To solve for $\displaystyle x$ in the equation $\displaystyle \sqrt{x^2+12x+36} =4$

Square both sides of the equation 

$\displaystyle \rightarrow (\sqrt{x^2+12x+36})^2 =(4)^2$

$\displaystyle \rightarrow x^2+12x+36 =16$

Set the equation equal to $\displaystyle 0$ by subtracting the constant $\displaystyle 16$ from both sides of the equation.

$\displaystyle \rightarrow (x^2+12x+36) - 16 =(16) -16$

$\displaystyle \rightarrow x^2+12x+20 =0$

Factor to find the zeros:

$\displaystyle \rightarrow (x+10)(x+2) = 0$

This gives the solutions 

$\displaystyle x = \left \{ -10, -2\right \}$.

Verify that these work in the original equation by substituting them in for $\displaystyle x$. This is especially important to do in equations involving radicals to ensure no imaginary numbers (square roots of negative numbers) are created.

Begin by subtracting $\displaystyle 2$ from each side of the equation:

$\displaystyle \sqrt{3x+1}+2=7$

$\displaystyle \sqrt{3x+1}=5$

Now, square the equation:

$\displaystyle (\sqrt{3x+1}=5)^2$

$\displaystyle 3x+1=25$

Solve the linear equation:

$\displaystyle 3x=24$

$\displaystyle x=8$

Begin by squaring both sides of the equation:

$\displaystyle \sqrt{2x+1}-2=\sqrt{2x-7}$

$\displaystyle (\sqrt{2x+1}-2)^2=(\sqrt{2x-7})^2$

$\displaystyle (2x+1)-4\sqrt{2x+1}+4=2x-7$

Combine like terms:

$\displaystyle -4\sqrt{2x+1}=-12$

$\displaystyle \sqrt{2x+1}=3$

Once again, square both sides of the equation:

$\displaystyle (\sqrt{2x+1}=3)^2$

$\displaystyle 2x+1=9$

Solve the linear equation:

$\displaystyle 2x=8$

$\displaystyle x=4$

Begin by squaring both sides of the equation:

$\displaystyle \sqrt{4x+29}=(x+6)$

$\displaystyle (\sqrt{4x+29})^2=(x+6)^2$

$\displaystyle 4x+29=x^2+12x+36$

Now, combine like terms:

$\displaystyle x^2+8x+7=0$

Factor the equation:

$\displaystyle (x+7)(x+1)=0$

$\displaystyle x=-1,-7$

However, when plugging in the values, $\displaystyle -7$ does not work. Therefore, there is only one solution:

$\displaystyle x=-1$

Begin by squaring both sides of the equation:

$\displaystyle \sqrt{y+12}+1=\sqrt{y+21}$

$\displaystyle (\sqrt{y+12}+1)^2=(\sqrt{y+21})^2$

$\displaystyle (y+12)+2\sqrt{y+12}+1=y+21$

Now, combine like terms and simplify:

$\displaystyle 2\sqrt{y+12}=8$

$\displaystyle \sqrt{y+12}=4$

Once again, take the square of both sides of the equation:

$\displaystyle (\sqrt{y+12}=4)^2$

$\displaystyle y+12=16$

Solve the linear equation:

$\displaystyle y=4$

Begin by taking the square of both sides:

$\displaystyle \sqrt{5+2x}=x-5$

$\displaystyle (\sqrt{5+2x})^2=(x-5)^2$

$\displaystyle 5+2x=x^2-10x+25$

Combine like terms:

$\displaystyle x^2-12x+20=0$

Factor the equation and solve:

$\displaystyle (x-10)(x-2)=0$

$\displaystyle x=10,2$

However, when plugging in the values, $\displaystyle 2$ does not work. Therefore, there is only one solution:

$\displaystyle x=10$

To solve the radical expression, begin by subtracting $\displaystyle x$ from each side of the equation:

$\displaystyle \sqrt{x+11}-x=-9$

$\displaystyle \sqrt{x+11}=x-9$

Now, square both sides of the equation:

$\displaystyle (\sqrt{x+11})^2=(x-9)^2$

$\displaystyle x+11=x^2-18x+81$

Combine like terms:

$\displaystyle x^2-19x+70=0$

Factor the expression and solve:

$\displaystyle (x-14)(x-5)=0$

$\displaystyle x=14,5$

However, when plugged into the original equation, $\displaystyle 5$ does not work because the radical cannot be negative. Therefore, there is only one solution:

$\displaystyle x=14$

$\displaystyle \small \sqrt{4x+9}-8=5$

Add $\displaystyle 8$ to both sides.

$\displaystyle \small \sqrt{4x+9}=13$

Square both sides.

$\displaystyle \small (\sqrt{4x+9})^2=13^2$

$\displaystyle \small 4x+9=169$

Isolate $\displaystyle x$.

$\displaystyle \small 4x=160$

$\displaystyle \small x=40$

Begin by cubing both sides:

$\displaystyle \sqrt[3]{y+1}=3$

$\displaystyle (\sqrt[3]{y+1}=3)^3$

$\displaystyle y+1=27$

Now we can easily solve:

$\displaystyle y=26$