplication by 3.
a^{2}-\frac{7}{3}a=2
Divide 6 by 3.
a^{2}-\frac{7}{3}a+\left(-\frac{7}{6}\right)^{2}=2+\left(-\frac{7}{6}\right)^{2}
Divide -\frac{7}{3}, the coefficient of the x term, by 2 to get -\frac{7}{6}. Then add the square of -\frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
a^{2}-\frac{7}{3}a+\frac{49}{36}=2+\frac{49}{36}
Square -\frac{7}{6} by squaring both the numerator and the denominator of the fraction.
a^{2}-\frac{7}{3}a+\frac{49}{36}=\frac{121}{36}
Add 2 to \frac{49}{36}.
\left(a-\frac{7}{6}\right)^{2}=\frac{121}{36}
Factor a^{2}-\frac{7}{3}a+\frac{49}{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(a-\frac{7}{6}\right)^{2}}=\sqrt{\frac{121}{36}}
Take the square root of both sides of the equation.
a-\frac{7}{6}=\frac{11}{6} a-\frac{7}{6}=-\frac{11}{6}
Simplify.
a=3 a=-\frac{2}{3}
Add \frac{7}{6} to both sides of the equation.
x ^ 2 -\frac{7}{3}x -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 = \frac{7}{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{7}{6} - u s = \frac{7}{6} + u
Two numbers r and s sum up to \frac{7}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{3} = \frac{7}{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{7}{6} - u) (\frac{7}{6} + u) = -2
To solve for unknown quantity u, substitute these in the product equation rs = -2
\frac{49}{36} - u^2 = -2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -2-\frac{49}{36} = -\frac{121}{36}
Simplify the expression by subtracting \frac{49}{36} on both sides
u^2 = \frac{121}{36} u = \pm\sqrt{\frac{121}{36}} = \pm \frac{11}{6}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{7}{6} - \frac{11}{6} = -0.667 s = \frac{7}{6} + \frac{11}{6} = 3
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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