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 andyhaus1057 Posts: 98, Reputation: 1 Junior Member #1 Oct 7, 2008, 08:13 PM
Solving problems in simplest form
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}
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #2 Oct 8, 2008, 01:07 AM

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-s...ml#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.
 andyhaus1057 Posts: 98, Reputation: 1 Junior Member #3 Oct 8, 2008, 11:49 AM
Thanks for your help.

12) The square root of 81 and the square root of 144.. what would that be?
 andyhaus1057 Posts: 98, Reputation: 1 Junior Member #4 Oct 8, 2008, 01:52 PM
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}
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #5 Oct 9, 2008, 12:41 AM

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.

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