Solve 1 over 36 = 6x−3. x = −5 x = negative 3 over 2 x = 7 over 2 x = 1
2 answers:
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X²-2x-3 = 0 let's start by grouping the "x"s
(x² - 2x + [?])² - 3 = 0
so, we have a missing fellow there, in order to make the group, a perfect square trinomial, namely to "complete the square", hmmm so the tell-tale fellow is the middle term.
from a perfect square trinomial we know that the middle term is a product of 2 and the "term on the left" and the "term on the right", like
aha!! so our missing fellow is 1.
now, let's keep in mind that all we're doing is borrowing from our very good friend Mr Zero, 0. So if we add 1², we also have to subtract 1².
(x² - 2x + 1² - 1²) - 3 = 0
(x² - 2x + 1) -1 -3 =0
(x - 1)² - 4 = 0
(x - 1)² = 4
(x² - 2x + [?])² - 3 = 0
so, we have a missing fellow there, in order to make the group, a perfect square trinomial, namely to "complete the square", hmmm so the tell-tale fellow is the middle term.
from a perfect square trinomial we know that the middle term is a product of 2 and the "term on the left" and the "term on the right", like
aha!! so our missing fellow is 1.
now, let's keep in mind that all we're doing is borrowing from our very good friend Mr Zero, 0. So if we add 1², we also have to subtract 1².
(x² - 2x + 1² - 1²) - 3 = 0
(x² - 2x + 1) -1 -3 =0
(x - 1)² - 4 = 0
(x - 1)² = 4