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Shahajiraje and Jijabai

If 2 divides ...

Question

If 2 divides $x^{2} + 6 x + 9$ then x + 3 would be a/an_____for any value of x, provided $x^{2} + 6 x + 9$ is a positive integer.

A

None of the above

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B

Odd Number

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C

Even Number

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D

Prime Number

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Solution

The correct option is C Even Number
2 divides $x^{2} + 6 x + 9 = (x + 3)^{2}$
$\Rightarrow 2$ divides $x + 3$ ( $∵$ For any prime number p and a positive integer b, if $p$ divides $b^{2}$ , then $p$ divides $b$ )
$\Rightarrow x + 3$ is even $(∵$ Numbers divisible by 2 are even)

So, option a is correct.

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Similar questions

f (x) = \sqrt{x^{2} + 6 x + 9} - \sqrt{x^{2} - 6 x + 9}

, then the value of

f (\frac{456}{123}) - f (- \frac{456}{123})

x \in R - {3},

then the possible value(s) of the expression

\frac{\sqrt{x^{2} - 6 x + 9}}{3 - x} + 2

A = \int \frac{d x}{x^{2} + 6 x + 25}

B = \int \frac{d x}{x^{2} - 6 x - 27}

12 (A + B) = λ . {tan}^{- 1} (\frac{x + 3}{4}) + μ . ln ∣ ∣ ∣ \frac{x - 9}{x + 3} ∣ ∣ ∣ + C

, then the value of

(λ + μ)

\dots

.

f (x) = \sqrt{x^{2} + 6 x + 9} - \sqrt{x^{2} - 6 x + 9}

, then the value of

f (\frac{456}{123}) - f (- \frac{456}{123})

(x^{2} + 1) (y^{2} + 1) + 9 = 6 (x + y)

x, y \in R

), then the value of

(x^{2} + y^{2})

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