Is x^2 + 9 prime?

[VJesus12]

Is x^2 + 9 prime?

(1) x is odd
(2) 3 ≤ x ≤ 7

[spoiler]OA=A[/spoiler]

Source: Veritas Prep

[Jay@ManhattanReview]

Is x^2 + 9 prime?

(1) x is odd
(2) 3 ≤ x ≤ 7

[spoiler]OA=A[/spoiler]

Source: Veritas Prep

Let's take each statement one by one.

(1) x is odd

So, we have x^2 + 9 => (Odd)^2 + 9 => Odd + Odd = Even > 9.

Note that all prime numbers greater than 2 are odd, thus, x^2 + 9 is not a prime number. Sufficient.

(2) 3 ≤ x ≤ 7

We must not assume that x is an integer and can take only 3, 4, 5, 6, and 7 as values. This can happen if you have a hangover of Statement 1.

So, x can take any value within 3 ≤ x ≤ 7.

Say x^2 + 9 = 23 a prime number

Thus, x^2 = 23 - 9 = 14 => x = √14 = 3... The value of x lies within 3 ≤ x ≤ 7. The answer is Yes, x is a prime number.

However, if x = 3, then 3 ≤ x ≤ 7 => x^2 + 9 = 3^2 + 9 = 9 + 9 = 18, not a prime number. The answer is No, x is not a prime number.

Insufficient.

The correct answer: A

Hope this helps!

-Jay
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[fskilnik@GMATH]

Is x^2 + 9 prime?

(1) x is odd
(2) 3 ≤ x ≤ 7
Source: Veritas Prep

$${x^2} + 9\,\,\,\mathop = \limits^? \,\,\,{\rm{prime}}$$
$$\left( 1 \right)\,\,\,x\,\,{\rm{odd}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{x^2} + 9\,\, \ge \,\,\,10\,\,\,{\rm{and}}\,\,{\rm{even}}\,\,\,\,\,\left( {{\rm{also}}\,\,{\rm{when}}\,\,x \le - 1} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.\,\,$$
$$\left( 2 \right)\,\,\,3 \le x \le 7\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,x = 3\,\,\,\, \Rightarrow \,\,\,{x^2} + 9 = 18\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,x = \,\,\sqrt {10} \,\,\,\left( {\sqrt 9 < \,\,\sqrt {10} < \sqrt {49} } \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{x^2} + 9 = 19\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.