Solve for all possible values of x.

Question

square root of, 6, x, plus, 40, end square root, equals, x, plus, 8
6x+40
​
=
x+8
Answer
Attempt 1 out of 2

Answer 1

To solve the equation

$$\sqrt{6x+40}=x+8,$$

we will first square both sides to eliminate the square root.

$$(\sqrt{6x+40}{)}^2=(x+8{)}^2.$$

This simplifies to:

$$6x+40=(x+8)(x+8).$$

Now expand the right side:

$$6x+40=x^2+16x+64.$$

Next, we will rearrange the equation to set it to zero:

$$0=x^2+16x+64-6x-40.$$

Combine like terms:

$$0=x^2+10x+24.$$

Now we have a quadratic equation:

$$x^2+10x+24=0.$$

We can factor this equation:

$$(x+4)(x+6)=0.$$

Setting each factor equal to zero gives us the possible solutions:

1. $x+4=0⟹x=-4$
2. $x+6=0⟹x=-6$

Now we have two possible solutions: $x=-4$ and $x=-6$.

However, we need to check each solution in the original equation to make sure they are valid (since squaring can introduce extraneous solutions).

1. Check $x=-4$:

$$\sqrt{6(-4)+40}=-4+8⟹\sqrt{-24+40}=4⟹\sqrt{16}=4,$$ which is true.

2. Check $x=-6$:

$$\sqrt{6(-6)+40}=-6+8⟹\sqrt{-36+40}=2⟹\sqrt{4}=2,$$ which is also true.

Both solutions are valid. Thus, the final solutions are:

$$\boxed{{\displaystyle -4\text{ and }-6}}.$$