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2. √2x-5-√x+6=0(solving the following radical equations and check for extraneous solutions)

Sagot :

LammettHash

I assume the equation is

√(2x - 5) - √(x + 6) = 0

Note the domains for the root expressions:

• √(2x - 5) : 2x - 5 ≥ 0   ⇒   x ≥ 5/2

• √(x + 6) : x + 6 ≥ 0   ⇒   x ≥ -6

So any valid solution we find must be at least 5/2.

Move one term to the other side.

√(2x - 5) = √(x + 6)

Take squares.

(√(2x - 5))² = (√(x + 6))²

2x - 5 = x + 6

Solve for x :

x = 11

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