In each case, state True if the trinomial is a perfect square else state False.
$16 x^{2} - 16 x y + y^{2}$

True

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False

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Solution

The correct option is B False
For this trinomial to be a perfect square, we should be able to write it in the form of $a^{2} - 2 a b + b^{2}$
$16 x^{2} - 16 x y + y^{2} = {(4 x)}^{2} - 2 \times (4 x) \times (2 y) + y^{2}$
The term $2 a b$ should have been, $8 x y = 2 \times (4 x) \times (y)$ , for the given trinomial to be a perfect square.
Since, it cannot be written in the given format, it is not a perfect square.

The answer is FALSE

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Similar questions

4 x^{2} + 4 x + 1

9 b^{2} + 12 b + 16

a^{2} - 10 a + 25

In each case, find whether the trinomial is a perfect square or not :

$(i) x^{2} + 14 x + 49$ $(i i) a^{2} - 10 a + 25$

$(i i i) 4 x^{2} + 4 x + 1$ $(i v) 9 b^{2} + 12 b + 16$

$(v) 16 x^{2} - 16 x y + y^{2}$ $(v i) x^{2} - 4 x + 16$

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