I know the answer to some of these, and the others I'm not sure of. Can you please explain to me how I can work them out in simplest form?

1) (-5 - √-9^2) answer is 30 + 16i?

2) x^2 + 5x + 2 = 0 answer is {-5 + √17 / 2, -5 - √17 / 2

3) √6x + 1 = x - 1 answer is {0, 8}?

4) 2x + 5 <17 answer is x<6?

5) 3x^2 - 1 = 47 answer is {-4, 4}

6) x^2 - 2x + 17 = 0 answer is {1 + 4i, 1 - 4i)

7) 4x^3 - 12x^2 = 9x - 27 answer is { -3/2, 3, 3/2}

8) |x| = 6 answer is {6}?

9) √x + 10 = x - 2 answer is {-1}?

10) (-7 + 5i) - (-9 -11i) answer is 2 + 16i

11) (2 + 7i) ( 2 - 7i) answer is 4 - 49i

12) √-81 - √-144 answer is -3

13) Solve x^2 - 13x + 36 = 0 answer is {4, 9}

14) Solve 5x^2 - 20x = 0 answer is {0, 4}

1) I don't unbderstand where the 'i' comes from iin your answer. What is \sqrt{9^2} ? Then solve.

2)You must know your formulae!!
ax^2 + bx + c = 0
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

3) Use the same formula, but here, you'll have the square root of x instead of x.

\sqrt{x}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

4) Solve normally, i.e. subtract 5 from both sides, then divide both sides by 2. Keep the '<' symbol.

5) Solve normally, i.e. add 1 to both sides, then divide both sides by 3. Lastly, put square root on both sides. But then, remember that the value of x may be positive or negative, so put both.

6) Same as 2)

7) That's more complicated. See my explanation that i gave you a while back

https://www.askmehelpdesk.com/math-sciences/factoring-producting-263608.html#post1293230

8) The two bars surrounding the x (modullus) show that for any value of x, the answer will be positive. As the result is 6, x must be either 6 or -6. Little mistake i think in your given answer

9) Same as in 3)

10) Expand normally then simplify. That's relatively easy!

11) Expand or use your formula:

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

12) What is \sqrt{81} and \sqrt{144}?

13) same as in 2) and 6)

14) Same as in 2), 6), and 13). But here, you can also factorise by putting the x outside 9better if you take the 5 also since both terms are divisible by 5x):

5x^2 - 20x = x(5x-20)

or

5x^2 - 20x = 5x(x-4)

Then solve.

Hope that helped. Just remember your basic formulae, and all will be fine.

Thanks for your help.

12) The square root of 81 and the square root of 144.. what would that be?

These are the ones I'm not sure of..

1) Solve √x + 10 = x - 2. I know that it has to be either
{-1, 6}
{-1}
{6}
or {0}

2) √6x + 1 = x - 1 I know that it has to be either...
{9, 18}
{0, 8}
{0}
or {8}

3) |x| = 6 I know that it has to be either..
{6}
{-6, 6}
{-6}
or {0, 6}

4)
√-81 - √-144 I know it has to be either..
-3, -3i, 3i, or 21i

5) x^2 = 11x - 10
{-10, -1}
{1, 10}
{5, 2}
or {10, -1}

I'll give you an easy away to be sure of your answers. REPLACE the answers you got into your question. By the way, i haven't done the square root of negatives yet... if you could provide me with the rule, i could perhaps help you more of the numbers requiring such formulae.

1) \sqrt{x} + 10 = x - 2

\sqrt{-1} + 10 = -1 - 2

\sqrt{6} + 10 = 6 - 2

Are they correct? Check it. Anyway, at first galnce, that cannot be 0 since it will be

\sqrt{0} + 10 = 0 - 2

10 = - 2 which is not correct.

2) Check it as for the 1)

3) Replace your answers.

|x| = 6

|6| = 6 √

|-6| = 6 √

|0| = 6 X

4) Give me that formula, asap.

5) x^2 = 11x - 10

x^2 - 11x + 10 = 0

(x-10)(x-1)= 0

So, x = 10 or 1

the other answers are wrong. You can check the others also.