Question

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

Answer 1

$\sqrt{2x-4}-x+6=0$

$\sqrt{2x-4}=x-6$

Square both sides

$2x-4=x^2-2.x.6+6^2$

$2x-4=x^2-12x+36$

$x^2-12x+36-2x+4=0$

$x^2-14x+40=0$

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 2

Answer:

Option D. is the answer

Step-by-step explanation:

The given expression is $\sqrt{2x-4}-x+6=0$

We further solve the given equation,

$\sqrt{2x-4}=(x-6)$

(2x - 4) = (x-6)²

$(2x-4)=x^{2}+36-12x$

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,

= $\sqrt{(2\times 4)-4}-4+6$

= $\sqrt{4}-4+6$

= 2 - 4 + 6

= 0

For x = 10

$\sqrt{2\times 10-4}-10+6$

= 4 - 10 + 6

= 0

Therefore, there are two solutions x = 4, 10

Option D will be the answer.