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Which statement best reflects the solution(s) of the equation?

(sqrt(2x−4))−x+6=0




There is only one solution: x = 10.
The solution x = 0 is an extraneous solution.


There is only one solution: x = 10.
The solution x = 4 is an extraneous solution.


There is only one solution: x = 4.
The solution x = 10 is an extraneous solution.


There are two solutions: x = 4 and x =10.

Respuesta :

a5904640

[tex]\sqrt{2x-4}-x+6=0[/tex]

[tex]\sqrt{2x-4}=x-6[/tex]

Square both sides

[tex]2x-4=x^2-2.x.6+6^2[/tex]

[tex]2x-4=x^2-12x+36[/tex]

[tex]x^2-12x+36-2x+4=0[/tex]

[tex]x^2-14x+40=0[/tex]

Solving

Δ = b² - 4.a.c

Δ = -14² - 4 . 1 . 40

Δ = 196 - 4. 1 . 40

Δ = 36

There are 2 real roots.

x = (-b +- √Δ)/2a

x' = (--14 + √36)/2.1    

x'' = (--14 - √36)/2.1

x' = 20 / 2    

x'' = 8 / 2

x' = 10    

x'' = 4

Answer:

Option D. is the answer

Step-by-step explanation:

The given expression is [tex]\sqrt{2x-4}-x+6=0[/tex]

We further solve the given equation,

[tex]\sqrt{2x-4}=(x-6)[/tex]

(2x - 4) = (x-6)²

[tex](2x-4)=x^{2}+36-12x[/tex]

x²- 12x + 36 - 2x + 4 = 0

x² - 14x + 40 = 0

x² - 10x - 4x + 40 = 0

x(x - 10) - 4(x - 10) = 0

(x - 4)(x - 10) = 0

(x - 4) = 0 ⇒ x = 4

Or (x - 10) = 0

⇒ x = 10

Now we put the value of x = 4 in the equation,

= [tex]\sqrt{(2\times 4)-4}-4+6[/tex]

= [tex]\sqrt{4}-4+6[/tex]

= 2 - 4 + 6

= 0

For x = 10

[tex]\sqrt{2\times 10-4}-10+6[/tex]

= 4 - 10 + 6

= 0

Therefore, there are two solutions x = 4, 10

Option D will be the answer.

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