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a+b=27 ab=25\times 2=50
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 25u^{2}+au+bu+2. To find a and b, set up a system to be solved.
1,50 2,25 5,10
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 50.
1+50=51 2+25=27 5+10=15
Calculate the sum for each pair.
a=2 b=25
The solution is the pair that gives sum 27.
\left(25u^{2}+2u\right)+\left(25u+2\right)
Rewrite 25u^{2}+27u+2 as \left(25u^{2}+2u\right)+\left(25u+2\right).
u\left(25u+2\right)+25u+2
Factor out u in 25u^{2}+2u.
\left(25u+2\right)\left(u+1\right)
Factor out common term 25u+2 by using distributive property.
u=-\frac{2}{25} u=-1
To find equation solutions, solve 25u+2=0 and u+1=0.
25u^{2}+27u+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.
u=\frac{-27±\sqrt{27^{2}-4\times 25\times 2}}{2\times 25}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, 27 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
u=\frac{-27±\sqrt{729-4\times 25\times 2}}{2\times 25}
Square 27.
u=\frac{-27±\sqrt{729-100\times 2}}{2\times 25}
Multiply -4 times 25.
u=\frac{-27±\sqrt{729-200}}{2\times 25}
Multiply -100 times 2.
u=\frac{-27±\sqrt{529}}{2\times 25}
Add 729 to -200.
u=\frac{-27±23}{2\times 25}
Take the square root of 529.
u=\frac{-27±23}{50}
Multiply 2 times 25.
u=-\frac{4}{50}
Now solve the equation u=\frac{-27±23}{50} when ± is plus. Add -27 to 23.
u=-\frac{2}{25}
Reduce the fraction \frac{-4}{50} to lowest terms by extracting and canceling out 2.
u=-\frac{50}{50}
Now solve the equation u=\frac{-27±23}{50} when ± is minus. Subtract 23 from -27.
u=-1
Divide -50 by 50.
u=-\frac{2}{25} u=-1
The equation is now solved.
25u^{2}+27u+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.
25u^{2}+27u+2-2=-2
Subtract 2 from both sides of the equation.
25u^{2}+27u=-2
Subtracting 2 from itself leaves 0.
\frac{25u^{2}+27u}{25}=-\frac{2}{25}
Divide both sides by 25.
u^{2}+\frac{27}{25}u=-\frac{2}{25}
Dividing by 25 undoes the multiplication by 25.
u^{2}+\frac{27}{25}u+\left(\frac{27}{50}\right)^{2}=-\frac{2}{25}+\left(\frac{27}{50}\right)^{2}
Divide \frac{27}{25}, the coefficient of the x term, by 2 to get \frac{27}{50}. Then add the square of \frac{27}{50} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
u^{2}+\frac{27}{25}u+\frac{729}{2500}=-\frac{2}{25}+\frac{729}{2500}
Square \frac{27}{50} by squaring both the numerator and the denominator of the fraction.
u^{2}+\frac{27}{25}u+\frac{729}{2500}=\frac{529}{2500}
Add -\frac{2}{25} to \frac{729}{2500} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(u+\frac{27}{50}\right)^{2}=\frac{529}{2500}
Factor u^{2}+\frac{27}{25}u+\frac{729}{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(u+\frac{27}{50}\right)^{2}}=\sqrt{\frac{529}{2500}}
Take the square root of both sides of the equation.
u+\frac{27}{50}=\frac{23}{50} u+\frac{27}{50}=-\frac{23}{50}
Simplify.
u=-\frac{2}{25} u=-1
Subtract \frac{27}{50} from both sides of the equation.
x ^ 2 +\frac{27}{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{27}{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{27}{50} - u s = -\frac{27}{50} + u
Two numbers r and s sum up to -\frac{27}{25} exactly when the average of the two numbers is \frac{1}{2}*-\frac{27}{25} = -\frac{27}{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{27}{50} - u) (-\frac{27}{50} + u) = \frac{2}{25}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{25}
\frac{729}{2500} - u^2 = \frac{2}{25}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{2}{25}-\frac{729}{2500} = -\frac{529}{2500}
Simplify the expression by subtracting \frac{729}{2500} on both sides
u^2 = \frac{529}{2500} u = \pm\sqrt{\frac{529}{2500}} = \pm \frac{23}{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{27}{50} - \frac{23}{50} = -1 s = -\frac{27}{50} + \frac{23}{50} = -0.080
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.