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a+b=-16 ab=60
To solve the equation, factor a^{2}-16a+60 using formula a^{2}+\left(a+b\right)a+ab=\left(a+a\right)\left(a+b\right). To find a and b, set up a system to be solved.
-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 60.
-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16
Calculate the sum for each pair.
a=-10 b=-6
The solution is the pair that gives sum -16.
\left(a-10\right)\left(a-6\right)
Rewrite factored expression \left(a+a\right)\left(a+b\right) using the obtained values.
a=10 a=6
To find equation solutions, solve a-10=0 and a-6=0.
a+b=-16 ab=1\times 60=60
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as a^{2}+aa+ba+60. To find a and b, set up a system to be solved.
-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 60.
-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16
Calculate the sum for each pair.
a=-10 b=-6
The solution is the pair that gives sum -16.
\left(a^{2}-10a\right)+\left(-6a+60\right)
Rewrite a^{2}-16a+60 as \left(a^{2}-10a\right)+\left(-6a+60\right).
a\left(a-10\right)-6\left(a-10\right)
Factor out a in the first and -6 in the second group.
\left(a-10\right)\left(a-6\right)
Factor out common term a-10 by using distributive property.
a=10 a=6
To find equation solutions, solve a-10=0 and a-6=0.
a^{2}-16a+60=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.
a=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 60}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -16 for b, and 60 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-\left(-16\right)±\sqrt{256-4\times 60}}{2}
Square -16.
a=\frac{-\left(-16\right)±\sqrt{256-240}}{2}
Multiply -4 times 60.
a=\frac{-\left(-16\right)±\sqrt{16}}{2}
Add 256 to -240.
a=\frac{-\left(-16\right)±4}{2}
Take the square root of 16.
a=\frac{16±4}{2}
The opposite of -16 is 16.
a=\frac{20}{2}
Now solve the equation a=\frac{16±4}{2} when ± is plus. Add 16 to 4.
a=10
Divide 20 by 2.
a=\frac{12}{2}
Now solve the equation a=\frac{16±4}{2} when ± is minus. Subtract 4 from 16.
a=6
Divide 12 by 2.
a=10 a=6
The equation is now solved.
a^{2}-16a+60=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.
a^{2}-16a+60-60=-60
Subtract 60 from both sides of the equation.
a^{2}-16a=-60
Subtracting 60 from itself leaves 0.
a^{2}-16a+\left(-8\right)^{2}=-60+\left(-8\right)^{2}
Divide -16, the coefficient of the x term, by 2 to get -8. Then add the square of -8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
a^{2}-16a+64=-60+64
Square -8.
a^{2}-16a+64=4
Add -60 to 64.
\left(a-8\right)^{2}=4
Factor a^{2}-16a+64. 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(a-8\right)^{2}}=\sqrt{4}
Take the square root of both sides of the equation.
a-8=2 a-8=-2
Simplify.
a=10 a=6
Add 8 to both sides of the equation.
x ^ 2 -16x +60 = 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 = 16 rs = 60
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 = 8 - u s = 8 + u
Two numbers r and s sum up to 16 exactly when the average of the two numbers is \frac{1}{2}*16 = 8. 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>
(8 - u) (8 + u) = 60
To solve for unknown quantity u, substitute these in the product equation rs = 60
64 - u^2 = 60
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 60-64 = -4
Simplify the expression by subtracting 64 on both sides
u^2 = 4 u = \pm\sqrt{4} = \pm 2
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =8 - 2 = 6 s = 8 + 2 = 10
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.