If |x² + 3x – 7| = 3, is x prime?

If |x² + 3x – 7| = 3, is x prime?

1. x² > 9
2. 3x ≤ 12

|x² + 3x – 7| = 3..
you can solve it by..
(a) Squaring as both sides are Non-negative..
(b) critical method

However a simple look and the prompt asking you if x is prime should tell you some points..
\(|x² + 3x – 7| = 3\).....For it to be prime, x can be 2, 3 and so on..
\(x=2....|x² + 3x – 7| = 2^2+3*2-7=4+6-7=3\)..Yes
\(x=3....|x² + 3x – 7| = 3^2+3*3-7=9+9-7=11\neq{3}\)..No
Any value greater would further increase the value of \(|x² + 3x – 7|\)

So the question finally becomes --- Is x=2?

1. x² > 9....|x|>|3|....x<-3 or x>3....\(x\neq{2}\)...Suff
2. 3x ≤ 12.... x ≤ 4.....x=2 or x =4 or x=-1... Insuff

A

If |x² + 3x – 7| = 3, is x prime?


|x² + 3x – 7| = 3
case1:x² + 3x – 7 = 3
x² + 3x – 10 = 0
x+5=0 or x -2=0 therefore -5,2
case1 x² + 3x – 7 = -3
x² + 3x – 4 = 0
x + 4=0 or x-1=0 therefore -4,1
so all values -5,-4,1,2

1. x² > 9 :range specifies between -3>x or x>+3 so values can be -4 or -5 therefore not prime : sufficient

2. 3x ≤ 12
x<=4: range gives values anything less than 4 so values can be 2 prime or 1 non prime : Insufficient
(A)

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
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The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Case 1) \(x^2+3x-7≥0\)
\(|x^2+3x-7| = 3\)
\(⇔ x^2+3x-7 = 3 since x^2+3x-7≥0\)
\(⇔ x^2+3x-10 = 0\)
\(⇔ (x-2)(x+5)=0\)
\(⇔ x=2\) or \(x=-5\)

Case 2) \(x^2+3x-7<0\)
\(|x^2+3x-7| = 3\)
\(⇔ -(x^2+3x-7) = 3 since x^2+3x-7<0\)
\(⇔ -x^2-3x+7 = 3\)
\(⇔ x^2+3x-4 = 0\)
\(⇔ (x-1)(x+4)=0\)
\(⇔ x=1\) or \(x=-4\)

The original condition tells that \(-5, -4, 1\) and \(2\) are all possible values of \(x\).

Since we have 1 variable (\(x\)) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
Since \(x^2 > 9\), the possible values of \(x\) are \(-5\) and \(-4\).
Both numbers are negative and they are not prime numbers.
The answer is 'no' and condition 1) is sufficient, since 'no' is also a unique answer by CMT (Common Mistake Type) 1.

Condition 2)
Since condition 2) tells \(x≤4\), the possible values are \(-5, -4, -1\) and \(2\).
If \(x = 2\), then \(x\) is a prime number and the answer is 'yes'.
If \(x = -1\), then \(x\) is not a prime number and the answer is 'no'
Since condition 2) does not yield a unique solution, it is not sufficient

Therefore, A is the answer.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.

|x² + 3x – 7| = 3

x² + 3x – 7=3 or x² + 3x – 7=-3

x= -5 or x= 2 x= 1 or x=-4


in 1) x^2 >9

so x cannot be 2, 3, 0, 1 and for both x =-5 and x=-4 are not prime ( sufficient )

2) 3x ≤ 12

x ≤ 4 ( dividing by 3 on both sides)
so x
values of x = {-5, -4, 1, 2}
x can be a prime or a non-prime ( insufficient )


please correct me if my approach for the first statement is wrong...