Question

Extended Problems, Extraneous Solutions
I need help asap please, this is due in 1 day!
100 POINTS

Answer 1

Answer:

1a. 2 is the solution

1b. -3 is because square root of 1 doesn't equal -1.

1c.-3

Step-by-step explanation:

1.

$(x + 2 = \sqrt{3x + 10}$

Square both sides

$(x + 2) {}^{2} = 3x + 10$

Remeber perfect square Rule

$(x + y) {}^{2} = {x}^{2} + 2xy + {y}^{2}$

So now we got

$x {}^{2} + 4x + 4 = 3x + 10$

${x}^{2} + x + 4 = 10$

${x}^{2} + x - 6 = 0$

Factor into binomial.

$(x + 3)(x - 2) = 0$

So our solution is 3 or 2. However, let's check our work.

$- 3 + 2 = \sqrt{3( - 3) + 10}$

$- 1 = \sqrt{1}$

However square root of 1 is 1. It could be 1 but in Math, We only take the principal square root which is positive always. So -3 isn't a solution.

So let try 2.

$2 + 2 = \sqrt{3(2) + 10} = 4 = \sqrt{16} = 4 = 4$

So this is indeed the solution.

Make up equation for 2 and 3. Ask me if you need help.