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a+b=-36 ab=260
To solve the equation, factor T^{2}-36T+260 using formula T^{2}+\left(a+b\right)T+ab=\left(T+a\right)\left(T+b\right). To find a and b, set up a system to be solved.
-1,-260 -2,-130 -4,-65 -5,-52 -10,-26 -13,-20
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 260.
-1-260=-261 -2-130=-132 -4-65=-69 -5-52=-57 -10-26=-36 -13-20=-33
Calculate the sum for each pair.
a=-26 b=-10
The solution is the pair that gives sum -36.
\left(T-26\right)\left(T-10\right)
Rewrite factored expression \left(T+a\right)\left(T+b\right) using the obtained values.
T=26 T=10
To find equation solutions, solve T-26=0 and T-10=0.
a+b=-36 ab=1\times 260=260
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as T^{2}+aT+bT+260. To find a and b, set up a system to be solved.
-1,-260 -2,-130 -4,-65 -5,-52 -10,-26 -13,-20
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 260.
-1-260=-261 -2-130=-132 -4-65=-69 -5-52=-57 -10-26=-36 -13-20=-33
Calculate the sum for each pair.
a=-26 b=-10
The solution is the pair that gives sum -36.
\left(T^{2}-26T\right)+\left(-10T+260\right)
Rewrite T^{2}-36T+260 as \left(T^{2}-26T\right)+\left(-10T+260\right).
T\left(T-26\right)-10\left(T-26\right)
Factor out T in the first and -10 in the second group.
\left(T-26\right)\left(T-10\right)
Factor out common term T-26 by using distributive property.
T=26 T=10
To find equation solutions, solve T-26=0 and T-10=0.
T^{2}-36T+260=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.
T=\frac{-\left(-36\right)±\sqrt{\left(-36\right)^{2}-4\times 260}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -36 for b, and 260 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
T=\frac{-\left(-36\right)±\sqrt{1296-4\times 260}}{2}
Square -36.
T=\frac{-\left(-36\right)±\sqrt{1296-1040}}{2}
Multiply -4 times 260.
T=\frac{-\left(-36\right)±\sqrt{256}}{2}
Add 1296 to -1040.
T=\frac{-\left(-36\right)±16}{2}
Take the square root of 256.
T=\frac{36±16}{2}
The opposite of -36 is 36.
T=\frac{52}{2}
Now solve the equation T=\frac{36±16}{2} when ± is plus. Add 36 to 16.
T=26
Divide 52 by 2.
T=\frac{20}{2}
Now solve the equation T=\frac{36±16}{2} when ± is minus. Subtract 16 from 36.
T=10
Divide 20 by 2.
T=26 T=10
The equation is now solved.
T^{2}-36T+260=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.
T^{2}-36T+260-260=-260
Subtract 260 from both sides of the equation.
T^{2}-36T=-260
Subtracting 260 from itself leaves 0.
T^{2}-36T+\left(-18\right)^{2}=-260+\left(-18\right)^{2}
Divide -36, the coefficient of the x term, by 2 to get -18. Then add the square of -18 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
T^{2}-36T+324=-260+324
Square -18.
T^{2}-36T+324=64
Add -260 to 324.
\left(T-18\right)^{2}=64
Factor T^{2}-36T+324. 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(T-18\right)^{2}}=\sqrt{64}
Take the square root of both sides of the equation.
T-18=8 T-18=-8
Simplify.
T=26 T=10
Add 18 to both sides of the equation.
x ^ 2 -36x +260 = 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 = 36 rs = 260
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 = 18 - u s = 18 + u
Two numbers r and s sum up to 36 exactly when the average of the two numbers is \frac{1}{2}*36 = 18. 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>
(18 - u) (18 + u) = 260
To solve for unknown quantity u, substitute these in the product equation rs = 260
324 - u^2 = 260
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 260-324 = -64
Simplify the expression by subtracting 324 on both sides
u^2 = 64 u = \pm\sqrt{64} = \pm 8
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =18 - 8 = 10 s = 18 + 8 = 26
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.