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Re: If there are two unique solutions to the equation x^2 + bx + 9 = 0, wh [#permalink]
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Bunuel wrote:
If there are two unique solutions to the equation x^2 + bx + 9 = 0, which of the following could not be a value of b?

A. -10
B. -6.5
C. -6
D. 6.5
E. 10


Looking at our answer choices, we see that b cannot be -6. If b were -6 we would have:

(x - 3)(x - 3) = 0

x^2 - 6x + 9 = 0

However, because there must be two unique solutions, -6 is not a possible value for b.

Answer: C
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Re: If there are two unique solutions to the equation x^2 + bx + 9 = 0, wh [#permalink]
Putting value of C in eqn
Makes eqn as (x-3)^2=0
Which is giving a unique solution

So C is answer

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If there are two unique solutions to the equation x^2 + bx + 9 = 0, wh [#permalink]
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Hey guys

Hopefully you have enough experience with quadratic expressions to eyeball the answer here

For shorthand, in a quadratic expression \(x^2 + bx + c = 0\) factored as \((x + y)(x + z) = 0\), b is the result of adding y and z, and c is the result of multiplying them

That is, the middle coefficient (the number that multiplies the variable x) is the result of adding y and z, and the last number is the result of multiplying y and z

Here is an example:

\(x^2 - 2x + 15 = 0\)

The b here is -2 and the c is 15

This factors into (x + 3)(x - 5) = 0

Multiplying back out we see \(x^2\) - 5x + 3x + (3 * -5) = 0

The middle coefficient -2 is formed by adding 3 and -5. They are added because they both have the same variable x

This is because they both multiply by x when multiplying the two factors together


So when you look at \(x^2 + bx + 9 = 0\), you should think "what two numbers can multiply to make 9?"

Start with the simplest answers that work, 3 * 3 or -3 * -3

The factors can therefore be (x + 3)(x + 3) = 0 or (x - 3)(x - 3) = 0

Both will lead to a 9 in the c position

In the first case, b would be 6

In the second case, b would be -6

Since we see that -6 is in the answer choices, and one of the conditions is that there be two distinct answers to the equation, we know that -6 cannot be the value of b

This is because a b of -6 would mean the factors are (x - 3)(x - 3) = 0, in which case there would be only one solution to the equation x = 3

The problem states that there must be two unique solutions and therefore b cannot be -6

The answer is (C)
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If there are two unique solutions to the equation x^2 + bx + 9 = 0, wh [#permalink]
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