what is the solution of sqrt(2x+4)-sqrt(x)=2 A. x = 0 B. x = 0 and x = 16 C. x = 0 and x = –16 D. x = 16 and x = –16

Mathematics

2 answers:

Charra [1.4K]3 years ago

5 0

We want to solve
√(2x+4) - √(x) = 2

Write equation as
√(2x+4) = √x + 2
Square each side.
2x + 4 = x + 4√x + 4
x = 4√x
x - 4√x = 0
√x (√x - 4) = 0

Either
√x = 0 => x = 0
or
√x = 4 => x = 16

Test for extraneous solutions.
When x = 0:
√(2x+4) - √x = 2 (Correct)
When x = 16:
√(2x+4) - √x = √(36) - √(16) = 6 - 4 = 2 (Correct)

A plot of f(x) = √(2x+4) - √x - 2 = 0 confirms hat the solutions are correct.

Answer: B. x = 0 and x = 16.

levacccp [35]3 years ago

3 0

The answer is a, x is 0

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Read 2 more answers

Find the first three terms in the expansion , in ascending power of x , of (2+x)^6 and obtain the coefficient of x^2 in the expa

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Answer:

The first 3 terms in the expansion of $(2 + x)^{6}$ , in ascending power of x are,

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Step-by-step explanation:

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= $\sum_{k=0}^{6}(6_{C_{k}} \times x^{k} \times 2^{6 - k})$

= $6_{C_{0}} \times x^{0} \times 2^{6} + 6_{C_{1}} \times x^{1} \times 2^{5} + 6_{C_{2}} \times x^{2} \times 2^{4}$ + terms involving higher powers of x