In future, please show your work, so we can "see" where you're needing help. Thank you!

You can check the solution to any "solving" problem by plugging it back into the original exercise. In this case:

. . . . .$\displaystyle \sqrt{16}\, -\, 2\, =\, 4\, -\, 2\, =\, 2$

. . . . .$\displaystyle \sqrt[4]{16}\, =\, \sqrt[4]{2^4}\, =\, 2$

So your answer "checks".

How did you arrive at your answer? Was the original equation as follows?

. . . . .$\displaystyle \sqrt{3\, -\, 3x}\, -\, \sqrt{3x\, +\, 2}\, =\, 3$

You started

**the solution** by squaring both sides:

. . . . .$\displaystyle (3\, -\, 3x)\, -\, 2\sqrt{6\, +\, 3x\, -\, 9x^2}\, +\, (3x\, +\, 2)\, =\, 9$

. . . . .$\displaystyle 5\, -\, 2\sqrt{6\, +\, 3x\, -\, 9x^2}\, =\, 9$

. . . . .$\displaystyle -2\sqrt{6\, +\, 3x\, -\, 9x^2}\, =\, 4$

Then you squared again:

. . . . .$\displaystyle 4(6\, +\, 3x\, -\, 9x^2)\, =\, 16$

. . . . .$\displaystyle 6\, +\, 3x\, -\, 9x^2\, =\, 4$

. . . . .$\displaystyle 0\, =\, 9x^2\, -\, 3x\, -\, 2$

. . . . .$\displaystyle 0\, =\, (3x\, -\, 2)(3x\, +\, 1)$

So x = 2/3 or x = -1/3.

Plugging these back into the original equation gives invalid results: neither 1 - 2 nor 2 - 1 equals 3. So there is "no solution".

To learn how to find a quadratic by working backwards from its zeroes, try

**here**.

Once you have learned the basic technique, set up the exercise as:

. . . . .$\displaystyle \left(x\, -\, \left(\frac{-1\, -\, \sqrt{-2}}{5}\right)\right)\left(x\, -\, \left(\frac{-1\, +\, \sqrt{-2}}{5}\right)\right)$

Note that, since you must have integer coefficients, you'll need to multiply the product above by "25" to clear the denominator.

From what you've learned about

**the Quadratic Formula**, you know that the only way to get two different real roots is for the value inside the square root to be positive. In other words:

. . . . .$\displaystyle b^2\, -\, 4ac\, =\, p^2\, -\, 4(p\, +\, 1)(-1)\, >\, 0$

So you need to

**solve the quadratic inequality**:

. . . . .$\displaystyle (p\, +\, 2)^2\, >\, 0$

By nature of squaring, (p + 2)^2 will be positive for every non-zero value of p + 2. So the only place the inequality will not hold will be for p + 2 = 0, or p = -2. So your solution is correct.

To learn how to find the vertex of a quadratic, try

**here**. If you're stuck on some other aspect of the exercise, please reply showing your work and reasoning so far.

What is the total area? (Hint: Multiply the given length by the given width.)

Draw a rectangle, and label its sides with the given dimensions.

Draw a smaller rectangle inside, and label the gap between the rectangles with "x", being the (unknown) width of the walkway.

What expression would

**stand for** the width of the inner rectangle? What expression would stand for the length of the inner rectangle?

What expression would then stand for the area of the inner rectangle?

If the area of the garden is half of the total park area, what is the garden's area? (Hint: Divide the total area by 2.)

You now have an expression, in terms of "x", for the area of the garden, and a numerical value for the area of the garden. Set these equal, and solve.

If you get stuck, please reply showing how far you have gotten in following the provided steps. Thank you!