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Senior Manager  Joined: 13 Jun 2013
Posts: 266
If both roots of the quadratic equation y^2 -63y +x =0 are prime numbe  [#permalink]

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Difficulty:   75% (hard)

Question Stats: 52% (02:12) correct 48% (02:25) wrong based on 106 sessions

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If both roots of the quadratic equation y^2 -63y +x =0 are prime numbers then number of possible value of x is

a) 0
b) 1
c) 2
d) 4
e) more than 4
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: If both roots of the quadratic equation y^2 -63y +x =0 are prime numbe  [#permalink]

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Hi All,

This question is interesting in that it combines Algebra rules with Number Property rules. While you don't have to do much advanced math to answer it, you DO need to recognize the patterns that the equation is built on.

We're given Y^2 -63Y + X = 0

At first glance, this looks a bit "scary", but it should remind you of a Quadratic.

We're told that there are 2 roots and that they are BOTH PRIMES. This severely limits the possibilities....

Y^2 - 63Y + X = 0

Since the "middle term" is -63Y, the two parentheses are either BOTH negative or 1 negative/1 positive...

(Y - ?)(Y - ?) = 0

or

(Y - ?)(Y + ?) = 0

-63Y is the SUM of the roots. BOTH roots are PRIME though, so we have to think about what happens when you add/subtract odds and evens

EVEN + EVEN = EVEN
EVEN - EVEN = EVEN
ODD + ODD = EVEN
ODD - ODD = EVEN

None of these outcomes matches -63, which is an ODD number. This means that the two primes MUST be 1 odd and 1 even. The ONLY even prime is 2....so now the possibilities are...

(Y - ?)(Y - 2) = 0

or

(Y - ?)(Y + 2) = 0

To end up with -63Y as the middle term, the missing value in each of the above 2 examples would have to be.... -61 or +65....but 65 is NOT prime, so the only option is -61....

(Y - 61)(Y - 2) = 0

FOILing this out, we get...

Y^2 - 63Y + 122 = 0

X can ONLY be +122, so there's only one answer to the question.

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Re: If both roots of the quadratic equation y^2 -63y +x =0 are prime numbe  [#permalink]

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B.... the sum of roots in the eq is 63(-b/a)... it has to be addition of an odd and an even number.... only possible value 2,61 so X becomes 2*61=122
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Re: If both roots of the quadratic equation y^2 -63y +x =0 are prime numbe  [#permalink]

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Answer = B = 1

Given that both roots are prime number, it means roots cannot be even EXCEPT for 2, which is a prime number as well as a even number

In this case, only possible value of x = 2*61 = 122
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Re: If both roots of the quadratic equation y^2 -63y +x =0 are prime numbe  [#permalink]

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Sum of roots in a quadratic equation : (-b/a)
thus here sum of roots = 63
only one case would be possible that is 61+2 = 63 ( since 63 is odd , we need odd+even, hence only one condition is possible. 2 is only even prime number)
Hence the value of x = 61*2 =122
The answer must be B
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Re: If both roots of the quadratic equation y^2 -63y +x =0 are prime numbe  [#permalink]

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manpreetsingh86 wrote:
If both roots of the quadratic equation y^2 -63y +x =0 are prime numbers then number of possible value of x is

a) 0
b) 1
c) 2
d) 4
e) more than 4

We see that we need two values that sum to 63. Since 63 is odd and each root is prime, one of them must be 2 (the only even prime number). In that case, the other root is 61. So x = 2(61) = 122. That is, there is only one value possible for x.

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If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Re: If both roots of the quadratic equation y^2 -63y +x =0 are prime numbe   [#permalink] 19 Mar 2019, 19:12
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