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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.

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.

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

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 .

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.
_________________

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.

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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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
_________________

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

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03 Jan 2015, 05:32

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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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