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If x - 5 = x + 5, which of followings is true?
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22 Jun 2023, 03:00
22 Jun 2023, 03:00
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A
B
C
D
E
Difficulty:
65%
(hard)
Question Stats:
61% (03:02) correct
39%
(02:56)
wrong
based on 61
sessions
If \(x-5=√x+√5\), which of followings is true?
A. \(x^2+12x+16 = 0\)
B. \(x^2-12x+16 = 0\)
C. \(x^2+6x+8 = 0\)
D. \(x^2-6x+8 = 0\)
E. \(x^2+6x-8 = 0\)
A. \(x^2+12x+16 = 0\)
B. \(x^2-12x+16 = 0\)
C. \(x^2+6x+8 = 0\)
D. \(x^2-6x+8 = 0\)
E. \(x^2+6x-8 = 0\)
If x - 5 = x + 5, which of followings is true?
[#permalink]
22 Jun 2023, 06:23
22 Jun 2023, 06:23
1
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Option B
Replacing \(\sqrt{x}\) with A and \(\sqrt{5}\) with B,
\(x−5 = √x+√5 => A^2 - B^2 = A + B\)
=> \(A^2 - B^2 = A + B\)
=> \((A+B)(A-B) = A + B\)
=> \((A+B)(A-B) -1*(A + B) = 0\)
=> \((A + B)(A - B - 1) = 0\)
=> \(\sqrt{x} = \sqrt{5} + 1.\) ( A cannot be = - B because \(\sqrt{x}\) cannot be negative )
=> \((\sqrt{x})^2 = (\sqrt{5} + 1)^2\)
=> \(x = 6 + 2*\sqrt{5}\)
Comparing the options, only option B when solved using \((-b +- \sqrt{b^2 - 4*a*c}) / 2*a\) will have one of its roots as \(6 + 2*\sqrt{5}\)
Replacing \(\sqrt{x}\) with A and \(\sqrt{5}\) with B,
\(x−5 = √x+√5 => A^2 - B^2 = A + B\)
=> \(A^2 - B^2 = A + B\)
=> \((A+B)(A-B) = A + B\)
=> \((A+B)(A-B) -1*(A + B) = 0\)
=> \((A + B)(A - B - 1) = 0\)
=> \(\sqrt{x} = \sqrt{5} + 1.\) ( A cannot be = - B because \(\sqrt{x}\) cannot be negative )
=> \((\sqrt{x})^2 = (\sqrt{5} + 1)^2\)
=> \(x = 6 + 2*\sqrt{5}\)
Comparing the options, only option B when solved using \((-b +- \sqrt{b^2 - 4*a*c}) / 2*a\) will have one of its roots as \(6 + 2*\sqrt{5}\)
Quant Chat Moderator
Joined: 22 Dec 2016
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Location: India
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Re: If x - 5 = x + 5, which of followings is true?
[#permalink]
22 Jun 2023, 23:26
22 Jun 2023, 23:26
Bunuel wrote:
If \(x-5=√x+√5\), which of followings is true?
A. \(x^2+12x+16 = 0\)
B. \(x^2-12x+16 = 0\)
C. \(x^2+6x+8 = 0\)
D. \(x^2-6x+8 = 0\)
E. \(x^2+6x-8 = 0\)
A. \(x^2+12x+16 = 0\)
B. \(x^2-12x+16 = 0\)
C. \(x^2+6x+8 = 0\)
D. \(x^2-6x+8 = 0\)
E. \(x^2+6x-8 = 0\)
\(x-5=√x+√5\)
\((√x+√5)(√x-√5)=√x+√5\)
The value (√x+√5) cannot be equal to 0, hence we can divide both sides of the equation by (√x+√5)
\((√x-√5)=1\)
\(√x = 1 + √5\)
Squaring both sides
\(x = 6 + 2√5\)
Squaring both sides
\(x^2 = 56 + 24√5\)
Let's review the options
We can eliminate A, and C as each term in the sum is positive. Hence, the sum cannot be equal to 0.
Option B
B. \(x^2-12x+16 = 0\)
56 + 24√5 - 12(6 + 2√5) + 16
56 + 24√5 - 72 - 24√5 + 16
= 0
RHS = LHS
Option B
