View Full Version : Solving problems in simplest form
andyhaus1057
Oct 7, 2008, 08:13 PM
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
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-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.
andyhaus1057
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
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
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.