Solve for n
n=-413
n=12
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a+b=401 ab=-4956
To solve the equation, factor n^{2}+401n-4956 using formula n^{2}+\left(a+b\right)n+ab=\left(n+a\right)\left(n+b\right). To find a and b, set up a system to be solved.
-1,4956 -2,2478 -3,1652 -4,1239 -6,826 -7,708 -12,413 -14,354 -21,236 -28,177 -42,118 -59,84
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 -4956.
-1+4956=4955 -2+2478=2476 -3+1652=1649 -4+1239=1235 -6+826=820 -7+708=701 -12+413=401 -14+354=340 -21+236=215 -28+177=149 -42+118=76 -59+84=25
Calculate the sum for each pair.
a=-12 b=413
The solution is the pair that gives sum 401.
\left(n-12\right)\left(n+413\right)
Rewrite factored expression \left(n+a\right)\left(n+b\right) using the obtained values.
n=12 n=-413
To find equation solutions, solve n-12=0 and n+413=0.
a+b=401 ab=1\left(-4956\right)=-4956
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn-4956. To find a and b, set up a system to be solved.
-1,4956 -2,2478 -3,1652 -4,1239 -6,826 -7,708 -12,413 -14,354 -21,236 -28,177 -42,118 -59,84
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 -4956.
-1+4956=4955 -2+2478=2476 -3+1652=1649 -4+1239=1235 -6+826=820 -7+708=701 -12+413=401 -14+354=340 -21+236=215 -28+177=149 -42+118=76 -59+84=25
Calculate the sum for each pair.
a=-12 b=413
The solution is the pair that gives sum 401.
\left(n^{2}-12n\right)+\left(413n-4956\right)
Rewrite n^{2}+401n-4956 as \left(n^{2}-12n\right)+\left(413n-4956\right).
n\left(n-12\right)+413\left(n-12\right)
Factor out n in the first and 413 in the second group.
\left(n-12\right)\left(n+413\right)
Factor out common term n-12 by using distributive property.
n=12 n=-413
To find equation solutions, solve n-12=0 and n+413=0.
n^{2}+401n-4956=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.
n=\frac{-401±\sqrt{401^{2}-4\left(-4956\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 401 for b, and -4956 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
n=\frac{-401±\sqrt{160801-4\left(-4956\right)}}{2}
Square 401.
n=\frac{-401±\sqrt{160801+19824}}{2}
Multiply -4 times -4956.
n=\frac{-401±\sqrt{180625}}{2}
Add 160801 to 19824.
n=\frac{-401±425}{2}
Take the square root of 180625.
n=\frac{24}{2}
Now solve the equation n=\frac{-401±425}{2} when ± is plus. Add -401 to 425.
n=12
Divide 24 by 2.
n=-\frac{826}{2}
Now solve the equation n=\frac{-401±425}{2} when ± is minus. Subtract 425 from -401.
n=-413
Divide -826 by 2.
n=12 n=-413
The equation is now solved.
n^{2}+401n-4956=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.
n^{2}+401n-4956-\left(-4956\right)=-\left(-4956\right)
Add 4956 to both sides of the equation.
n^{2}+401n=-\left(-4956\right)
Subtracting -4956 from itself leaves 0.
n^{2}+401n=4956
Subtract -4956 from 0.
n^{2}+401n+\left(\frac{401}{2}\right)^{2}=4956+\left(\frac{401}{2}\right)^{2}
Divide 401, the coefficient of the x term, by 2 to get \frac{401}{2}. Then add the square of \frac{401}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
n^{2}+401n+\frac{160801}{4}=4956+\frac{160801}{4}
Square \frac{401}{2} by squaring both the numerator and the denominator of the fraction.
n^{2}+401n+\frac{160801}{4}=\frac{180625}{4}
Add 4956 to \frac{160801}{4}.
\left(n+\frac{401}{2}\right)^{2}=\frac{180625}{4}
Factor n^{2}+401n+\frac{160801}{4}. 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(n+\frac{401}{2}\right)^{2}}=\sqrt{\frac{180625}{4}}
Take the square root of both sides of the equation.
n+\frac{401}{2}=\frac{425}{2} n+\frac{401}{2}=-\frac{425}{2}
Simplify.
n=12 n=-413
Subtract \frac{401}{2} from both sides of the equation.
x ^ 2 +401x -4956 = 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 = -401 rs = -4956
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{401}{2} - u s = -\frac{401}{2} + u
Two numbers r and s sum up to -401 exactly when the average of the two numbers is \frac{1}{2}*-401 = -\frac{401}{2}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(-\frac{401}{2} - u) (-\frac{401}{2} + u) = -4956
To solve for unknown quantity u, substitute these in the product equation rs = -4956
\frac{160801}{4} - u^2 = -4956
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -4956-\frac{160801}{4} = -\frac{180625}{4}
Simplify the expression by subtracting \frac{160801}{4} on both sides
u^2 = \frac{180625}{4} u = \pm\sqrt{\frac{180625}{4}} = \pm \frac{425}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{401}{2} - \frac{425}{2} = -413 s = -\frac{401}{2} + \frac{425}{2} = 12
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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