DAT Destroyer QRT#37

[cookie13]

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I had a question about #37 in the QRT section of the DAT Destroyer:

The question wants you to solve for x:
sq. root of x+5= x-1

It says the answer is only 4 but technically when you take the sq. root of a number isn't it plus or minus that number? So shouldn't -1 be a solution too?

 

[casper22]

It could be a possible solution.. but if its not in the answer choice, its not to required..

I dont have my Destroyer with me right now so I dont know the answer choice, but I am guessing -1 is not in there...

 

[silveryhair]

Yes, technically it can be -/+ of the number, however in this question, it isn't because it does not satisfy the requirements.

To solve the problem

sqr(x=5) = x-1
X+5 = x^2-2X+1
x^2-3X-4 Than factor

(x+1)(x-4)

the solutions can be -1 or 4, but always remember to check your answer, so plug it back in and you'll notice that only 4 works and not negative one. Hope that helps.

 

[cookie13]

It could be a possible solution.. but if its not in the answer choice, its not to required..

I dont have my Destroyer with me right now so I dont know the answer choice, but I am guessing -1 is not in there...

It's actually number 38...

and yes it is included in the the answer choices: a. 4, b. -1, c. 1, d. {-1,4}, and e. {-4,1}

 

[casper22]

Yes, technically it can be -/+ of the number, however in this question, it isn't because it does not satisfy the requirements.

To solve the problem

sqr(x=5) = x-1
X+5 = x^2-2X+1
x^2-3X-4 Than factor

(x+1)(x-4)

the solutions can be -1 or 4, but always remember to check your answer, so plug it back in and you'll notice that only 4 works and not negative one. Hope that helps.

Ahh...makes sense...

 

[cookie13]

Yes, technically it can be -/+ of the number, however in this question, it isn't because it does not satisfy the requirements.

To solve the problem

sqr(x=5) = x-1
X+5 = x^2-2X+1
x^2-3X-4 Than factor

(x+1)(x-4)

the solutions can be -1 or 4, but always remember to check your answer, so plug it back in and you'll notice that only 4 works and not negative one. Hope that helps.

you are right... but when I plug it back in... I get sq.root of (-1+5) which is 4 and if I take the sq.root of that I get +/- 2 which is equal to the other side (-1-1= -2)

so should -1 be also in the solution? or am I not plugging it in right

 

[cookie13]

It made better sense when I graphed it.. Thanks!

 

[DemQ]

you are right... but when I plug it back in... I get sq.root of (-1+5) which is 4 and if I take the sq.root of that I get +/- 2 which is equal to the other side (-1-1= -2)

so should -1 be also in the solution? or am I not plugging it in right

when you plug back in -1 to check, you get sqrt-4 = -2. this cannot be true because you cannot get a negative number from sqrt of any number. so -1 is not an answer.

 

[cookie13]

when you plug back in -1 to check, you get sqrt-4 = -2. this cannot be true because you cannot get a negative number from sqrt of any number. so -1 is not an answer.

The question was sq. root (x+5)
-1+5= 4 not -4

 

[ZFT]

you are right... but when I plug it back in... I get sq.root of (-1+5) which is 4 and if I take the sq.root of that I get +/- 2 which is equal to the other side (-1-1= -2)

so should -1 be also in the solution? or am I not plugging it in right

sqrt(4) = 2, not +/- 2

on the other hand, x^2 = 4, means x = +/- 2

it's a matter of notation: the sqrt symbol mean "postive square root only"

sqrt(4) = 2 only

-sqrt(4) = -2 only

 

[Streetwolf]

sqrt(4) = 2, not +/- 2

on the other hand, x^2 = 4, means x = +/- 2

it's a matter of notation: the sqrt symbol mean "postive square root only"

sqrt(4) = 2 only

-sqrt(4) = -2 only

Correct. The only way sqrt(x) is a function is if you only consider y > or = 0. Otherwise it would violate the vertical line rule.

 

[ZFT]

Also, y=x^2 will pass the vertical line test on its entire domain, this is why x can take on +/- values.