Danou
Bunuel
AnkurGMAT20
Given that b > 0 and x^2 + bx + 5 = 0 has exactly 1 solution for x, what is the value of b?
(A) √5/2
(B) √5
(C) √10
(D) 2√5
(E) 2√10
Since \(x^2 + bx + 5 = 0\) has exactly 1 solution for x (and b > 0), then \(x^2 + bx + 5\) must be the square of a sum: \(x^2 + bx + (√5)^2 = (x + √5)^2 = x^2 + 2√5x + (√5)^2\). Therefore, \(b = 2√5\).
Answer: D.
Would it be possible to elaborate? I am not understanding the relationship / solution at all. Thanks so much
Usually, quadratic equations have two solutions. Here, we are given that the equation \(x^2 + bx + 5 = 0\) has exactly one solution. This happens only when the left side is a perfect square, meaning it can be written as \((x + √5)^2 = 0\). In this case, it has only one solution: \(x = -√5\).
Expanding \((x + √5)^2\), we get: \(x^2 + 2√5x + 5\).
Comparing this with \(x^2 + bx + 5\), we see that \(b = 2√5\).
Thus, the value of \(b\) is \(2√5\), and the answer is D.