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a+b=-23 ab=25\left(-2\right)=-50
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 25x^{2}+ax+bx-2. To find a and b, set up a system to be solved.
1,-50 2,-25 5,-10
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -50.
1-50=-49 2-25=-23 5-10=-5
Calculate the sum for each pair.
a=-25 b=2
The solution is the pair that gives sum -23.
\left(25x^{2}-25x\right)+\left(2x-2\right)
Rewrite 25x^{2}-23x-2 as \left(25x^{2}-25x\right)+\left(2x-2\right).
25x\left(x-1\right)+2\left(x-1\right)
Factor out 25x in the first and 2 in the second group.
\left(x-1\right)\left(25x+2\right)
Factor out common term x-1 by using distributive property.
x=1 x=-\frac{2}{25}
To find equation solutions, solve x-1=0 and 25x+2=0.
25x^{2}-23x-2=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(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 25\left(-2\right)}}{2\times 25}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, -23 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-23\right)±\sqrt{529-4\times 25\left(-2\right)}}{2\times 25}
Square -23.
x=\frac{-\left(-23\right)±\sqrt{529-100\left(-2\right)}}{2\times 25}
Multiply -4 times 25.
x=\frac{-\left(-23\right)±\sqrt{529+200}}{2\times 25}
Multiply -100 times -2.
x=\frac{-\left(-23\right)±\sqrt{729}}{2\times 25}
Add 529 to 200.
x=\frac{-\left(-23\right)±27}{2\times 25}
Take the square root of 729.
x=\frac{23±27}{2\times 25}
The opposite of -23 is 23.
x=\frac{23±27}{50}
Multiply 2 times 25.
x=\frac{50}{50}
Now solve the equation x=\frac{23±27}{50} when ± is plus. Add 23 to 27.
x=1
Divide 50 by 50.
x=-\frac{4}{50}
Now solve the equation x=\frac{23±27}{50} when ± is minus. Subtract 27 from 23.
x=-\frac{2}{25}
Reduce the fraction \frac{-4}{50} to lowest terms by extracting and canceling out 2.
x=1 x=-\frac{2}{25}
The equation is now solved.
25x^{2}-23x-2=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.
25x^{2}-23x-2-\left(-2\right)=-\left(-2\right)
Add 2 to both sides of the equation.
25x^{2}-23x=-\left(-2\right)
Subtracting -2 from itself leaves 0.
25x^{2}-23x=2
Subtract -2 from 0.
\frac{25x^{2}-23x}{25}=\frac{2}{25}
Divide both sides by 25.
x^{2}-\frac{23}{25}x=\frac{2}{25}
Dividing by 25 undoes the multiplication by 25.
x^{2}-\frac{23}{25}x+\left(-\frac{23}{50}\right)^{2}=\frac{2}{25}+\left(-\frac{23}{50}\right)^{2}
Divide -\frac{23}{25}, the coefficient of the x term, by 2 to get -\frac{23}{50}. Then add the square of -\frac{23}{50} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{23}{25}x+\frac{529}{2500}=\frac{2}{25}+\frac{529}{2500}
Square -\frac{23}{50} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{23}{25}x+\frac{529}{2500}=\frac{729}{2500}
Add \frac{2}{25} to \frac{529}{2500} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{23}{50}\right)^{2}=\frac{729}{2500}
Factor x^{2}-\frac{23}{25}x+\frac{529}{2500}. 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{23}{50}\right)^{2}}=\sqrt{\frac{729}{2500}}
Take the square root of both sides of the equation.
x-\frac{23}{50}=\frac{27}{50} x-\frac{23}{50}=-\frac{27}{50}
Simplify.
x=1 x=-\frac{2}{25}
Add \frac{23}{50} to both sides of the equation.
x ^ 2 -\frac{23}{25}x -\frac{2}{25} = 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 25
r + s = \frac{23}{25} rs = -\frac{2}{25}
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{23}{50} - u s = \frac{23}{50} + u
Two numbers r and s sum up to \frac{23}{25} exactly when the average of the two numbers is \frac{1}{2}*\frac{23}{25} = \frac{23}{50}. 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{23}{50} - u) (\frac{23}{50} + u) = -\frac{2}{25}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{25}
\frac{529}{2500} - u^2 = -\frac{2}{25}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{2}{25}-\frac{529}{2500} = -\frac{729}{2500}
Simplify the expression by subtracting \frac{529}{2500} on both sides
u^2 = \frac{729}{2500} u = \pm\sqrt{\frac{729}{2500}} = \pm \frac{27}{50}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{23}{50} - \frac{27}{50} = -0.080 s = \frac{23}{50} + \frac{27}{50} = 1
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.