Mastering Hard Thinking Questions: Solving for a, p, and q with Given Equations

[I Like Pi]

Homework Statement

a) given, a + 5 and 2a-1, one is 40% greater than the other one, solve for a.

b) given, 5x$$^{4}$$ + 4x$$^{3}$$ + 3x$$^{2}$$ + Px + Q. When divided by x$$^{2}$$ - 1, you get a remainder of 0, solve for p and q.

c) given csc(6b + $$\frac{pi}{8}$$) = sec(2b - $$\frac{pi}{8}$$), solve for b.

Homework Equations

The Attempt at a Solution

please help, i can't get started on these three at all...

edit:

c) i know that cscx = 1/sinx and secx = 1/cosx, i then cross multiplied to get $$cos(2b-pi/8) = sin(6b+pi/8)...$$

 

[Quinzio]

Question a) shouldn't be a problem at all.
Any attempt by you ?

 

[I Like Pi]

Question a) shouldn't be a problem at all.
Any attempt by you ?

Would it be (a+5) = .4(2a-1) + 2a-1 ? or .4(a+5) + (a+5) = 2a - 1 ?

 

[Quinzio]

Would it be (a+5) = .4(2a-1) + 2a-1 ? or .4(a+5) + (a+5) = 2a - 1 ?

It's not clear from the text. Choose one and go on, or solve both in sequence.

 

[I Like Pi]

It's not clear from the text. Choose one and go on, or solve both in sequence.

well then you get a+5 = 0.8a - 0.4 + 2a -1
a+5 = 2.8a - 1.4
1.8a - 6.4 = 0
1.8a = 6.4
a = 3.55555556

or

2a - 1 = .4a + 2 + a + 5
2a - 1 = 1.4a + 7
0.6a - 8 = 0
0.6a = 8
a = 13.3333333

therefore a can be 13.333 or 3.555 ?

 

[Quinzio]

yes.

 

[I Like Pi]

yes.

thanks could you help me get started on b) ?

Thanks for your time!

 

[Quinzio]

b)
Make this multiplication $$(x^2-1)(Ax^2+Bx+C)$$

 

[I Like Pi]

b)
Make this multiplication $$(x^2-1)(Ax^2+Bx+C)$$

that would give you $$ax^4 + bx^3 + (c - a)x^2 - bx - c$$. Is that what you mean?

 

[Quinzio]

that would give you $$ax^4 + bx^3 + (c - a)x^2 - bx - c$$. Is that what you mean?

Ok, so now give the solution. It's really easy.

 

[I Like Pi]

Ok, so now give the solution. It's really easy.

do you use the original equation? so a = 5, b = 4, and c = 3?

but that doesn't work... cause then you get:$$5x^4 + 4x^3 + (3 - 5)x^2 - 4x - 3$$

 

[Quinzio]

You realize by yourself there's something wrong. Ok.

You should come alone to the solution. The next step would be giving you the solution.

 

[I Like Pi]

You realize by yourself there's something wrong. Ok.

You should come alone to the solution. The next step would be giving you the solution.

I don't see what I could possibly do to get the answer I know that c - a = 3, that's all... but c itself is 3

 

[Quinzio]

Why you say "c itself is 3"... ?
Tell me where is written that ?

 

[I Like Pi]

Why you say "c itself is 3"... ?
Tell me where is written that ?

oh, well i based it on the original... a = 5, b = 4, c = 3...

well, then if that's the case,

$$ax^4 + bx^3 + (c - a)x^2 - bx - c$$
$$=5x^4 + 4x^3 + (c - 5)x^2 - 4x - c$$
therefore
$$c = 8$$?
and therefore:
$$P = -4$$
$$Q = -8$$

is that what you meant?

Thanks for your help! Means a lot!

 

[zgozvrm]

well then you get a+5 = 0.8a - 0.4 + 2a -1
a+5 = 2.8a - 1.4
1.8a - 6.4 = 0
1.8a = 6.4
a = 3.55555556

or

2a - 1 = .4a + 2 + a + 5
2a - 1 = 1.4a + 7
0.6a - 8 = 0
0.6a = 8
a = 13.3333333

therefore a can be 13.333 or 3.555 ?

