If x^2 - 12x + k = 28 and one root is 2, find the other root and the value of k.

x^2 - 12x + k = 28
x^2 - 12x + k - 28 = 0
x = (12 ± √(144 - 4(k - 28))) / 2
x = (12 ± √(144 - 4k + 112)) / 2
x = (12 ± √(256 - 4k)) / 2
x = (12 ± 2√(64 - k)) / 2
x = 6 ± √(64 - k)
As 2 is a root and 2 < 6, we need 6 - √(64 - k) = 2.
√(64 - k) = 4
64 - k = 16
k = 48
Then the other root is 6 + √(64 - 48) = 6 + √16 = 6 + 4 = 10.

That is the really long way to do it. You should use the style that I did, short and simple. No square roots involved at all.

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Let f(x) = x^2 - 12x + k - 28.

We know f(2) = 0.

Divide f(x) by (x-2):

2 | 1 -12 k-28
-------- 2 -20
---------------------------
1 -10 k-48

Since f(2) is a solution, k-48 = 0 -> k = 48.

We also know that x - 10 = 0

x = 10 is the other solution.

x = 2 or x = 10
k = 48

The other root is 10 and k is 48.

Say other root is y,
sor: -b/a=12=2+y so y=10!! (OTHER ROOT)

por:c/a=k-28=2y so, k=28+2(10)=48!!!

Short and simple.