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1 vote
Find the value of n such that x^2-19x n is a perfect square trinomial.

Answer choices:
A. -19/2
B. 361/4
C. 361
D. 361/2

2 Answers

3 votes
B. 361/4
you get: x^2-19x+361/4=(x-19/2)^2
User Johnnymire
by
8.3k points
6 votes

Answer: The correct option is (b)
(361)/(4).

Step-by-step explanation: We are given to select the correct value of 'n' such that
x^2-19x+n becomes a perfect square trinomial.

The standard form of a perfect square trinomial is


(x+a)^2=x^2+2a+a^2.

Now, we can write


x^2-19x+n\\\\=x^2-2* x* (19)/(2)+(361)/(4)+n-(361)/(4)\\\\\\=(x-(19)/(2))^2+n-(361)/(4).

So, for the given expression to be perfect trinomial,


n-(361)/(4)=0\\\\\Rightarrow n=(361)/(4).

Thus, (b) is the correct option.

User Carnez Davis
by
8.5k points

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