Not quite...
You solve for the cases where one term is 40% of the other, not 40% greater then the other, as the question asked!

In other words, suppose you have a number (let's say 20) and you want to find the value that is 40% greater.
0.4 * 20 = 8 which is clearly not 40% greater. In fact, it is less!

What would you do to determine the value that is 40% greater than 20?

 

[I Like Pi]

Not quite...
You solve for the cases where one term is 40% of the other, not 40% greater then the other, as the question asked!

In other words, suppose you have a number (let's say 20) and you want to find the value that is 40% greater.
0.4 * 20 = 8 which is clearly not 40% greater. In fact, it is less!

What would you do to determine the value that is 40% greater than 20?

Well you would do .4 * 20 + 20, a value 40% greater then 20 is 28?

 

[zgozvrm]

Yes, but that simplifies...

0.4 * 20 + 20 = (0.4 + 1) * 20 = 1.4 * 20

Therefore, a number that is 40% larger than x is 1.4 * x

 

[94JZA80]

Not quite...
You solve for the cases where one term is 40% of the other, not 40% greater then the other, as the question asked!

In other words, suppose you have a number (let's say 20) and you want to find the value that is 40% greater.
0.4 * 20 = 8 which is clearly not 40% greater. In fact, it is less!

What would you do to determine the value that is 40% greater than 20?

well at first glance (and i do mean i only "glanced" at it for a split second), i thought the same thing...here are the equations he set up:

(a+5) = .4(2a-1) + 2a-1
.4(a+5) + (a+5) = 2a - 1

look closer at the right-hand side of the first equation. 0.4(2a-1) + 2a-1 = 1.4(2a-1), and the original equation becomes a+5 = 1.4(2a-1), which certainly implies that a+5 is 40% larger than 2a-1. likewise, we can see that the left-hand side of the 2nd equation, 0.4(a+5) + (a+5), equals 1.4(a+5). and so equation 2 becomes 1.4(a+5) = 2a-1, which certainly implies that 2a-1 is 40% greater than a+5.

Well you would do .4 * 20 + 20, a value 40% greater then 20 is 28?

your calculations are correct - you just chose to represent the quantity on one side ofthe equation as 0.4x + x instead of 1.4x.

 

[tiny-tim]

Hi I Like Pi!

(since you like them so much, have a pi: π )

c) given csc(6b + $$\frac{pi}{8}$$) = sec(2b - $$\frac{pi}{8}$$), solve for b.
…
c) i know that cscx = 1/sinx and secx = 1/cosx, i then cross multiplied to get $$cos(2b-pi/8) = sin(6b+pi/8)...$$

$$<br /> Yes, and now use cosθ = sin(π/2 - θ) to get that in the form sinA = sinB. <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" />$$

 

[zgozvrm]

My bad. I both misread your equations and miscalculated mine.

You are absolutely right!

 

[I Like Pi]

Hi I Like Pi!

(since you like them so much, have a pi: π )Yes, and now use cosθ = sin(π/2 - θ) to get that in the form sinA = sinB.

haha, thank you !

well i tried that... i used cos instead:
$$cos(2b-pi/8) = cos[pi/2-(6b+pi/8)]$$
$$cos(2b-pi/8) - cos[pi/2-(6b+pi/8)] = 0$$
$$cos[(2b-pi/8)-(pi/2-6b-pi/8)] = 0$$
$$8b-pi/2 = cos-inverse(0)$$
$$8b = pi$$
$$b = pi/8$$

it doesn't work

edit: I used geometer's sketchpad to graph the two, and the first point of intersection is (1.76715.., -1)...

 

[94JZA80]

what confuses me about the first question is that i am not so sure that a should have two distinct values. sure, if the text (or wherever the OP got the problem from) literally does not give any more information than what he posted (i.e. doesn't tell you which quantity is 40% greater than the other), then i can see how you would have to solve for a in two different situations, yielding two disticnt values for a (that is, when a+5 is 40% greater than 2a-1, AND when 2a-1 is 40% greater than a+5). either way, the OP has clearly shown us that he knows how to solve for a variable using arithmetic and algebraic manipulations, and has been able to determine a value for a regardless of which quantity is 40% larger than the other.

