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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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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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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

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

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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 \(b=(\frac{a}{2})^2=\frac{a^2}{4}\).

Answer: E.

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: \(a^2-4b=0\) --> \(b=\frac{a^2}{4}\).

Answer: E.

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.

Answer: E.

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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New post 04 Mar 2012, 04:12
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 \(b=(\frac{a}{2})^2=\frac{a^2}{4}\).

Answer: E.

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: \(a^2-4b=0\) --> \(b=\frac{a^2}{4}\).

Answer: E.

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.

Answer: E.

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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New post 10 Aug 2013, 08:36
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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New post 10 Aug 2013, 08:44
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.

:(

for quadratic equation:
\(ax^2+bx+c=0\)
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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New post 10 Aug 2013, 14:55
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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If there is exactly one root of the equation x^2 + ax + b [#permalink]

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New post 14 Jun 2017, 07:29
Quote:
For given expression \(x^2+ax+b=0\) discriminant is \(a^2-4b\)


Can someone explain where this came from?
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Re: If there is exactly one root of the equation x^2 + ax + b [#permalink]

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New post 14 Jun 2017, 07:50
Smokeybear00 wrote:
Quote:
For given expression \(x^2+ax+b=0\) discriminant is \(a^2-4b\)


Can someone explain where this came from?


Check the links below:

Factoring Quadratics: http://www.purplemath.com/modules/factquad.htm
Solving Quadratic Equations: http://www.purplemath.com/modules/solvquad.htm

Theory on Algebra: http://gmatclub.com/forum/algebra-101576.html
Algebra - Tips and hints: http://gmatclub.com/forum/algebra-tips- ... 75003.html

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PS Algebra Questions to practice: http://gmatclub.com/forum/search.php?se ... &tag_id=50

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

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New post 14 Jun 2017, 15:47
Thanks, I figured it out. I got confused when you plugged in the terms from the equation into the formula for the discriminant. Same variable names threw me off.
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Re: If there is exactly one root of the equation x^2 + ax + b   [#permalink] 14 Jun 2017, 15:47
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