Manhattan GMAT Challenge Problem of the Week - 30 Aug 2011

by Manhattan Prep, Aug 30, 2011

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Question

If there is exactly one root of the equation [pmath]x^2[/pmath] + 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. [pmath]a^2[/pmath]/2

E. [pmath]a^2[/pmath]/4

Answer

For the given equation to have exactly one root, the quadratic expression on the left must factor into [pmath](x + k)^2[/pmath], so that the equation becomes [pmath](x + k)^2[/pmath] = 0. The single root of this equation would be x = k.

Expand [pmath](x + k)^2[/pmath]. You get [pmath]x^2[/pmath] + 2xk + [pmath]k^2[/pmath]. Now match that up to the given expression [pmath]x^2[/pmath] + ax + b. For these expressions to match, the coefficients of the x terms must be the same, and the constants must be the same as well.

This means that 2xk = ax and that [pmath]k^2[/pmath]= b. Solve the first equation for k in terms of a. You get k = a/2. Plug into the second equation. You get [pmath]a^2/4[/pmath] = b. This is the answer.

You can also test numbers to solve this problem. For instance, make up an a and a b so that the equation has one root.

[pmath]x^2[/pmath] + 6x + 9 = 0 has just one root, because the left side factors to [pmath](x + 3)^2[/pmath]. The solution is x = 3.

Likewise, [pmath]x^2[/pmath] + 8x + 16 = 0 has just one root, because the left side factors to [pmath](x + 4)^2[/pmath]. The solution is x = 4.

In the first case, a = 6 and b = 9. In the second case, a = 8 and b = 16. Only answer choice E works for both cases.

The correct answer is E.

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