 

[I Like Pi]

well at first glance (and i do mean i only "glanced" at it for a split second), i thought the same thing...here are the equations he set up:


look closer at the right-hand side of the first equation. 0.4(2a-1) + 2a-1 = 1.4(2a-1), and the original equation becomes a+5 = 1.4(2a-1), which certainly implies that a+5 is 40% larger than 2a-1. likewise, we can see that the left-hand side of the 2nd equation, 0.4(a+5) + (a+5), equals 1.4(a+5). and so equation 2 becomes 1.4(a+5) = 2a-1, which certainly implies that 2a-1 is 40% greater than a+5.




your calculations are correct - you just chose to represent the quantity on one side ofthe equation as 0.4x + x instead of 1.4x.

what confuses me about the first question is that i am not so sure that a should have two distinct values. sure, if the text (or wherever the OP got the problem from) literally does not give any more information than what he posted (i.e. doesn't tell you which quantity is 40% greater than the other), then i can see how you would have to solve for a in two different situations, yielding two disticnt values for a (that is, when a+5 is 40% greater than 2a-1, AND when 2a-1 is 40% greater than a+5). either way, the OP has clearly shown us that he knows how to solve for a variable using arithmetic and algebraic manipulations, and has been able to determine a value for a regardless of which quantity is 40% larger than the other.

Thanks very much for your help! and yes, I guess it's just a question that could go both ways because it is written like that. Though I was in a rush to write it down, i hope i didn't write it wrong, but either way, it is pretty straight forward now that i think of it..

 

[zgozvrm]

oh, well i based it on the original... a = 5, b = 4, c = 3...

well, then if that's the case,

$$ax^4 + bx^3 + (c - a)x^2 - bx - c$$
$$=5x^4 + 4x^3 + (c - 5)x^2 - 4x - c$$
therefore
$$c = 8$$?
and therefore:
$$P = -4$$
$$Q = -8$$

is that what you meant?

Thanks for your help! Means a lot!

That's correct!

You can also attack this problem by doing the long division:

The dividend is [itex]5x^4 + 4x^3 + 3x^2 + Px + Q[/tex]<br /> The divisor is [itex]x^2 - 1[/tex]<br /> <br /> [itex]x^2[/tex] goes into [itex]5x^4[/tex]<br /> [itex]5x^2[/tex] times.<br /> (That is, [itex]x^2 \times 5x^2 = 5x^4[/tex])<br /> <br /> So multiply the divisor by [itex]5x^2[/tex] and you get [itex](5x^4 - 5x^2)[/tex]<br /> Subtract this from the dividend and you get [itex]4x^3 + 8x^2 + Px + Q[/tex]<br /> <br /> Now, determine how many times [itex]x^2[/tex] goes into [itex]4x^3[/tex]: that would be [itex]4x[/tex] times.<br /> Multiply [itex]4x \times (x^2 - 1) = (4x^3 - 4x)[/tex]<br /> <br /> Subtract this from [itex]4x^3 + 8x^2 + Px + Q[/tex] and you get [itex]8x^2 + (P + 4)x + Q[/tex]<br /> <br /> Continuing...<br /> [itex]8 \times (x^2 - 1) = (8x^2 - 8)[/tex]<br /> [itex]8x^2 + (P + 4)x + Q - (8x^2 - 8) = (P + 4)x + (Q + 8)[/tex]<br /> <br /> Since there is no remainder, (P + 4) and (Q + 8) must both be equal to 0.<br /> <br /> <br /> Quinzio's way is MUCH easier, I just wanted to offer an alternative way.[/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex]

 

[tiny-tim]

You're saying cos(A) - cos(B) = 0 if cos(A-B) = 0, which isn't correct.

Start again … draw a graph and look at it …

when does cos(A) = cos(B)?​

 

[94JZA80]

haha, thank you !

well i tried that... i used cos instead:
$$cos(2b-pi/8) = cos[pi/2-(6b+pi/8)]$$
$$cos(2b-pi/8) - cos[pi/2-(6b+pi/8)] = 0$$
$$cos[(2b-pi/8)-(pi/2-6b-pi/8)] = 0$$
$$8b-pi/2 = cos-inverse(0)$$
$$8b = pi$$
$$b = pi/8$$

it doesn't work

i think you're trying to manipulate the equation more than you have to. leave it in the following form:

sin(6b+pi/8) = cos(2b-pi/8)


then, using the identity cosθ = sin(π/2 - θ), you have sin(6b+pi/8) = sin(pi/2-(2b-pi/8)). therefore, 6b+pi/8 = pi/2-2b+pi/8. now solve for b.

