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p+q=10 pq=9\times 1=9
Factor the expression by grouping. First, the expression needs to be rewritten as 9a^{2}+pa+qa+1. To find p and q, set up a system to be solved.
1,9 3,3
Since pq is positive, p and q have the same sign. Since p+q is positive, p and q are both positive. List all such integer pairs that give product 9.
1+9=10 3+3=6
Calculate the sum for each pair.
p=1 q=9
The solution is the pair that gives sum 10.
\left(9a^{2}+a\right)+\left(9a+1\right)
Rewrite 9a^{2}+10a+1 as \left(9a^{2}+a\right)+\left(9a+1\right).
a\left(9a+1\right)+9a+1
Factor out a in 9a^{2}+a.
\left(9a+1\right)\left(a+1\right)
Factor out common term 9a+1 by using distributive property.
9a^{2}+10a+1=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
a=\frac{-10±\sqrt{10^{2}-4\times 9}}{2\times 9}
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{-10±\sqrt{100-4\times 9}}{2\times 9}
Square 10.
a=\frac{-10±\sqrt{100-36}}{2\times 9}
Multiply -4 times 9.
a=\frac{-10±\sqrt{64}}{2\times 9}
Add 100 to -36.
a=\frac{-10±8}{2\times 9}
Take the square root of 64.
a=\frac{-10±8}{18}
Multiply 2 times 9.
a=-\frac{2}{18}
Now solve the equation a=\frac{-10±8}{18} when ± is plus. Add -10 to 8.
a=-\frac{1}{9}
Reduce the fraction \frac{-2}{18} to lowest terms by extracting and canceling out 2.
a=-\frac{18}{18}
Now solve the equation a=\frac{-10±8}{18} when ± is minus. Subtract 8 from -10.
a=-1
Divide -18 by 18.
9a^{2}+10a+1=9\left(a-\left(-\frac{1}{9}\right)\right)\left(a-\left(-1\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{9} for x_{1} and -1 for x_{2}.
9a^{2}+10a+1=9\left(a+\frac{1}{9}\right)\left(a+1\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
9a^{2}+10a+1=9\times \frac{9a+1}{9}\left(a+1\right)
Add \frac{1}{9} to a by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
9a^{2}+10a+1=\left(9a+1\right)\left(a+1\right)
Cancel out 9, the greatest common factor in 9 and 9.
x ^ 2 +\frac{10}{9}x +\frac{1}{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{10}{9} rs = \frac{1}{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{5}{9} - u s = -\frac{5}{9} + u
Two numbers r and s sum up to -\frac{10}{9} exactly when the average of the two numbers is \frac{1}{2}*-\frac{10}{9} = -\frac{5}{9}. 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}{9} - u) (-\frac{5}{9} + u) = \frac{1}{9}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{9}
\frac{25}{81} - u^2 = \frac{1}{9}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{9}-\frac{25}{81} = -\frac{16}{81}
Simplify the expression by subtracting \frac{25}{81} on both sides
u^2 = \frac{16}{81} u = \pm\sqrt{\frac{16}{81}} = \pm \frac{4}{9}
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}{9} - \frac{4}{9} = -1 s = -\frac{5}{9} + \frac{4}{9} = -0.111
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.