Question

What is the solution to the equation shown below

Answer 1

Plug values in!
3/(3-2) + 6 = sqrt(3-2) + 8
3 + 6 = 1 + 8
9=9 so the first one

Answer 2

Answer:

x = 3

Option 1 is correct.

Step-by-step explanation:

Given: $\dfrac{3}{x-2}+6=\sqrt{x-2}+8$

First we check restriction on x from equation.

Denominator can't be 0

So,  x-2 ≠0

         x ≠ 2

Inside the square root value always positive.

Therefore, x-2 ≥ 0

                     x ≥ 2

Now simplify the equation.

$\dfrac{3}{x-2}+6=\sqrt{x-2}+8$

$\dfrac{3}{x-2}-2=\sqrt{x-2}$

squaring both sides

$(\dfrac{3}{x-2}-2)^2=(\sqrt{x-2})^2$

$\dfrac{9}{(x-2)^2}+4-\dfrac{12}{x-2}=x-2$

$9+4(x-2)^2-12(x-2)=(x-2)^3$

$9+4x^2+16-16x-12x+24=x^3-8-6x^2+12x$

$x^3-10x^2+40x-57=0$

$(x-3)(x^2-7x+19)=0$

x=3 and two are imaginary

Now we take intersection of all two condition and one solution.

x ≠ 2 and x ≥ 2  and x=3

Intersection of all three, x = 3

Hence, The solution of the given equation is 3