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P and Q are roots

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P and Q are roots [#permalink] New post 04 Feb 2018, 15:05
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P and Q are roots of \(x^2 - 9x = 36\)

Quantity A
Quantity B
\(P+Q\)
\(P*Q\)



A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

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[Reveal] Spoiler: OA

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Re: P and Q are roots [#permalink] New post 04 Feb 2018, 20:14
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This is a quadratic equation. The best thing to do with these is shove all terms to one side, setting it equal to zero, and then factor. Usually. In this case we don't need to factor. But let's shove. So

x^2 - 9x = 36

becomes

x^2 - 9x - 36 = 0

This is now in the same format as the classic ax^2 - bx + c equation. Now it's good to know that the roots of an equation will always multiply to make c. And they will always add to make
negative b. Notice the negative! Crucial! Anyway since Quantity A is adding the roots, that should be --9, which is 9. And since Quantity B is multiplying them, we'll get c, which is -36. So A is the answer.

In case you have doubts, let's actually factor this thing to show why the roots multiply to c but add to negative b. Factoring x^2 - 9x - 36 gets us (x - 12)(x + 3). Thus, the roots are 12 and -3. Notice that they multiply to -36, (just like -12 and 3 would multiply to -36), but they add to 9, which is negative b. On the other hand -12 and 3 would add to -9, which is b. Important distinction between the roots and the numbers you see in the factored quadratic.
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Re: P and Q are roots [#permalink] New post 05 Feb 2018, 02:42
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Carcass wrote:


P and Q are roots of \(x^2 - 9x = 36\)

Quantity A
Quantity B
\(P+Q\)
\(P*Q\)




Here let us solve the equation \(x^2 - 9x = 36\)

or it can be written as \(x^2-12x-3x-36=0\)

or (x-12)(x+3)=0

Hence 12 and -3 are the roots of the equation.

Now compare QTY A and QTY B

QTY A = 12 + (-3) = 9

QTY B = 12 * (-3) = -36

Hence QTY A > QTY B. i.e option A
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Re: P and Q are roots [#permalink] New post 07 Feb 2018, 02:10
Answer: A
Re: P and Q are roots   [#permalink] 07 Feb 2018, 02:10
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