Help with 5 problems...

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Help with 5 problems...

Mathematics
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2 [7- 3/2x] +5=21 [] are absolute value lines
7- 3/2x =8
Not sure what to do next

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you're on the right track, but when we have absolute value signs, we must consider the positive and negative solutions to the equation so, going by what you have so far, we have: | 7 - 3/2x | = 8, where | | are the absolute value signs, with me so far?
Yes
So, going from there, we need to write two different equations 7 - 3/2x = 8 and 7 - 3/2x = -8
then we solve each one separately so, let's tackle the first one 7 - 3/2x = 8 first, subtract 7 from both sides and tell me what you get
- 3/2x= 1
yup, now we divide both sides by (-3/2)
um -2/3
yup, so x = -2/3 now, we're only halfway done, there's another value of x that works for the equation. now we solve 7 - (3/2)x = -8 using the same process as before. start by subtracting 7 from both sides
x=10
yup, so x = 10 and x = -2/3 are your two solutions, good work ready for the next one?
Yes thank you
Solve. [x+8] +9=7 same as the one before
I got x=-10
ah, not quite, there's an important rule to remember when it comes to absolute values. the absolute value of a mathematical expression cannot be negative as an example, if we have |x| = a, then a must be a positive number in order to have a solution so, looking at our problem, we first subtract 9 from each side to get |x + 8| = -2 the absolute value of any quantity must be positive, so this has no solution
Okay
does that make sense? are you ready to move to the next one?
Yes it does. I thought it was but wanted to make sure.. Solve using appropriate substitution. (x^2 -6x)^2 -11(x^2 -6x) -30=0
I got 3+- square root 14 and 3+- square root 15 But its wrong...
huh, this is a tricky one, hold on a moment
Okay
well, I'm not sure if this is the right approach but...I'm guessing it wants us to substitute a variable for x^2 -6x so, if we let a = x^2 -6x, then our equation becomes a^2 - 11a - 30 after we apply the quadratic formula we get |dw:1440800955390:dw|
I factored (w-5)(w-6)
I'm not really sure if my thinking is correct and that factoring doesn't work, that would only work for positive 30, not -30 (w-5)(w-6) = w^2 - 11w + 30
Oops
ah, I wish I could be of more help on this one :( I feel like I'm missing an obvious trick here @welshfella @SithsAndGiggles
Its okay really. Your being super helpful :)
Given that \(a^2 - 11a - 30=0\) yields roots \(a=\dfrac{11\pm\sqrt{241}}{2}\), you can find the roots to \(x^2-x-a=0\) by applying the QF again.
typo: \(x^2-\color{red}6x-a=0\)
Um okay..
w(1) = 5 or w(2) = 6
So... 1+- square root 5 over 2
2 solution w(1) = 5 and w(2) = 6
Okay but how did you get to that?? Id like to know how to solve this
\[\Delta = 1\]
u should find that
The triangle is square root correct?
What?? O.o
\[w =\frac{ 11 \pm \sqrt{121-120} }{ 2} \]
Yes I have that
Except I have a positive 120
121+120=241
Now what?
u have mistake
Where?
b^2-4ac = \[(-11^2) - 4*1*30\]
u looking for this w^2-11w+30 @Destinyyyy
its a negative 30 not positive...
The original problem contains a -30
what is your problem could u tell me what is equation
plz
Solve using appropriate substitution. (x^2 -6x)^2 -11(x^2 -6x) -30=0
w^2 -11w -30=0 w^2--> (x^2-6x)^2 w--> x^2 -6x
are u sure its -30 ? cuz like that we dont have good numbers
Its actually -80 ... I miss read. So sorry
Everything else is correct
so give me right equation without mistakes
ok
Solve using appropriate substitution. (x^2 -6x)^2 -11(x^2 -6x) -80=0 I can screenshot it if you want
ok so w(1) = 5 and w(2) = 16
??
i solve it when i do w^2-11w-80=0
we have 2 solution w(1) = -5 and w(2) = 16
u have too check it cuz i am sure about it
@Destinyyyy did find like that or no ?
\[\frac{ 11\pm \sqrt{121+320} }{ 2 }\]
I havent read a thing youve said... x= 5,1 x=-2,8
Correct?
I factored
???
u have 2 find 6 nomber to solve this problem 8;-2;5;1;16;-5
Why 6 numbers?
The last one I did like this was 2 numbers
I dont count the w only the x's
x= 5,-2,8 I should have one more step
w^2-11w-80 = 0 i find 2 numbers is -5 and 16 this only solution of W w----> x^2-6x we have find x now w=16 and w=x^2-6x------> x^2-6x=16 and also for -5 we solve this equation x^2-6x-16=0 and x^2-6x+5=0 and we find x
Yeah I already found x
x= 5 and 1 x= -2 and 8
I should have another step where I do x^2
yup this
did u understand now
Are my answers correct? Are they the final answer?
yup u right
And there are no more steps?
yeah
They are right.. I have two more problems
u still did understand or what ??o.O
I understand how to solve that last one.
Can you help me with two more?
Im going to post them in a new question and tag you.
my time is 1 :00 Am

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