Solve for a
a = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333
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a+b=24 ab=9\times 16=144
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9a^{2}+aa+ba+16. To find a and b, set up a system to be solved.
1,144 2,72 3,48 4,36 6,24 8,18 9,16 12,12
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 144.
1+144=145 2+72=74 3+48=51 4+36=40 6+24=30 8+18=26 9+16=25 12+12=24
Calculate the sum for each pair.
a=12 b=12
The solution is the pair that gives sum 24.
\left(9a^{2}+12a\right)+\left(12a+16\right)
Rewrite 9a^{2}+24a+16 as \left(9a^{2}+12a\right)+\left(12a+16\right).
3a\left(3a+4\right)+4\left(3a+4\right)
Factor out 3a in the first and 4 in the second group.
\left(3a+4\right)\left(3a+4\right)
Factor out common term 3a+4 by using distributive property.
\left(3a+4\right)^{2}
Rewrite as a binomial square.
a=-\frac{4}{3}
To find equation solution, solve 3a+4=0.
9a^{2}+24a+16=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{-24±\sqrt{24^{2}-4\times 9\times 16}}{2\times 9}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 24 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-24±\sqrt{576-4\times 9\times 16}}{2\times 9}
Square 24.
a=\frac{-24±\sqrt{576-36\times 16}}{2\times 9}
Multiply -4 times 9.
a=\frac{-24±\sqrt{576-576}}{2\times 9}
Multiply -36 times 16.
a=\frac{-24±\sqrt{0}}{2\times 9}
Add 576 to -576.
a=-\frac{24}{2\times 9}
Take the square root of 0.
a=-\frac{24}{18}
Multiply 2 times 9.
a=-\frac{4}{3}
Reduce the fraction \frac{-24}{18} to lowest terms by extracting and canceling out 6.
9a^{2}+24a+16=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.
9a^{2}+24a+16-16=-16
Subtract 16 from both sides of the equation.
9a^{2}+24a=-16
Subtracting 16 from itself leaves 0.
\frac{9a^{2}+24a}{9}=-\frac{16}{9}
Divide both sides by 9.
a^{2}+\frac{24}{9}a=-\frac{16}{9}
Dividing by 9 undoes the multiplication by 9.
a^{2}+\frac{8}{3}a=-\frac{16}{9}
Reduce the fraction \frac{24}{9} to lowest terms by extracting and canceling out 3.
a^{2}+\frac{8}{3}a+\left(\frac{4}{3}\right)^{2}=-\frac{16}{9}+\left(\frac{4}{3}\right)^{2}
Divide \frac{8}{3}, the coefficient of the x term, by 2 to get \frac{4}{3}. Then add the square of \frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
a^{2}+\frac{8}{3}a+\frac{16}{9}=\frac{-16+16}{9}
Square \frac{4}{3} by squaring both the numerator and the denominator of the fraction.
a^{2}+\frac{8}{3}a+\frac{16}{9}=0
Add -\frac{16}{9} to \frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(a+\frac{4}{3}\right)^{2}=0
Factor a^{2}+\frac{8}{3}a+\frac{16}{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{4}{3}\right)^{2}}=\sqrt{0}
Take the square root of both sides of the equation.
a+\frac{4}{3}=0 a+\frac{4}{3}=0
Simplify.
a=-\frac{4}{3} a=-\frac{4}{3}
Subtract \frac{4}{3} from both sides of the equation.
a=-\frac{4}{3}
The equation is now solved. Solutions are the same.
x ^ 2 +\frac{8}{3}x +\frac{16}{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{8}{3} rs = \frac{16}{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{4}{3} - u s = -\frac{4}{3} + u
Two numbers r and s sum up to -\frac{8}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{8}{3} = -\frac{4}{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{4}{3} - u) (-\frac{4}{3} + u) = \frac{16}{9}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{16}{9}
\frac{16}{9} - u^2 = \frac{16}{9}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{16}{9}-\frac{16}{9} = 0
Simplify the expression by subtracting \frac{16}{9} on both sides
u^2 = 0 u = 0
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r = s = -\frac{4}{3} = -1.333
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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