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a+b=100 ab=1\times 2500=2500
Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+2500. To find a and b, set up a system to be solved.
1,2500 2,1250 4,625 5,500 10,250 20,125 25,100 50,50
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 2500.
1+2500=2501 2+1250=1252 4+625=629 5+500=505 10+250=260 20+125=145 25+100=125 50+50=100
Calculate the sum for each pair.
a=50 b=50
The solution is the pair that gives sum 100.
\left(x^{2}+50x\right)+\left(50x+2500\right)
Rewrite x^{2}+100x+2500 as \left(x^{2}+50x\right)+\left(50x+2500\right).
x\left(x+50\right)+50\left(x+50\right)
Factor out x in the first and 50 in the second group.
\left(x+50\right)\left(x+50\right)
Factor out common term x+50 by using distributive property.
\left(x+50\right)^{2}
Rewrite as a binomial square.
factor(x^{2}+100x+2500)
This trinomial has the form of a trinomial square, perhaps multiplied by a common factor. Trinomial squares can be factored by finding the square roots of the leading and trailing terms.
\sqrt{2500}=50
Find the square root of the trailing term, 2500.
\left(x+50\right)^{2}
The trinomial square is the square of the binomial that is the sum or difference of the square roots of the leading and trailing terms, with the sign determined by the sign of the middle term of the trinomial square.
x^{2}+100x+2500=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-100±\sqrt{100^{2}-4\times 2500}}{2}
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{-100±\sqrt{10000-4\times 2500}}{2}
Square 100.
x=\frac{-100±\sqrt{10000-10000}}{2}
Multiply -4 times 2500.
x=\frac{-100±\sqrt{0}}{2}
Add 10000 to -10000.
x=\frac{-100±0}{2}
Take the square root of 0.
x^{2}+100x+2500=\left(x-\left(-50\right)\right)\left(x-\left(-50\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -50 for x_{1} and -50 for x_{2}.
x^{2}+100x+2500=\left(x+50\right)\left(x+50\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
x ^ 2 +100x +2500 = 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.
r + s = -100 rs = 2500
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 = -50 - u s = -50 + u
Two numbers r and s sum up to -100 exactly when the average of the two numbers is \frac{1}{2}*-100 = -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>
(-50 - u) (-50 + u) = 2500
To solve for unknown quantity u, substitute these in the product equation rs = 2500
2500 - u^2 = 2500
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 2500-2500 = 0
Simplify the expression by subtracting 2500 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 = -50
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.