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Kamila [148]
3 years ago
12

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

Mathematics
2 answers:
Marianna [84]3 years ago
6 0
B. 361/4
you get: x^2-19x+361/4=(x-19/2)^2
Alex17521 [72]3 years ago
3 0

Answer:  The correct option is (b) \dfrac{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\times x\times \dfrac{19}{2}+\dfrac{361}{4}+n-\dfrac{361}{4}\\\\\\=(x-\dfrac{19}{2})^2+n-\dfrac{361}{4}.

So, for the given expression to be perfect trinomial,

n-\dfrac{361}{4}=0\\\\\Rightarrow n=\dfrac{361}{4}.

Thus, (b) is the correct option.

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