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# If (x - 6)(x - m) = x^2 + rx + k. What is the value of m?

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Re: If (x - 6)(x - m) = x^2 + rx + k. What is the value of m? [#permalink]
Bunuel wrote:
If $$(x - 6)(x - m) = x^2 + rx + k$$. What is the value of m?

(1) $$k = 18$$

(2) $$r = -9$$

Simplifying the equation:

x^2 - mx - 6x +6m=x^2+rx+k
=> (-m-6)x+6m=rx+k
Statement 1: if k=18, then from the above equation, 6m=18 => m=3 Sufficient
Statement 2: if r=9, then from the above equation, (-m-6)=9 => m=3 - Sufficient

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Re: If (x - 6)(x - m) = x^2 + rx + k. What is the value of m? [#permalink]
Given the following form of a quadratic equation:

a(x)^2 + b(x) + c

(1) the SUM of the Roots of the quadratic is given by = -(b) / (a)

(2) the PRODUCT of the Roots of the quadratic is given by = (c) / (a)

The ROOTS of the quadratic are the values of the Input (here X) that make the quadratic equal 0

We are given the factored form of the quadratic as:

(x - 6) (x - m)

The ROOTS are therefore: x = 6 …..and…. x = m

SUM of Roots = (6 + m) = -(r)

PRODUCT of Roots = (6) (m) = k

What is the value of m = ?

D - each statement is sufficient alone as we can plug in the value of k or r and find that

m = 3

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Re: If (x - 6)(x - m) = x^2 + rx + k. What is the value of m? [#permalink]
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