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Which is greater x or 4
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25 Dec 2018, 15:30
25 Dec 2018, 15:30
Expert Reply
00:00
Question Stats:
70% (00:30) correct
29% (00:27) wrong based on 94 sessions
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\(\sqrt{x^2 + 39} = 8\)
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.
Quantity A |
Quantity B |
x |
4 |
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.
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: Which is greater x or 4
[#permalink]
04 Jan 2019, 07:05
04 Jan 2019, 07:05
1
Carcass wrote:
\(\sqrt{x^2 + 39} = 8\)
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.
Quantity A |
Quantity B |
x |
4 |
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.
Given: \(\sqrt{x^2 + 39} = 8\)
Square both sides to get: \((\sqrt{x^2 + 39})^2 = 8^2\)
Simplify to get: \(x^2 + 39 = 64\)
Subtract 39 from both sides to get: \(x^2 = 25\)
Solve: x = 5 OR x = -5
Let's test each case for EXTRANEOUS ROOTS
When plug x = 5 and x = -5 into the original equation, the equation holds true.
If x = 5, then Quantity A is greater than Quantity B
If x = -5, then Quantity B is greater than Quantity A
Answer: D
Cheers,
Brent
Re: Which is greater x or 4
[#permalink]
04 Jan 2019, 19:45
04 Jan 2019, 19:45
Expert Reply
Carcass wrote:
\(\sqrt{x^2 + 39} = 8\)
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.
Quantity A |
Quantity B |
x |
4 |
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.
If we are looking at x^2 and no other term in x, it will surely give us two values of X, one negative and other positive..
But the value could be between 0and 4 and in that case 4 will always be greater..
So let us check the values..
\(\sqrt{x^2 + 39} = 8\)
Square both sides..
\(x^2+39=8^2=64....x^2=64-39=25\) so X can be 5 or -5..
Thus it may be less than 4 or greater than 4..
D
