Asked by Elssy
y^2-7y+p. Find the value of p that it becomes a perfect square algebra
Answers
Answered by
Damon
(y-a)^2 = y^2 - 7 y + p
y^2 - 2 a + a^2 = y^2 - 7 y + p
2 a = 7
a = 7/2
so
p = a^2 = 49/4
y^2 - 2 a + a^2 = y^2 - 7 y + p
2 a = 7
a = 7/2
so
p = a^2 = 49/4
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