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If one root of x^2+px+12=0 is 4, and the equation x^2+px+q=0 [#permalink]
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09 Dec 2010, 07:39
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If one root of x^2+px+12=0 is 4, and the equation x^2+px+q=0 has equal roots then the value of q is A. 49/4 B. 4/49 C. 4 D. 1/4 E. 12
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Re: equation puzzle [#permalink]
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09 Dec 2010, 08:06
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Re: equation puzzle [#permalink]
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16 Dec 2010, 08:31
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Hi
I guess the answer is 12.
X^2+px+12=0 with one root as 4
ie 4+b=p where b is another root and 4b=12,then b=3
From second equation X^2+px+q=0 ,as roots are equal ,then first root ie 4 and second root ie b=3 are the roots for second equation as well,making 4*3=q ie 12 Ans E



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Re: equation puzzle [#permalink]
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16 Dec 2010, 08:42



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Re: equation puzzle [#permalink]
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16 Dec 2010, 22:22
Bunuel wrote: Eshika wrote: Hi
I guess the answer is 12.
X^2+px+12=0 with one root as 4
ie 4+b=p where b is another root and 4b=12,then b=3
From second equation X^2+px+q=0 ,as roots are equal ,then first root ie 4 and second root ie b=3 are the roots for second equation as well,making 4*3=q ie 12 Ans E It's given that the roots of x^2+px+q=0 are equal (so given that it has double root), not that the roots of this equation equal to the roots of another: x^2+px+12=0. If it were as you say then we would have q=12 right away: x^2+px+ q=0; x^2+px+ 12=0. OA for this question is A. Yeah I gt that..My mistake...



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Re: equation puzzle [#permalink]
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18 Dec 2010, 08:25
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Eshika wrote: Hi
I guess the answer is 12.
X^2+px+12=0 with one root as 4
ie 4+b=p where b is another root and 4b=12,then b=3
From second equation X^2+px+q=0 ,as roots are equal ,then first root ie 4 and second root ie b=3 are the roots for second equation as well,making 4*3=q ie 12 Ans E so the 2 roots of equation are 4 & 3 p =  (4+3) =7 so 2 eq becomes x^27x+q it is given 2 roots of the equation are same a=b a+b =7, 2a=7 a=7/2 q = a^2 = 49/4



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Re: equation puzzle [#permalink]
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19 Dec 2010, 22:07
feruz77 wrote: If one root of x^2+px+12=0 is 4, and the equation x^2+px+q=0 has equal roots then the value of q is a) 49/4 b) 4/49 c) 4 d) 1/4 e) 12 Thanks feruz77,I just started using GMAT club and got to know the concept of kudos after getting a mail that I have received 1 kudos.Will start using them now onwards.



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Re: If one root of x^2+px+12=0 is 4, and the equation x^2+px+q=0 [#permalink]
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27 Sep 2013, 08:59
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If one root of x^2+px+12=0 is 4, and the equation x^2+px+q=0 [#permalink]
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12 Sep 2015, 03:13
Bunuel wrote: feruz77 wrote: If one root of x^2+px+12=0 is 4, and the equation x^2+px+q=0 has equal roots then the value of q is a) 49/4 b) 4/49 c) 4 d) 1/4 e) 12 As one of the roots of \(x^2+px+12=0\) is \(x=4\) then substituting we'll get: \(16+4p+12=0\) > \(p=7\). Now, the second equation becomes \(x^27x+q=0\). As it has equal roots then its discriminant must equal to zero: \(d=494q=0\) > \(q=\frac{49}{4}\). Answer: A. can you please explain this As it has equal roots then its discriminant must equal to zero i can not understand



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If one root of x^2+px+12=0 is 4, and the equation x^2+px+q=0 [#permalink]
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12 Sep 2015, 06:21
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anik19890 wrote: Bunuel wrote: feruz77 wrote: If one root of x^2+px+12=0 is 4, and the equation x^2+px+q=0 has equal roots then the value of q is a) 49/4 b) 4/49 c) 4 d) 1/4 e) 12 As one of the roots of \(x^2+px+12=0\) is \(x=4\) then substituting we'll get: \(16+4p+12=0\) > \(p=7\). Now, the second equation becomes \(x^27x+q=0\). As it has equal roots then its discriminant must equal to zero: \(d=494q=0\) > \(q=\frac{49}{4}\). Answer: A. can you please explain this As it has equal roots then its discriminant must equal to zero i can not understand Do not post multiple posts for the same doubt. Give the experts/posters some time to reply to your earlier post. As for your question, refer to the text below: For a quadratic equation \(ax^2+bx+c=0\), with a\(\neq\)0 Discriminant, \(D = b^24ac\) For a quadratic equation to have 2 distinct real roots: \(D > 0\) For a quadratic equation to have 2 equal real roots: \(D = 0\) For a quadratic equation to have 0 distinct real roots: \(D < 0\) Coming back to the quadratic equation, \(x^27x+q=0\) > for equal roots, \(D =0\) > \((7)^24*q*1 = 0\) > \(49=4q\)> \(q= 49/4 = 12.25\) Hope this helps.
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Re: If one root of x^2+px+12=0 is 4, and the equation x^2+px+q=0 [#permalink]
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10 Oct 2016, 15:54
Use quadratic formula.
> 7 + [sqrt(49(4)(1)(c))]/2
Try values in for C. The correct one should give you the same two values once the equation is solved



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Re: If one root of x^2+px+12=0 is 4, and the equation x^2+px+q=0 [#permalink]
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22 Oct 2016, 09:12
x²+px+12=0
if 4 is a root, then
(x  4) ( x +r2) = 0 ; R2 being the other root. So, 4*r2 = 12 => r2 = 7
So, x²+px+q= x²7x+q=0
We know that this equation has only one root, thus : (x+r)²=0=x²+2xr+r²=0
So, q=r² and 2r=7 r=7/2, and q=r²=49/4
Answer choice A



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Re: If one root of x^2+px+12=0 is 4, and the equation x^2+px+q=0 [#permalink]
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22 Oct 2016, 12:17
x^2+px+12=0 is 4
One root is given as 4 => Substituting the values
4p +28 = 0
p = 7
Second eqn x^2+px+q=0
x^2 7x + q = 0
q = 49/4 A




Re: If one root of x^2+px+12=0 is 4, and the equation x^2+px+q=0
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