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3 years ago

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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