x - 1 = -1
+1 +1
x = 0.
I AM A GENIUS.
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throughthefire wrote:
3x^2 + 6x + 6 = 0
no solution!
There actually is a real solution, but it's irrational so you can't find it out by factoring. To know if you can factor a quadratic trinomial easily, use the discriminant test: b² - 4ac where the constants abc are present in ax² + bx + c. Note that it's the part of the quadratic formula that's under the square root
Anyway, the answer to the equation is
S = { (-2 +/- 2root2) / 2 }
if you can read that
and if it's wrong, I'm sorry. I didn't even do it properly.
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meew0 wrote:
x-1 = b+1
x*2 = b*-2
What is x and what is b?
I'll be happy to solve it, just to keep me busy (school exams are over, I have nothing to do).
Note that I did it on paper and then transferred the working to here, so it's a bit messy. Oh, and I did it my way. I know it's pretty long but it's easy.
x - 1 = b + 1
x = b + 2
2x = -2b
x = -b
Join the two and get rid of x for now
b + 2 = -b
2b = -2
b = -1
Plug b back into the original
x - 1 = (-1) + 1
x = 1
S = {x = 1, b = -1}
Check it
1 - 1 = -1 + 1
0 = 0
yay!Offline
LS97 wrote:
meew0 wrote:
x-1 = b+1
x*2 = b*-2
What is x and what is b?I'll be happy to solve it, just to keep me busy (school exams are over, I have nothing to do).
Note that I did it on paper and then transferred the working to here, so it's a bit messy. Oh, and I did it my way. I know it's pretty long but it's easy.Code:
x - 1 = b + 1 x = b + 2 2x = -2b x = -b Join the two and get rid of x for now b + 2 = -b 2b = -2 b = -1 Plug b back into the original x - 1 = (-1) + 1 x = 1 S = {x = 1, b = -1} Check it 1 - 1 = -1 + 1 0 = 0 yay!
Right!
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Here's a (relatively) hard question for you:
solve x³ + 6x² - x = -6 (2x + 1), given that one of the factors of x³ + 6x² + 11x + 6 is (x+ 1).
Oh, and do remember it's cubic, there's 3 solutions.
Last edited by LS97 (2011-06-17 13:32:34)
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^^ anyone?
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LS97 wrote:
^^ anyone?
Me, but I'm not allowed to answer.
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meew0 wrote:
-6.309
At least MSM says that.
To which one? If it's to mine, no way.
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LS97 wrote:
throughthefire wrote:
3x^2 + 6x + 6 = 0
no solution!There actually is a real solution, but it's irrational so you can't find it out by factoring. To know if you can factor a quadratic trinomial easily, use the discriminant test: b² - 4ac where the constants abc are present in ax² + bx + c. Note that it's the part of the quadratic formula that's under the square root
![]()
Anyway, the answer to the equation is
S = { (-2 +/- 2root2) / 2 }
if you can read thatand if it's wrong, I'm sorry. I didn't even do it properly.
Actually, it's not technically a real solution, it's an imaginary solution. That is, it contains the square root of a negative number (in this case, it comes out as sqrt(-4) or 2sqrt(-1) which is 2i)
@LS97: I can do it, but I'll need to get some paper and I really don't feel like doing that.
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meew0 wrote:
LS97 wrote:
meew0 wrote:
-6.309
At least MSM says that.
To which one? If it's to mine, no way.
To yours...
As I said, I didn't solve it, Microsoft Mathematics solved it...
Yeah -- Really weird. I've had good experiences with MSM until now...
EDIT: I put the equation in and it did solve it well. Try again. But don't post the answer if you didn't work it out by hand!
Last edited by LS97 (2011-06-17 14:53:15)
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LS97 wrote:
throughthefire wrote:
3x^2 + 6x + 6 = 0
no solution!There actually is a real solution, but it's irrational so you can't find it out by factoring. To know if you can factor a quadratic trinomial easily, use the discriminant test: b² - 4ac where the constants abc are present in ax² + bx + c. Note that it's the part of the quadratic formula that's under the square root
![]()
Anyway, the answer to the equation is
S = { (-2 +/- 2root2) / 2 }
if you can read thatand if it's wrong, I'm sorry. I didn't even do it properly.
I got { +/- i + 1}. How did you get that? The answer should have been a complex number. (Part imaginary and part real.)
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LS97 wrote:
meew0 wrote:
LS97 wrote:
To which one? If it's to mine, no way.
To yours...
As I said, I didn't solve it, Microsoft Mathematics solved it...Yeah -- Really weird. I've had good experiences with MSM until now...
EDIT: I put the equation in and it did solve it well. Try again. But don't post the answer if you didn't work it out by hand!
I tried it again and it gave me the same result...
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OK, let me write it down and work it out properly
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Oh, so sorry, I somehow imagined a 4 in that original equation. You see:
x = -b +/- √b²-4ac all over 2a
i thought b in the b² was 4...
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meew0 wrote:
LS97 wrote:
meew0 wrote:
To yours...
As I said, I didn't solve it, Microsoft Mathematics solved it...Yeah -- Really weird. I've had good experiences with MSM until now...
EDIT: I put the equation in and it did solve it well. Try again. But don't post the answer if you didn't work it out by hand!I tried it again and it gave me the same result...
Remember there's 3 answers to it -- can't be 6000 whatever!
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Okay, I know this sounds weird, but for LS97's question I got -11 and -1
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Oh heckle! No no no! Never mind!
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Ugh... I never learned how to solve that kind of problem... I admit defeat.
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@skyline: sorry, but you missed a ton... A factorial (9!) is multiplied, not added
A square is a number times itself, not times 2. look at my answers (above yours)
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16Skittles wrote:
@skyline: sorry, but you missed a ton... A factorial (9!) is multiplied, not added
A square is a number times itself, not times 2. look at my answers (above yours)
Why do you have ssss' name in your signature?
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wulfmaster wrote:
16Skittles wrote:
@skyline: sorry, but you missed a ton... A factorial (9!) is multiplied, not added
A square is a number times itself, not times 2. look at my answers (above yours)Why do you have ssss' name in your signature?
you don't play minecraft do you...
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