 

[I Like Pi]

That's correct!

You can also attack this problem by doing the long division:

The dividend is [itex]5x^4 + 4x^3 + 3x^2 + Px + Q[/tex]<br /> The divisor is [itex]x^2 - 1[/tex]<br /> <br /> [itex]x^2[/tex] goes into [itex]5x^4[/tex]<br /> [itex]5x^2[/tex] times.<br /> (That is, [itex]x^2 \times 5x^2 = 5x^4[/tex])<br /> <br /> So multiply the divisor by [itex]5x^2[/tex] and you get [itex](5x^4 - 5x^2)[/tex]<br /> Subtract this from the dividend and you get [itex]4x^3 + 8x^2 + Px + Q[/tex]<br /> <br /> Now, determine how many times [itex]x^2[/tex] goes into [itex]4x^3[/tex]: that would be [itex]4x[/tex] times.<br /> Multiply [itex]4x \times (x^2 - 1) = (4x^3 - 4x)[/tex]<br /> <br /> Subtract this from [itex]4x^3 + 8x^2 + Px + Q[/tex] and you get [itex]8x^2 + (P + 4)x + Q[/tex]<br /> <br /> Continuing...<br /> [itex]8 \times (x^2 - 1) = (8x^2 - 8)[/tex]<br /> [itex]8x^2 + (P + 4)x + Q - (8x^2 - 8) = (P + 4)x + (Q + 8)[/tex]<br /> <br /> Since there is no remainder, (P + 4) and (Q + 8) must both be equal to 0.<br /> <br /> <br /> Quinzio's way is MUCH easier, I just wanted to offer an alternative way.[/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex]

[itex][itex][itex][itex][itex][itex][itex][itex][itex][itex][itex][itex][itex][itex][itex][itex]$<br /> Hey, thanks for the alternative! It makes much sense to me, especially with this, again, thank you for your time! <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" /> its amazing the help i get here <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" /><br /> <blockquote data-attributes="" data-quote="tiny-tim" data-source="post: 2945634" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> tiny-tim said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> You're saying cos(A) - cos(B) = 0 if cos(A-B) = 0, which isn't correct.<br /> <br /> Start again … draw a graph and look at it …<br /> <br /> <i><div style="margin-left: 20px">when does cos(A) = cos(B)?​</div></i> </div> </div> </blockquote><br /> well, i got that cos(A) = cos(B) (i think that's right)<br /> then you bring that over? So, cos(A)-cos(B) = 0 (am I right here?) and then you group similar terms, so cos(A-B) = 0? why isn't it correct? <br /> <br /> Thanks tim <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" /> i really appreciate your time!$[/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex][/itex]

 

[I Like Pi]

i think you're trying to manipulate the equation more than you have to. leave it in the form sin(6b+pi/8) = cos(2b-pi/8). then, sin(6b+pi/8) = sin(pi/2+(2b-pi/8)). if the sine of those quantities are equal, then the quantities themselves are equal. that is, if sin A = sin B, then A = B. likewise, if cos A = cos B, then A = B. therefore, 6b+pi/8 = pi/2+2b-pi/8. now just solve for b.

hey! Thank you so much! It makes a lot of sense! would this be the case any time you have same trig = same trig, you would cancel it out? so if tan(x) = tan(x), x = x and you would solve? Don't you have to do the inverse to get rid of it

again, i thank you dearly!

 

[tiny-tim]

well, i got that cos(A) = cos(B) (i think that's right)
then you bring that over? So, cos(A)-cos(B) = 0 (am I right here?) and then you group similar terms, so cos(A-B) = 0? why isn't it correct?

Well, for a start, if A = B, cos(A-B) = 1, isn't it?

Do what I said … draw a graph, and see when cosA = cosB !

(btw, it isn't only when A = B)

(and now I'm off to bed :zzz: …)