Anonymous
Anonymous asked in Science & MathematicsMathematics · 7 years ago

# Solve √(2x+5)-√(x-1)=2?

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√(2x + 5) - √(x - 1) = 2

3x + 4 - 2√(2x + 5)√(x - 1) = 4

9x^2 = 8x^2 + 12x - 20

x^2 - 12x + 20 = 0

(x - 10)(x - 2) = 0

Solutions:

x = 2

x = 10

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• Isolate one of the square roots:

√(2x + 5) = 2 + √(x - 1)

Square both sides to cancel out the square root:

2x + 5 = 4 + 4√(x - 1) + x - 1

Isolate the square root term again:

x + 2 = 4√(x - 1)

Square both sides to cancel out the square root:

x^2 + 4x + 4 = 16(x - 1)

Expand the brackets:

x^2 + 4x + 4 = 16x - 16

Move everything to one side:

x^2 - 12x + 20 = 0

Factor it out:

(x - 10)(x - 2) = 0

x = 2, 10

Whenever you square both sides always check your solutions because sometimes a false/extraneous solution is created. In this case both solutions fit.

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• Anonymous
7 years ago

√(2x+5)-√(x-1)=2

(2x+5)-(x-1)=4

x+4= 4

x=0

Source(s): batman
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