Solve for x
x=-75
x=55
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a+b=20 ab=-4125
To solve the equation, factor x^{2}+20x-4125 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.
-1,4125 -3,1375 -5,825 -11,375 -15,275 -25,165 -33,125 -55,75
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 -4125.
-1+4125=4124 -3+1375=1372 -5+825=820 -11+375=364 -15+275=260 -25+165=140 -33+125=92 -55+75=20
Calculate the sum for each pair.
a=-55 b=75
The solution is the pair that gives sum 20.
\left(x-55\right)\left(x+75\right)
Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.
x=55 x=-75
To find equation solutions, solve x-55=0 and x+75=0.
a+b=20 ab=1\left(-4125\right)=-4125
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-4125. To find a and b, set up a system to be solved.
-1,4125 -3,1375 -5,825 -11,375 -15,275 -25,165 -33,125 -55,75
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 -4125.
-1+4125=4124 -3+1375=1372 -5+825=820 -11+375=364 -15+275=260 -25+165=140 -33+125=92 -55+75=20
Calculate the sum for each pair.
a=-55 b=75
The solution is the pair that gives sum 20.
\left(x^{2}-55x\right)+\left(75x-4125\right)
Rewrite x^{2}+20x-4125 as \left(x^{2}-55x\right)+\left(75x-4125\right).
x\left(x-55\right)+75\left(x-55\right)
Factor out x in the first and 75 in the second group.
\left(x-55\right)\left(x+75\right)
Factor out common term x-55 by using distributive property.
x=55 x=-75
To find equation solutions, solve x-55=0 and x+75=0.
x^{2}+20x-4125=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{-20±\sqrt{20^{2}-4\left(-4125\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 20 for b, and -4125 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-20±\sqrt{400-4\left(-4125\right)}}{2}
Square 20.
x=\frac{-20±\sqrt{400+16500}}{2}
Multiply -4 times -4125.
x=\frac{-20±\sqrt{16900}}{2}
Add 400 to 16500.
x=\frac{-20±130}{2}
Take the square root of 16900.
x=\frac{110}{2}
Now solve the equation x=\frac{-20±130}{2} when ± is plus. Add -20 to 130.
x=55
Divide 110 by 2.
x=-\frac{150}{2}
Now solve the equation x=\frac{-20±130}{2} when ± is minus. Subtract 130 from -20.
x=-75
Divide -150 by 2.
x=55 x=-75
The equation is now solved.
x^{2}+20x-4125=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}+20x-4125-\left(-4125\right)=-\left(-4125\right)
Add 4125 to both sides of the equation.
x^{2}+20x=-\left(-4125\right)
Subtracting -4125 from itself leaves 0.
x^{2}+20x=4125
Subtract -4125 from 0.
x^{2}+20x+10^{2}=4125+10^{2}
Divide 20, the coefficient of the x term, by 2 to get 10. Then add the square of 10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+20x+100=4125+100
Square 10.
x^{2}+20x+100=4225
Add 4125 to 100.
\left(x+10\right)^{2}=4225
Factor x^{2}+20x+100. 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+10\right)^{2}}=\sqrt{4225}
Take the square root of both sides of the equation.
x+10=65 x+10=-65
Simplify.
x=55 x=-75
Subtract 10 from both sides of the equation.
x ^ 2 +20x -4125 = 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 = -20 rs = -4125
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 = -10 - u s = -10 + u
Two numbers r and s sum up to -20 exactly when the average of the two numbers is \frac{1}{2}*-20 = -10. 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>
(-10 - u) (-10 + u) = -4125
To solve for unknown quantity u, substitute these in the product equation rs = -4125
100 - u^2 = -4125
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -4125-100 = -4225
Simplify the expression by subtracting 100 on both sides
u^2 = 4225 u = \pm\sqrt{4225} = \pm 65
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-10 - 65 = -75 s = -10 + 65 = 55
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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