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555-605 (Medium)|   Algebra|      
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rxs0005
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Is x^2-8x+15 = 0? Updated on: 24 Jul 2013, 08:17
Originally posted by rxs0005 on 16 Aug 2010, 05:43.
Last edited by Bunuel on 24 Jul 2013, 08:17, edited 2 times in total.
Renamed the topic and edited the question.
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C
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Difficulty:

35% (medium)

Question Stats:

65% (01:06) correct 35% (01:12) wrong based on 382 sessions

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Is x^2-8x+15 = 0?

(1) x not equal to 3
(2) x-5 not equal to 0
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C
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Bunuel
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Is x^2-8x+15 = 0? 16 Aug 2010, 05:52
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rxs0005
is x^2-8x+15 = 0

1 x not equal to 3

2 x-5 not equal to 0
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Solve \(x^2-8x+15=0\) for \(x\) --> \(x=3\) or \(x=5\) --> so \(x^2-8x+15=0\) is true if \(x=3\) or \(x=5\). Basically the question asks: is \(x=3\) or \(x=5\)?

(1) \(x\neq{3}\). \(x\) can still be 5. Not sufficient.

(2) \(x-5\neq{0}\) --> \(x\neq{5}\). \(x\) can still be 3. Not sufficient.

(1)+(2) \(x\neq{3}\) and \(x\neq{5}\), so the answer to the question is NO. Sufficient.

Answer: C.
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Is x^2-8x+15 = 0? 24 Jul 2013, 06:58
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Bunuel
rxs0005
is x^2-8x+15 = 0

1 x not equal to 3

2 x-5 not equal to 0

Solve \(x^2-8x+15=0\) for \(x\) --> \(x=3\) or \(x=5\) --> so \(x^2-8x+15=0\) is true if \(x=3\) or \(x=5\). Basically the question asks: is \(x=3\) or \(x=5\)?

(1) \(x\neq{3}\). \(x\) can still be 5. Not sufficient.

(2) \(x-5\neq{0}\) --> \(x\neq{5}\). \(x\) can still be 3. Not sufficient.

(1)+(2) \(x\neq{3}\) and \(x\neq{5}\), so the answer to the question is NO. Sufficient.

Answer: C.
Show more

Bunuel,

I did not understand the logic.

Stm1 - X can be 5, so x^2-8x+15 = 0 , Sufficient

Stm 2- X can be 3, So x^2-8x+15 = 0 , Sufficient..

Ans- D

Whats wrong in this.
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Is x^2-8x+15 = 0? 24 Jul 2013, 07:10
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rxs0005
is x^2-8x+15 = 0

1 x not equal to 3

2 x-5 not equal to 0

Solve \(x^2-8x+15=0\) for \(x\) --> \(x=3\) or \(x=5\) --> so \(x^2-8x+15=0\) is true if \(x=3\) or \(x=5\). Basically the question asks: is \(x=3\) or \(x=5\)?

(1) \(x\neq{3}\). \(x\) can still be 5. Not sufficient.

(2) \(x-5\neq{0}\) --> \(x\neq{5}\). \(x\) can still be 3. Not sufficient.

(1)+(2) \(x\neq{3}\) and \(x\neq{5}\), so the answer to the question is NO. Sufficient.

Answer: C.

Bunuel,

I did not understand the logic.

Stm1 - X can be 5, so x^2-8x+15 = 0 , Sufficient

Stm 2- X can be 3, So x^2-8x+15 = 0 , Sufficient..

Ans- D

Whats wrong in this.
Show more

QUESTION IS whether x^2-8x+15 = 0
statement 1:==> \(x\neq{3}\).===>means \(x\) can be \(4,5,6,7\)...anthing except \(3\)
when \(x=5\)..the given equation equals to zero but with other numbers let say \(4...16-32+15=-1\)
so insufficient.

statement 2:similarly here\(x\) cant be \(5.\)....but at \(x=3\)==>equation equals to \(0\) BUT AT other numbers it is not equal to zero..
hence insufficient.

combining both you are neglecting both the roots of X hence that equation will never be zero .
hence sufficient C

HOPE IT HELPS
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Is x^2-8x+15 = 0? 24 Jul 2013, 08:23
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Mountain14
Bunuel
rxs0005
is x^2-8x+15 = 0

1 x not equal to 3

2 x-5 not equal to 0

Solve \(x^2-8x+15=0\) for \(x\) --> \(x=3\) or \(x=5\) --> so \(x^2-8x+15=0\) is true if \(x=3\) or \(x=5\). Basically the question asks: is \(x=3\) or \(x=5\)?

(1) \(x\neq{3}\). \(x\) can still be 5. Not sufficient.

(2) \(x-5\neq{0}\) --> \(x\neq{5}\). \(x\) can still be 3. Not sufficient.

(1)+(2) \(x\neq{3}\) and \(x\neq{5}\), so the answer to the question is NO. Sufficient.

Answer: C.

Bunuel,

I did not understand the logic.

Stm1 - X can be 5, so x^2-8x+15 = 0 , Sufficient

Stm 2- X can be 3, So x^2-8x+15 = 0 , Sufficient..

Ans- D

Whats wrong in this.
Show more

The question asks is x=3 or x=5?

(1) says that x is not 3. If x=5, then we have an YES answer, if x is any other values, then we have a NO answer. Not sufficient.
(2) says that x is not 5. If x=3, then we have an YES answer, if x is any other values, then we have a NO answer. Not sufficient.

(1)+(2) We know that x is neither 3 nor 5, so we have a NO answer to the question. Sufficient.

Answer: E.

Hope it's clear.
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Is x^2-8x+15 = 0? 25 Jul 2013, 18:51
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Thanks both of you...... Got it.... :-D
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Is x^2-8x+15 = 0? 18 Sep 2016, 09:24
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C is correct. Here's why:

eqn factors to be (x-3)(x-5) = 0

(1) we know from this that x is not equal to 3, but that's it. We have (x-5) = 0 still to deal with

INSUFFICIENT

(2) from this x is not equal to 5, but we have the problem again of (x-3) = 0 remaining

INSUFFICIENT

Together - (1) + (2) = tells us that x cannot be both 3 and 5, which are the only values that would make the orig eqn equal to zero, thus the eqn cannot equal zero
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Is x^2-8x+15 = 0? 16 Jul 2024, 10:49
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x^2-8x+15=0

From a conceptual understanding pov if you solve the above quadratic equation you get

(x-5)(x-3)=0
x=3,x=5

That means the value of the equation x^2-8x+15 can be zero only either x=3 or x=5

We have been asked if x^2-8x+15 = 0 ???

(1) x not equal to 3

Okay, if x is not equal to 3 maybe then x = 2 or 4 or 5 or 11 etc etc etc

Here if we put x=5 then yes the eqn x^2-8x+15 becomes 0 but for any other value of x [2,7,6,11,43..etc]

The eqn x^2-8x+15 will never be equal to 0

So we ultimately have 2 cases and cannot give a definite answer as to whether x^2-8x+15=0

Hence (1) not sufficient

(2) x not equal to 5

similar reasoning as above for x=3 we will get x^2-8x+15=0 but for any other value of x we will not get x^2-8x+15 = 0

Hence (2) not sufficient

Now combining (1) and (2)

now we know x is not equal to 3 and is not equal to 5

and those are the only values of x which make the eqn x^2-8x+15 = 0

so for every other value of x, the eqn x^2-8x+15 will not be equal to 0 and we can definitely say that, hence, (1) and (2) together sufficient
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