Solve for a
a=-\frac{2}{3}\approx -0.666666667
a = \frac{3}{2} = 1\frac{1}{2} = 1.5
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a+b=-5 ab=6\left(-6\right)=-36
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6a^{2}+aa+ba-6. To find a and b, set up a system to be solved.
1,-36 2,-18 3,-12 4,-9 6,-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 -36.
1-36=-35 2-18=-16 3-12=-9 4-9=-5 6-6=0
Calculate the sum for each pair.
a=-9 b=4
The solution is the pair that gives sum -5.
\left(6a^{2}-9a\right)+\left(4a-6\right)
Rewrite 6a^{2}-5a-6 as \left(6a^{2}-9a\right)+\left(4a-6\right).
3a\left(2a-3\right)+2\left(2a-3\right)
Factor out 3a in the first and 2 in the second group.
\left(2a-3\right)\left(3a+2\right)
Factor out common term 2a-3 by using distributive property.
a=\frac{3}{2} a=-\frac{2}{3}
To find equation solutions, solve 2a-3=0 and 3a+2=0.
6a^{2}-5a-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(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6\left(-6\right)}}{2\times 6}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -5 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-\left(-5\right)±\sqrt{25-4\times 6\left(-6\right)}}{2\times 6}
Square -5.
a=\frac{-\left(-5\right)±\sqrt{25-24\left(-6\right)}}{2\times 6}
Multiply -4 times 6.
a=\frac{-\left(-5\right)±\sqrt{25+144}}{2\times 6}
Multiply -24 times -6.
a=\frac{-\left(-5\right)±\sqrt{169}}{2\times 6}
Add 25 to 144.
a=\frac{-\left(-5\right)±13}{2\times 6}
Take the square root of 169.
a=\frac{5±13}{2\times 6}
The opposite of -5 is 5.
a=\frac{5±13}{12}
Multiply 2 times 6.
a=\frac{18}{12}
Now solve the equation a=\frac{5±13}{12} when ± is plus. Add 5 to 13.
a=\frac{3}{2}
Reduce the fraction \frac{18}{12} to lowest terms by extracting and canceling out 6.
a=-\frac{8}{12}
Now solve the equation a=\frac{5±13}{12} when ± is minus. Subtract 13 from 5.
a=-\frac{2}{3}
Reduce the fraction \frac{-8}{12} to lowest terms by extracting and canceling out 4.
a=\frac{3}{2} a=-\frac{2}{3}
The equation is now solved.
6a^{2}-5a-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.
6a^{2}-5a-6-\left(-6\right)=-\left(-6\right)
Add 6 to both sides of the equation.
6a^{2}-5a=-\left(-6\right)
Subtracting -6 from itself leaves 0.
6a^{2}-5a=6
Subtract -6 from 0.
\frac{6a^{2}-5a}{6}=\frac{6}{6}
Divide both sides by 6.
a^{2}-\frac{5}{6}a=\frac{6}{6}
Dividing by 6 undoes the multiplication by 6.
a^{2}-\frac{5}{6}a=1
Divide 6 by 6.
a^{2}-\frac{5}{6}a+\left(-\frac{5}{12}\right)^{2}=1+\left(-\frac{5}{12}\right)^{2}
Divide -\frac{5}{6}, the coefficient of the x term, by 2 to get -\frac{5}{12}. Then add the square of -\frac{5}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
a^{2}-\frac{5}{6}a+\frac{25}{144}=1+\frac{25}{144}
Square -\frac{5}{12} by squaring both the numerator and the denominator of the fraction.
a^{2}-\frac{5}{6}a+\frac{25}{144}=\frac{169}{144}
Add 1 to \frac{25}{144}.
\left(a-\frac{5}{12}\right)^{2}=\frac{169}{144}
Factor a^{2}-\frac{5}{6}a+\frac{25}{144}. 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{5}{12}\right)^{2}}=\sqrt{\frac{169}{144}}
Take the square root of both sides of the equation.
a-\frac{5}{12}=\frac{13}{12} a-\frac{5}{12}=-\frac{13}{12}
Simplify.
a=\frac{3}{2} a=-\frac{2}{3}
Add \frac{5}{12} to both sides of the equation.
x ^ 2 -\frac{5}{6}x -1 = 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 6
r + s = \frac{5}{6} rs = -1
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{5}{12} - u s = \frac{5}{12} + u
Two numbers r and s sum up to \frac{5}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{6} = \frac{5}{12}. 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{5}{12} - u) (\frac{5}{12} + u) = -1
To solve for unknown quantity u, substitute these in the product equation rs = -1
\frac{25}{144} - u^2 = -1
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -1-\frac{25}{144} = -\frac{169}{144}
Simplify the expression by subtracting \frac{25}{144} on both sides
u^2 = \frac{169}{144} u = \pm\sqrt{\frac{169}{144}} = \pm \frac{13}{12}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{5}{12} - \frac{13}{12} = -0.667 s = \frac{5}{12} + \frac{13}{12} = 1.500
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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