Solve for t
t=\frac{1}{9}\approx 0.111111111
t=81
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a+b=-730 ab=9\times 81=729
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9t^{2}+at+bt+81. To find a and b, set up a system to be solved.
-1,-729 -3,-243 -9,-81 -27,-27
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 729.
-1-729=-730 -3-243=-246 -9-81=-90 -27-27=-54
Calculate the sum for each pair.
a=-729 b=-1
The solution is the pair that gives sum -730.
\left(9t^{2}-729t\right)+\left(-t+81\right)
Rewrite 9t^{2}-730t+81 as \left(9t^{2}-729t\right)+\left(-t+81\right).
9t\left(t-81\right)-\left(t-81\right)
Factor out 9t in the first and -1 in the second group.
\left(t-81\right)\left(9t-1\right)
Factor out common term t-81 by using distributive property.
t=81 t=\frac{1}{9}
To find equation solutions, solve t-81=0 and 9t-1=0.
9t^{2}-730t+81=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.
t=\frac{-\left(-730\right)±\sqrt{\left(-730\right)^{2}-4\times 9\times 81}}{2\times 9}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -730 for b, and 81 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-\left(-730\right)±\sqrt{532900-4\times 9\times 81}}{2\times 9}
Square -730.
t=\frac{-\left(-730\right)±\sqrt{532900-36\times 81}}{2\times 9}
Multiply -4 times 9.
t=\frac{-\left(-730\right)±\sqrt{532900-2916}}{2\times 9}
Multiply -36 times 81.
t=\frac{-\left(-730\right)±\sqrt{529984}}{2\times 9}
Add 532900 to -2916.
t=\frac{-\left(-730\right)±728}{2\times 9}
Take the square root of 529984.
t=\frac{730±728}{2\times 9}
The opposite of -730 is 730.
t=\frac{730±728}{18}
Multiply 2 times 9.
t=\frac{1458}{18}
Now solve the equation t=\frac{730±728}{18} when ± is plus. Add 730 to 728.
t=81
Divide 1458 by 18.
t=\frac{2}{18}
Now solve the equation t=\frac{730±728}{18} when ± is minus. Subtract 728 from 730.
t=\frac{1}{9}
Reduce the fraction \frac{2}{18} to lowest terms by extracting and canceling out 2.
t=81 t=\frac{1}{9}
The equation is now solved.
9t^{2}-730t+81=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.
9t^{2}-730t+81-81=-81
Subtract 81 from both sides of the equation.
9t^{2}-730t=-81
Subtracting 81 from itself leaves 0.
\frac{9t^{2}-730t}{9}=-\frac{81}{9}
Divide both sides by 9.
t^{2}-\frac{730}{9}t=-\frac{81}{9}
Dividing by 9 undoes the multiplication by 9.
t^{2}-\frac{730}{9}t=-9
Divide -81 by 9.
t^{2}-\frac{730}{9}t+\left(-\frac{365}{9}\right)^{2}=-9+\left(-\frac{365}{9}\right)^{2}
Divide -\frac{730}{9}, the coefficient of the x term, by 2 to get -\frac{365}{9}. Then add the square of -\frac{365}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}-\frac{730}{9}t+\frac{133225}{81}=-9+\frac{133225}{81}
Square -\frac{365}{9} by squaring both the numerator and the denominator of the fraction.
t^{2}-\frac{730}{9}t+\frac{133225}{81}=\frac{132496}{81}
Add -9 to \frac{133225}{81}.
\left(t-\frac{365}{9}\right)^{2}=\frac{132496}{81}
Factor t^{2}-\frac{730}{9}t+\frac{133225}{81}. 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(t-\frac{365}{9}\right)^{2}}=\sqrt{\frac{132496}{81}}
Take the square root of both sides of the equation.
t-\frac{365}{9}=\frac{364}{9} t-\frac{365}{9}=-\frac{364}{9}
Simplify.
t=81 t=\frac{1}{9}
Add \frac{365}{9} to both sides of the equation.
x ^ 2 -\frac{730}{9}x +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{730}{9} rs = 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{365}{9} - u s = \frac{365}{9} + u
Two numbers r and s sum up to \frac{730}{9} exactly when the average of the two numbers is \frac{1}{2}*\frac{730}{9} = \frac{365}{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{365}{9} - u) (\frac{365}{9} + u) = 9
To solve for unknown quantity u, substitute these in the product equation rs = 9
\frac{133225}{81} - u^2 = 9
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 9-\frac{133225}{81} = -\frac{132496}{81}
Simplify the expression by subtracting \frac{133225}{81} on both sides
u^2 = \frac{132496}{81} u = \pm\sqrt{\frac{132496}{81}} = \pm \frac{364}{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{365}{9} - \frac{364}{9} = 0.111 s = \frac{365}{9} + \frac{364}{9} = 81
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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