In each case, state True if the trinomial is a perfect square else state False.
$4 x^{2} + 4 x + 1$

True

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False

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Solution

The correct option is A True
$4 x^{2} + 4 x + 1 = {(2 x)}^{2} + 2 \times (2 x) + 1$
Applying the formula $a^{2} + b^{2} + 2 a b = {(a + b)}^{2}$
$4 x^{2} + 4 x + 1 = {(2 x)}^{2} + 2 \times (2 x) + 1 = {(2 x + 1)}^{2}$
Hence, the given trinomial is a perfect square.

The answer is TRUE

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Similar questions

9 b^{2} + 12 b + 16

16 x^{2} - 16 x y + y^{2}

a^{2} - 10 a + 25

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

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

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

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

The given trinomial $4 x^{2} + 12 x y + 9 y^{2}$

is not a perfect square trinomial.

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