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

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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
[Reveal] Spoiler: OA
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09 Dec 2010, 08:06
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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 $$x^2-7x+q=0$$. As it has equal roots then its discriminant must equal to zero: $$d=49-4q=0$$ --> $$q=\frac{49}{4}$$.

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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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16 Dec 2010, 08:42
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.
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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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18 Dec 2010, 08:25
1
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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^2-7x+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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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 $$x^2-7x+q=0$$. As it has equal roots then its discriminant must equal to zero: $$d=49-4q=0$$ --> $$q=\frac{49}{4}$$.

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
Expert's post
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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 $$x^2-7x+q=0$$. As it has equal roots then its discriminant must equal to zero: $$d=49-4q=0$$ --> $$q=\frac{49}{4}$$.

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, $$D = b^2-4ac$$

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, $$x^2-7x+q=0$$ ---> for equal roots, $$D =0$$ ---> $$(-7)^2-4*q*1 = 0$$ ---> $$49=4q$$---> $$q= 49/4 = 12.25$$

Hope this helps.
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If one root of x^2+px+12=0 is 4, and the equation x^2+px+q=0   [#permalink] 12 Sep 2015, 06:21
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