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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, 06:39
4
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Difficulty:

75% (hard)

Question Stats:

55% (02:00) correct 45% (01:59) wrong based on 403 sessions

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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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Joined: 02 Sep 2009
Posts: 51233

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09 Dec 2010, 07:06
3
3
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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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, 05:21
2
4
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.
##### General Discussion
Intern
Joined: 14 Dec 2010
Posts: 43

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16 Dec 2010, 07:31
1
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
Math Expert
Joined: 02 Sep 2009
Posts: 51233

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16 Dec 2010, 07:42
1
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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Joined: 14 Dec 2010
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16 Dec 2010, 21:22
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...
Manager
Joined: 07 Feb 2010
Posts: 129

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18 Dec 2010, 07:25
2
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
Intern
Joined: 14 Dec 2010
Posts: 43

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19 Dec 2010, 21:07
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.
Intern
Joined: 03 Jul 2015
Posts: 31
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, 02:13
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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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, 14:54

--> 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, 08:12
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= x²-7x+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

Manager
Joined: 11 Jul 2016
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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, 11:17
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
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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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29 Jun 2018, 05:05
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Re: If one root of x^2+px+12=0 is 4, and the equation x^2+px+q=0 &nbs [#permalink] 29 Jun 2018, 05:05
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