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505-555 Level|   Algebra|                     
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Ans. D

x^2 -2x - 15=0
x^2 - 5x + 3x -15=0
(x+3)(x-5)=0
x=5 (x=-3 is not possible since x>0)

II and III statements have (x-5) as factors therefore will be '0'.
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Solve x^2 - 2x - 15 = 0:
(x-5)(x+3) = 0;
x = -3, 5.

Since x is given as positive(x > 0), x = 5 is the valid one.

Substitute the value of x = 5 in the equations:
I. 25 - 30 + 9 = 4#0; Reject.
II. 25 - 35 + 10 = 0; Correct.
III. 25 - 50 + 25 = 0; Correct.

II and III only.

Ans is (D).
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Bunuel
If \(x^2-2x-15=0\) and \(x>0\), which of the following must be equal to zero ?

I. \(x^2-6x+9\)

II. \(x^2-7x+10\)

III. \(x^2-10x+25\)

A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III

Given: \(x^2-2x-15=0\)
Factor: \((x-5)(x+2)=0\)
So, either \(x = 5\) or \(x = -3\)
Since we're told \(x>0\), we know that it must be the case that \(x = 5\)

Now we can plug \(x = 5\) and to each of the given expressions to see which one(s) evaluate to be zero...

I. \(x^2-6x+9\)
We get: \(5^2-6(5)+9=25-30+9=4\)

II. \(x^2-7x+10\)
We get: \(5^2-7(5)+10=25-35+10=0\)

III. \(x^2-10x+25\)
We get: \(5^2-10(5)+25=25-50+25=0\)

Answer: D
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