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81x^{2}+90x+25=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.
x=\frac{-90±\sqrt{90^{2}-4\times 81\times 25}}{2\times 81}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 81 for a, 90 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-90±\sqrt{8100-4\times 81\times 25}}{2\times 81}
Square 90.
x=\frac{-90±\sqrt{8100-324\times 25}}{2\times 81}
Multiply -4 times 81.
x=\frac{-90±\sqrt{8100-8100}}{2\times 81}
Multiply -324 times 25.
x=\frac{-90±\sqrt{0}}{2\times 81}
Add 8100 to -8100.
x=-\frac{90}{2\times 81}
Take the square root of 0.
x=-\frac{90}{162}
Multiply 2 times 81.
x=-\frac{5}{9}
Reduce the fraction \frac{-90}{162} to lowest terms by extracting and canceling out 18.
81x^{2}+90x+25=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.
81x^{2}+90x+25-25=-25
Subtract 25 from both sides of the equation.
81x^{2}+90x=-25
Subtracting 25 from itself leaves 0.
\frac{81x^{2}+90x}{81}=-\frac{25}{81}
Divide both sides by 81.
x^{2}+\frac{90}{81}x=-\frac{25}{81}
Dividing by 81 undoes the multiplication by 81.
x^{2}+\frac{10}{9}x=-\frac{25}{81}
Reduce the fraction \frac{90}{81} to lowest terms by extracting and canceling out 9.
x^{2}+\frac{10}{9}x+\left(\frac{5}{9}\right)^{2}=-\frac{25}{81}+\left(\frac{5}{9}\right)^{2}
Divide \frac{10}{9}, the coefficient of the x term, by 2 to get \frac{5}{9}. Then add the square of \frac{5}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{10}{9}x+\frac{25}{81}=\frac{-25+25}{81}
Square \frac{5}{9} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{10}{9}x+\frac{25}{81}=0
Add -\frac{25}{81} to \frac{25}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{5}{9}\right)^{2}=0
Factor x^{2}+\frac{10}{9}x+\frac{25}{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(x+\frac{5}{9}\right)^{2}}=\sqrt{0}
Take the square root of both sides of the equation.
x+\frac{5}{9}=0 x+\frac{5}{9}=0
Simplify.
x=-\frac{5}{9} x=-\frac{5}{9}
Subtract \frac{5}{9} from both sides of the equation.
x=-\frac{5}{9}
The equation is now solved. Solutions are the same.
x ^ 2 +\frac{10}{9}x +\frac{25}{81} = 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 81
r + s = -\frac{10}{9} rs = \frac{25}{81}
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{25}{81}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{25}{81}
\frac{25}{81} - u^2 = \frac{25}{81}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{25}{81}-\frac{25}{81} = 0
Simplify the expression by subtracting \frac{25}{81} 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{5}{9} = -0.556
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.