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a+b=-30 ab=9\times 25=225
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx+25. To find a and b, set up a system to be solved.
-1,-225 -3,-75 -5,-45 -9,-25 -15,-15
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 225.
-1-225=-226 -3-75=-78 -5-45=-50 -9-25=-34 -15-15=-30
Calculate the sum for each pair.
a=-15 b=-15
The solution is the pair that gives sum -30.
\left(9x^{2}-15x\right)+\left(-15x+25\right)
Rewrite 9x^{2}-30x+25 as \left(9x^{2}-15x\right)+\left(-15x+25\right).
3x\left(3x-5\right)-5\left(3x-5\right)
Factor out 3x in the first and -5 in the second group.
\left(3x-5\right)\left(3x-5\right)
Factor out common term 3x-5 by using distributive property.
\left(3x-5\right)^{2}
Rewrite as a binomial square.
x=\frac{5}{3}
To find equation solution, solve 3x-5=0.
9x^{2}-30x+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{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 9\times 25}}{2\times 9}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -30 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-30\right)±\sqrt{900-4\times 9\times 25}}{2\times 9}
Square -30.
x=\frac{-\left(-30\right)±\sqrt{900-36\times 25}}{2\times 9}
Multiply -4 times 9.
x=\frac{-\left(-30\right)±\sqrt{900-900}}{2\times 9}
Multiply -36 times 25.
x=\frac{-\left(-30\right)±\sqrt{0}}{2\times 9}
x=-\frac{-30}{2\times 9}
Take the square root of 0.
x=\frac{30}{2\times 9}
The opposite of -30 is 30.
x=\frac{30}{18}
Multiply 2 times 9.
x=\frac{5}{3}
Reduce the fraction \frac{30}{18}\approx 1.666666667 to lowest terms by extracting and canceling out 6.
9x^{2}-30x+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.
9x^{2}-30x+25-25=-25
Subtract 25 from both sides of the equation.
9x^{2}-30x=-25
Subtracting 25 from itself leaves 0.
\frac{9x^{2}-30x}{9}=\frac{-25}{9}
Divide both sides by 9.
x^{2}+\frac{-30}{9}x=\frac{-25}{9}
Dividing by 9 undoes the multiplication by 9.
x^{2}-\frac{10}{3}x=\frac{-25}{9}
Reduce the fraction \frac{-30}{9}\approx -3.333333333 to lowest terms by extracting and canceling out 3.
x^{2}-\frac{10}{3}x=-\frac{25}{9}
Divide -25 by 9.
x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=-\frac{25}{9}+\left(-\frac{5}{3}\right)^{2}
Divide -\frac{10}{3}\approx -3.333333333, the coefficient of the x term, by 2 to get -\frac{5}{3}\approx -1.666666667. Then add the square of -\frac{5}{3}\approx -1.666666667 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{-25+25}{9}
Square -\frac{5}{3}\approx -1.666666667 by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{10}{3}x+\frac{25}{9}=0
Add -\frac{25}{9}\approx -2.777777778 to \frac{25}{9}\approx 2.777777778 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{5}{3}\right)^{2}=0
Factor x^{2}-\frac{10}{3}x+\frac{25}{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(x-\frac{5}{3}\right)^{2}}=\sqrt{0}
Take the square root of both sides of the equation.
x-\frac{5}{3}=0 x-\frac{5}{3}=0
Simplify.
x=\frac{5}{3} x=\frac{5}{3}
Add \frac{5}{3}\approx 1.666666667 to both sides of the equation.
x=\frac{5}{3}
The equation is now solved. Solutions are the same.
x ^ 2 -\frac{10}{3}x +\frac{25}{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}{3} rs = \frac{25}{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}{3} - u s = \frac{5}{3} + u
Two numbers r and s sum up to \frac{10}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{10}{3} = \frac{5}{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{5}{3} - u) (\frac{5}{3} + u) = \frac{25}{9}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{25}{9}
\frac{25}{9} - u^2 = \frac{25}{9}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{25}{9}-\frac{25}{9} = 0
Simplify the expression by subtracting \frac{25}{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{5}{3} = 1.667
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.