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Hello.

A trinomial is a perfect square if the square root of the first term times the square root of the third term times 2 equals the middle term.

[tex]\boxed{\mathsf{x^{2} + \dfrac{2}{3} x}}[/tex]

a) Adding 1/9:

[tex]\cdot \: \mathsf{x^{2} + \dfrac{2}{3} x + \dfrac{1}{9}} \\
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\mathsf{\sqrt{x^{2}} \times \sqrt{\dfrac{1}{9}} \times 2 =} \\
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\mathsf{x \times \dfrac{1}{3} \times 2 =} \\
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\mathsf{\dfrac{2}{3} x \rightarrow it \: is \: a \: perfect \: square \: trinomial}[/tex]

b) Adding 4/9:

[tex]\cdot \: \mathsf{x^{2} + \dfrac{2}{3} x + \dfrac{4}{9}} \\
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\mathsf{\sqrt{x^{2}} \times \sqrt{\dfrac{4}{9}} \times 2 =} \\
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\mathsf{x \times \dfrac{2}{3} \times 2 =} \\
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\mathsf{\dfrac{4}{3} x \rightarrow it \: is \: not \: a \: perfect \: square \: trinomial}[/tex]

c) Adding 4 and 1/9:

[tex]\cdot \: \mathsf{x^{2} + \dfrac{2}{3} x + \dfrac{1}{9} + 4 = x^{2} + \dfrac{2}{3} x + \dfrac{37}{9}} \\
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\mathsf{\sqrt{x^{2}} \times \sqrt{\dfrac{37}{9}} \times 2 =} \\
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\mathsf{x \times \dfrac{\sqrt{37}}{3} \times 2 =} \\
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\mathsf{\dfrac{2\sqrt{37}}{3} x \rightarrow it \: is \: not \: a \: perfect \: square \: trinomial}[/tex]

Hope I helped.

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