Solve for x
x=-150
x=50
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a+b=100 ab=-7500
To solve the equation, factor x^{2}+100x-7500 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.
-1,7500 -2,3750 -3,2500 -4,1875 -5,1500 -6,1250 -10,750 -12,625 -15,500 -20,375 -25,300 -30,250 -50,150 -60,125 -75,100
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -7500.
-1+7500=7499 -2+3750=3748 -3+2500=2497 -4+1875=1871 -5+1500=1495 -6+1250=1244 -10+750=740 -12+625=613 -15+500=485 -20+375=355 -25+300=275 -30+250=220 -50+150=100 -60+125=65 -75+100=25
Calculate the sum for each pair.
a=-50 b=150
The solution is the pair that gives sum 100.
\left(x-50\right)\left(x+150\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=50 x=-150
To find equation solutions, solve x-50=0 and x+150=0.
a+b=100 ab=1\left(-7500\right)=-7500
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-7500. To find a and b, set up a system to be solved.
-1,7500 -2,3750 -3,2500 -4,1875 -5,1500 -6,1250 -10,750 -12,625 -15,500 -20,375 -25,300 -30,250 -50,150 -60,125 -75,100
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -7500.
-1+7500=7499 -2+3750=3748 -3+2500=2497 -4+1875=1871 -5+1500=1495 -6+1250=1244 -10+750=740 -12+625=613 -15+500=485 -20+375=355 -25+300=275 -30+250=220 -50+150=100 -60+125=65 -75+100=25
Calculate the sum for each pair.
a=-50 b=150
The solution is the pair that gives sum 100.
\left(x^{2}-50x\right)+\left(150x-7500\right)
Rewrite x^{2}+100x-7500 as \left(x^{2}-50x\right)+\left(150x-7500\right).
x\left(x-50\right)+150\left(x-50\right)
Factor out x in the first and 150 in the second group.
\left(x-50\right)\left(x+150\right)
Factor out common term x-50 by using distributive property.
x=50 x=-150
To find equation solutions, solve x-50=0 and x+150=0.
x^{2}+100x-7500=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{-100±\sqrt{100^{2}-4\left(-7500\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 100 for b, and -7500 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-100±\sqrt{10000-4\left(-7500\right)}}{2}
Square 100.
x=\frac{-100±\sqrt{10000+30000}}{2}
Multiply -4 times -7500.
x=\frac{-100±\sqrt{40000}}{2}
Add 10000 to 30000.
x=\frac{-100±200}{2}
Take the square root of 40000.
x=\frac{100}{2}
Now solve the equation x=\frac{-100±200}{2} when ± is plus. Add -100 to 200.
x=50
Divide 100 by 2.
x=-\frac{300}{2}
Now solve the equation x=\frac{-100±200}{2} when ± is minus. Subtract 200 from -100.
x=-150
Divide -300 by 2.
x=50 x=-150
The equation is now solved.
x^{2}+100x-7500=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.
x^{2}+100x-7500-\left(-7500\right)=-\left(-7500\right)
Add 7500 to both sides of the equation.
x^{2}+100x=-\left(-7500\right)
Subtracting -7500 from itself leaves 0.
x^{2}+100x=7500
Subtract -7500 from 0.
x^{2}+100x+50^{2}=7500+50^{2}
Divide 100, the coefficient of the x term, by 2 to get 50. Then add the square of 50 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+100x+2500=7500+2500
Square 50.
x^{2}+100x+2500=10000
Add 7500 to 2500.
\left(x+50\right)^{2}=10000
Factor x^{2}+100x+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+50\right)^{2}}=\sqrt{10000}
Take the square root of both sides of the equation.
x+50=100 x+50=-100
Simplify.
x=50 x=-150
Subtract 50 from both sides of the equation.
x ^ 2 +100x -7500 = 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 = -7500
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) = -7500
To solve for unknown quantity u, substitute these in the product equation rs = -7500
2500 - u^2 = -7500
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -7500-2500 = -10000
Simplify the expression by subtracting 2500 on both sides
u^2 = 10000 u = \pm\sqrt{10000} = \pm 100
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-50 - 100 = -150 s = -50 + 100 = 50
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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