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If x is an integer, how many possible values [#permalink]
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07 Aug 2017, 07:24
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If x is an integer, how many possible values of x exist for \(x^2+5x+6=0 ?\) A. 4 B. 2 C. 3 D. 1 E. 0
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Re: If x is an integer, how many possible values [#permalink]
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07 Aug 2017, 08:11
E When x <0 The equation will become x^2 5x+6=0. The two roots are 3 and 2 going against the range of x . If we take x>0, x^2 +5x+6=0 and two roots will be 3, and 2. This too goes against the range of x. So zero solutions. Another way of looking at this will be to observe that all the three terms of the equation are each greater than zero. So the equation will never be zero and hence no or zero solution. Sent from my Moto G (5) Plus using GMAT Club Forum mobile app



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Re: If x is an integer, how many possible values [#permalink]
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07 Aug 2017, 08:17
By range of x what are you suggesting that? X is an integer. Sent from my Redmi 4 using GMAT Club Forum mobile app
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If x is an integer, how many possible values [#permalink]
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DH99 wrote: If x is an integer, how many possible values of x exist for \(x^2+5x+6=0 ?\)
A. 4 B. 2 C. 3 D. 1 E. 0 Looking at other posts, I might be oversimplifying here . . . Please correct me if I'm mistaken. One method: check the signs of the terms. The squared term is positive (or nonnegative if x=0). The term whose product is (+5) * (some nonnegative or positive number because absolute value nonnegative or positive), is positive or nonnegative (if x = 0). The constant is positive. You cannot sum three positive numbers, or two zeros (if x=0) and a positive, and get zero. No values will work. Answer E Another way: If "check the signs method" doesn't occur to you, try factoring the quadratic as if there were no absolute value bars around the x in the second term. \(x^2+5x+6=0\) (x + 3)(x + 2) So x = 3 or 2 Check the values. Plug 3 and 2 into original equation. Neither works: 3: 9 + 15 + 6 does not equal zero. From the pattern of positive terms that result when plugging in a negative number (squared term is positive, absolute value term is positive, constant is positive), 2 will not work either. From the (+) + (+) + (+) pattern: you cannot get to 0 with three positive numbers. No values will work. Answer E
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Originally posted by generis on 07 Aug 2017, 10:46.
Last edited by generis on 07 Aug 2017, 12:08, edited 1 time in total.



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If x is an integer, how many possible values [#permalink]
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genxer123 wrote: DH99 wrote: If x is an integer, how many possible values of x exist for \(x^2+5x+6=0 ?\)
A. 4 B. 2 C. 3 D. 1 E. 0 Looking at other posts, I might be oversimplifying here . . . Please correct me if I'm mistaken. One method: check the signs of the terms. The squared term is positive. The term whose product is (+5) * (some positive number because absolute value is positive), is positive. The constant is positive. You cannot sum three positive numbers and get zero. No values will work. Answer E Another way: If "check the signs method" doesn't occur to you, try factoring the quadratic as if there were no absolute value bars around the x in the second term. \(x^2+5x+6=0\) (x + 3)(x + 2) So x = 3 or 2 Check the values. Plug 3 and 2 into original equation. Neither works: 3: 9 + 15 + 6 does not equal zero. From the pattern of positive terms that result when plugging in a negative number (squared term is positive, absolute value term is positive, constant is positive), 2 will not work either. From the (+) + (+) + (+) pattern: you cannot get to 0 with three positive numbers. No values will work. Answer E genxer123I like your "One method: check the signs of the terms." very much.+1 kudos given. So, it will be true for any quadratic equation in the form ax^2+bx+c=0 as long a,b and c are positive?



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If x is an integer, how many possible values [#permalink]
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07 Aug 2017, 12:05
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DH99 wrote: genxer123I like your "One method: check the signs of the terms." very much.+1 kudos given. So, it will be true for any quadratic equation in the form ax^2+bx+c=0 as long a,b and c are positive? Yes  but also nonnegative (x = 0 especially, or a AND b = 0) . . . as long as c is positive. In other words, if your first two terms result in the nonnegative 0, check c. Positive? No solution. You can't add a positive number to zero and get zero. Sometimes you will see posters insist that there is no such thing as the absolute value of 0. Because absolute value is a distance (from point of origin, often 0) there is such a thing: 0 is 0. Zero is 0 distance away from zero. Come to think of it, though the coefficients wouldn't work in the factor method part, I'm going to amend the first part my answer to include nonnegative! Thanks.
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Re: If x is an integer, how many possible values [#permalink]
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07 Aug 2017, 12:48
a tricky one x^2+5x+6=0 can be written as (x + 3) (x + 2) = 0 as x^2 always = x^2 then no solution for 0 as there is no negative value for x



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Re: If x is an integer, how many possible values [#permalink]
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28 Jan 2018, 00:57
\(x^2 = xx\) given \(x^2 + 5x + 6 = 0\)
Method1: Let a = x a^2 + 5a + 6 = 0 roots: a = 2, 3 x = 2, x = 3 modulus can never be negative, so no solution exists
Method 2: \(x^2 + 5x + 6 = 0\) => to hold this, \(x^2 + 5x = 6\), but modulus can never be negative as the minimum value is 0, so solution exists



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Re: If x is an integer, how many possible values [#permalink]
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28 Jan 2018, 10:51
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DHAR wrote: If x is an integer, how many possible values of x exist for \(x^2+5x+6=0 ?\)
A. 4 B. 2 C. 3 D. 1 E. 0 Humbly I think that it´s much easier to think in the way that \(x^2+5x+6=0\) is gonna have to suffer a check for strenuous roots afterwards; thus, it´s not necessary to solve or to do absolutely anything with the equation, and it´s just enough that your bubble lights up with the strenuous nuance and say to youself: "okay, the absolutevalue term is always positive, and so is the quadratic term, so there is simply no way that a solution exists that can make the left side side equal to the right". Conclusion: 0 solutions. 
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If x is an integer, how many possible values [#permalink]
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29 Jan 2018, 22:50
RooIgle wrote: DHAR wrote: If x is an integer, how many possible values of x exist for \(x^2+5x+6=0 ?\)
A. 4 B. 2 C. 3 D. 1 E. 0 Humbly I think that it´s much easier to think in the way that \(x^2+5x+6=0\) is gonna have to suffer a check for strenuous roots afterwards; thus, it´s not necessary to solve or to do absolutely anything with the equation, and it´s just enough that your bubble lights up with the strenuous nuance and say to youself: "okay, the absolutevalue term is always positive, and so is the quadratic term, so there is simply no way that a solution exists that can make the left side side equal to the right". Conclusion: 0 solutions.  RooIgleSurely, a few travelers will linger, then halt, when a collection of oddly graceful words unfurls. Surely, those stockstill few will wonder: of what stuff is this "strenuous nuance" made? Is it balletic, like a thought? Or sensible, like a nod? Surely, as they wander away, they will dare to see beyond content (zero does equal zero, after all). Surely, they will know enough to murmur, "Who gets that close to the oxymoronic but does not collide with it?" Brave soul. Perilous territory. No. Not "surely." But this time, yes. At least one. Nicely done.
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Re: If x is an integer, how many possible values [#permalink]
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15 Apr 2018, 12:06
+1 for option E. The equation can be rewritten as x^2+5x+6=0. Solving we get x=3,2. This clearly not possible. Hence no value of x will yield the required value. Hence option E.
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