If there is exactly one root of the equation x^2 + ax + b : GMAT Problem Solving (PS)
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# If there is exactly one root of the equation x^2 + ax + b

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If there is exactly one root of the equation x^2 + ax + b [#permalink]

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03 Mar 2012, 17:14
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If there is exactly one root of the equation x^2 + ax + b, where a and b are positive constants, what is b in terms of a?

A. a/2
B. a
C. 3a/2
D. a^2/2
E. a^2/4

I'm not sure how to solve this problem. It take me almost five minutes of brainstorming but nothing.

The only thing on how I 'm triyng to attack the same is : (x+b)^2 whre the only root is x=-b.

[Reveal] Spoiler: OA

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Re: If there is exactly one root of the equation x^2 + ax + b [#permalink]

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03 Mar 2012, 20:52
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If there is only one root then discriminant b^2-4ac=0 i.e both roots are equal

so here from the equation we get a^2-4b=0 so solving we get b=a^2/4

Hope it helps

+1 Kudos if my post is valuable
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Re: If there is exactly one root of the equation x^2 + ax + b [#permalink]

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03 Mar 2012, 21:14
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carcass wrote:
If there is exactly one root of the equation x^2 + ax + b, where a and b are positive constants, what is b in terms of a?

A. a/2
B. a
C. 3a/2
D. a^2/2
E. a^2/4

I'm not sure how to solve this problem. It take me almost five minutes of brainstorming but nothing.

The only thing on how I 'm triyng to attack the same is : (x+b)^2 whre the only root is x=-b.

Yes, you can solve it this way: $$x^2+ax+b=0$$ will have only one root if it can be factored as $$(x+n)^2=0$$, in this case the root will be $$x=-n$$. $$(x+n)^2=x^2+2nx+n=0$$ --> $$a=2n$$ and $$b=n^2$$. Now, since $$n=\frac{a}{2}$$, then $$b=(\frac{a}{2})^2=\frac{a^2}{4}$$.

Or: a quadratic function is $$ax^2+bx+c=0$$ will have only one root (only one intercept with x-axis) if discriminant is zero, so when $$discriminant=b^2-4ac=0$$.

For given expression $$x^2+ax+b=0$$ discriminant is $$a^2-4b$$, so it must equal to zero: $$a^2-4b=0$$ --> $$b=\frac{a^2}{4}$$.

Or: try number plugging, $$x^2+ax+b=0$$ will have only one root if it can be factored for example as $$(x+4)^2=0$$ --> $$x^2+8x+16=0$$ --> $$a=8$$ and $$b=16$$. Now, plug $$a=8$$ in the answer choices and see which one gives $$b=16$$, only answer choice E works.

Note that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only. For example if you pick $$(x+2)^2=0$$ then you get two "correct" options B and E.

Hope it helps.
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Re: If there is exactly one root of the equation x^2 + ax + b [#permalink]

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04 Mar 2012, 03:12
Bunuel wrote:
carcass wrote:
If there is exactly one root of the equation x^2 + ax + b, where a and b are positive constants, what is b in terms of a?

A. a/2
B. a
C. 3a/2
D. a^2/2
E. a^2/4

I'm not sure how to solve this problem. It take me almost five minutes of brainstorming but nothing.

The only thing on how I 'm triyng to attack the same is : (x+b)^2 whre the only root is x=-b.

Yes, you can solve it this way: $$x^2+ax+b=0$$ will have only one root if it can be factored as $$(x+n)^2=0$$, in this case the root will be $$x=-n$$. $$(x+n)^2=x^2+2nx+n=0$$ --> $$a=2n$$ and $$b=n^2$$. Now, since $$n=\frac{a}{2}$$, then $$b=(\frac{a}{2})^2=\frac{a^2}{4}$$.

Or: a quadratic function is $$ax^2+bx+c=0$$ will have only one root (only one intercept with x-axis) if discriminant is zero, so when $$discriminant=b^2-4ac=0$$.

For given expression $$x^2+ax+b=0$$ discriminant is $$a^2-4b$$, so it must equal to zero: $$a^2-4b=0$$ --> $$b=\frac{a^2}{4}$$.

Or: try number plugging, $$x^2+ax+b=0$$ will have only one root if it can be factored for example as $$(x+4)^2=0$$ --> $$x^2+8x+16=0$$ --> $$a=8$$ and $$b=16$$. Now, plug $$a=8$$ in the answer choices and see which one gives $$b=16$$, only answer choice E works.

Note that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only. For example if you pick $$(x+2)^2=0$$ then you get two "correct" options B and E.

Hope it helps.

OMG in red is the best part. Why I didn't think to the quadratic formula where if positive we have 2 solutions, = 0 ONE solution, < 0 NO solution. In less than 10 seconds .

Thanks Bunuel.
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Re: If there is exactly one root of the equation x^2 + ax + b [#permalink]

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29 May 2013, 06:36
Bumping for review and further discussion.
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Re: If there is exactly one root of the equation x^2 + ax + b [#permalink]

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10 Aug 2013, 07:36
In any Quadratic Equation $$ax^2 + bx + c = 0$$ if,

Discriminant (i.e. $$b^2 - 4ac$$) > 0 then Roots are real and unequal.
Discriminant = 0 then roots are equal
Discriminant < 0 then roots are imaginary.
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Re: If there is exactly one root of the equation x^2 + ax + b [#permalink]

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10 Aug 2013, 07:44
carcass wrote:
If there is exactly one root of the equation x^2 + ax + b, where a and b are positive constants, what is b in terms of a?

A. a/2
B. a
C. 3a/2
D. a^2/2
E. a^2/4

I'm not sure how to solve this problem. It take me almost five minutes of brainstorming but nothing.

The only thing on how I 'm triyng to attack the same is : (x+b)^2 whre the only root is x=-b.

$$ax^2+bx+c=0$$
sum of roots in quadratic equation = $$\frac{-b}{a}$$
product of roots = $$\frac{c}{a}$$
now let root = $$x$$
for the give question
sum of root = $$2x = -a$$==>$$x=\frac{-a}{2}$$......1
product of roots = $$x^2 = b$$ ==>$$x = \sqrt{b}.$$.....2
equating 1 and 2
$$b= \frac{a^2}{4}$$
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Re: If there is exactly one root of the equation x^2 + ax + b [#permalink]

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10 Aug 2013, 13:55
carcass wrote:
If there is exactly one root of the equation x^2 + ax + b, where a and b are positive constants, what is b in terms of a?

A. a/2
B. a
C. 3a/2
D. a^2/2
E. a^2/4

I'm not sure how to solve this problem. It take me almost five minutes of brainstorming but nothing.

The only thing on how I 'm triyng to attack the same is : (x+b)^2 whre the only root is x=-b.

Procedure 1:
only one root, but x^2 indicates there must be two roots. so two root are equal = x
now, x+x = -a
or, x = -a/2
And x.x= b
or, x^2 = b
or, a^2/4 = b (so (E) answer)

procedure 2: (the standard equation for all the equation that have the highest power that is 2, is below=
x^2 - (addition of two root)x + (multiplication of two roots) = 0
now compare it with
x^2 + ax + b
or, x^2 - (-a/2 - a/2)x+ (-a/2)(-a/2) = 0
so, b = a^2 /4
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Re: If there is exactly one root of the equation x^2 + ax + b [#permalink]

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03 Jan 2015, 05:32
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If there is exactly one root of the equation x^2 + ax + b [#permalink]

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03 Apr 2016, 23:45
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If there is exactly one root of the equation x^2 + ax + b   [#permalink] 03 Apr 2016, 23:45
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