Question

What is the solution to the equation below?

Answer 1

Answer: Option C.

Step-by-step explanation:

First, we need to square both sides of the equation:

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

We know that:

$(a-b)^2=a^2-2ab+b^2$

Then, applying this, we get:

$x+3=x^2-2(x)(3)+3^2\\\\x+3=x^2-6x+9$

Now we need to subtract "x" and 3 from both sides of the equation:

$x+3-(x)-(3)=x^2-6x+9-(x)-(3)\\\\0=x^2-6x+9-x-3$

Adding like terms:

$0=x^2-7x+6$

Factor the quadratic equation.  Find two numbers whose sum be -7 and whose product be 6. These numbers are: -1 and -6. Then:

$(x-1)(x-6)=0$

Then:

$x_1=1\\x_2=6$

Checking the first solution is correct:

$\sqrt{1+3}=1-3\\ 2=-2 \ (False)$

Checking the second solution is correct:

$\sqrt{6+3}=6-3\\ 3=3 \ (True)$

Answer 2

Answer:

C  x=6

Step-by-step explanation:

sqrt(x+3) = x-3

Square each side

(sqrt(x+3))^2 = (x-3)^2

x+3 = (x-3)^2

x+3 = (x-3)(x-3)

FOIL

x+3 = x^2 -3x-3x+9

Combine like terms

x+3 = x^2 -6x+9

Subtract x from each side

x-x+3 = x^2 -6x-x +9

3 =  x^2 -7x +9

Subtract 3 from each side

3-3 =  x^2 -7x +9-3

0 = x^2 -7x+6

Factor

0 = (x-6)(x-1)

Using the zero product property

x-6=0   x-1 =0

x=6  x=1

Since we squared we need to check for extraneous solutions

x=1

sqrt(1+3) = 1-3

sqrt(4) = -2

2=-2

False

Extraneous

x=6

sqrt(6+3) = 6-3

sqrt(9) = 3

3=3

True   solutions