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a+b=-8 ab=12\left(-15\right)=-180
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 12a^{2}+aa+ba-15. To find a and b, set up a system to be solved.
1,-180 2,-90 3,-60 4,-45 5,-36 6,-30 9,-20 10,-18 12,-15
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 -180.
1-180=-179 2-90=-88 3-60=-57 4-45=-41 5-36=-31 6-30=-24 9-20=-11 10-18=-8 12-15=-3
Calculate the sum for each pair.
a=-18 b=10
The solution is the pair that gives sum -8.
\left(12a^{2}-18a\right)+\left(10a-15\right)
Rewrite 12a^{2}-8a-15 as \left(12a^{2}-18a\right)+\left(10a-15\right).
6a\left(2a-3\right)+5\left(2a-3\right)
Factor out 6a in the first and 5 in the second group.
\left(2a-3\right)\left(6a+5\right)
Factor out common term 2a-3 by using distributive property.
a=\frac{3}{2} a=-\frac{5}{6}
To find equation solutions, solve 2a-3=0 and 6a+5=0.
12a^{2}-8a-15=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(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 12\left(-15\right)}}{2\times 12}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -8 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-\left(-8\right)±\sqrt{64-4\times 12\left(-15\right)}}{2\times 12}
Square -8.
a=\frac{-\left(-8\right)±\sqrt{64-48\left(-15\right)}}{2\times 12}
Multiply -4 times 12.
a=\frac{-\left(-8\right)±\sqrt{64+720}}{2\times 12}
Multiply -48 times -15.
a=\frac{-\left(-8\right)±\sqrt{784}}{2\times 12}
Add 64 to 720.
a=\frac{-\left(-8\right)±28}{2\times 12}
Take the square root of 784.
a=\frac{8±28}{2\times 12}
The opposite of -8 is 8.
a=\frac{8±28}{24}
Multiply 2 times 12.
a=\frac{36}{24}
Now solve the equation a=\frac{8±28}{24} when ± is plus. Add 8 to 28.
a=\frac{3}{2}
Reduce the fraction \frac{36}{24} to lowest terms by extracting and canceling out 12.
a=-\frac{20}{24}
Now solve the equation a=\frac{8±28}{24} when ± is minus. Subtract 28 from 8.
a=-\frac{5}{6}
Reduce the fraction \frac{-20}{24} to lowest terms by extracting and canceling out 4.
a=\frac{3}{2} a=-\frac{5}{6}
The equation is now solved.
12a^{2}-8a-15=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.
12a^{2}-8a-15-\left(-15\right)=-\left(-15\right)
Add 15 to both sides of the equation.
12a^{2}-8a=-\left(-15\right)
Subtracting -15 from itself leaves 0.
12a^{2}-8a=15
Subtract -15 from 0.
\frac{12a^{2}-8a}{12}=\frac{15}{12}
Divide both sides by 12.
a^{2}+\left(-\frac{8}{12}\right)a=\frac{15}{12}
Dividing by 12 undoes the multiplication by 12.
a^{2}-\frac{2}{3}a=\frac{15}{12}
Reduce the fraction \frac{-8}{12} to lowest terms by extracting and canceling out 4.
a^{2}-\frac{2}{3}a=\frac{5}{4}
Reduce the fraction \frac{15}{12} to lowest terms by extracting and canceling out 3.
a^{2}-\frac{2}{3}a+\left(-\frac{1}{3}\right)^{2}=\frac{5}{4}+\left(-\frac{1}{3}\right)^{2}
Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
a^{2}-\frac{2}{3}a+\frac{1}{9}=\frac{5}{4}+\frac{1}{9}
Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.
a^{2}-\frac{2}{3}a+\frac{1}{9}=\frac{49}{36}
Add \frac{5}{4} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(a-\frac{1}{3}\right)^{2}=\frac{49}{36}
Factor a^{2}-\frac{2}{3}a+\frac{1}{9}. 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{1}{3}\right)^{2}}=\sqrt{\frac{49}{36}}
Take the square root of both sides of the equation.
a-\frac{1}{3}=\frac{7}{6} a-\frac{1}{3}=-\frac{7}{6}
Simplify.
a=\frac{3}{2} a=-\frac{5}{6}
Add \frac{1}{3} to both sides of the equation.
x ^ 2 -\frac{2}{3}x -\frac{5}{4} = 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 12
r + s = \frac{2}{3} rs = -\frac{5}{4}
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}{3} - u s = \frac{1}{3} + u
Two numbers r and s sum up to \frac{2}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{2}{3} = \frac{1}{3}. 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}{3} - u) (\frac{1}{3} + u) = -\frac{5}{4}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{4}
\frac{1}{9} - u^2 = -\frac{5}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{5}{4}-\frac{1}{9} = -\frac{49}{36}
Simplify the expression by subtracting \frac{1}{9} on both sides
u^2 = \frac{49}{36} u = \pm\sqrt{\frac{49}{36}} = \pm \frac{7}{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{1}{3} - \frac{7}{6} = -0.833 s = \frac{1}{3} + \frac{7}{6} = 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.