roots, finding angle (1 Viewer)

[adidasboy]

Got some questions I need help with:

A) If A and B are roots of 2x^2-2x-6=0, find the value of (A+1)^-1 + (B+1)^-1.

B) Factorise fully 1-64d^6f^6.

C) Solve 2tan x+1 =0 for 0_<x_<360 degrees.

D) Find the acute angle between y=2x-6 and x-3y+7=0. << are you ment to graph it to work it out?



thanks in advance

 

[Aerath]

A) Roots are 1 and -3 (by using alpha + beta = -b/a and alpha*beta = c/a)
B) Not sure
C) tanx = -1/2. Use your calculator
D) No, no need for graph, use the gradients, and the angle between two lines formula. Gradient of the first line is 2, and the second is 1/3.

 

[lyounamu]

Got some questions I need help with:

A) If A and B are roots of 2x^2-2x-6=0, find the value of (A+1)^-1 + (B+1)^-1.

B) Factorise fully 1-64d^6f^6.


C) Solve 2tan x+1 =0 for 0_<X_<360 p degrees.<>
D) Find the acute angle between y=2x-6 and x-3y+7=0. << are you ment to graph it to work it out?



thanks in advance

a) 2x^2 - 2x - 6 =0
Therefore, a=2, b=-2 and c=-6
A+B = -b/a = 2/2 = 1
AB = c/a = -6/2 = -3

Now, (A+1)^-1 + (B+1)^-1 = 1/(A+1) + 1(B+1)
= (B+1)/((A+1)(B+1)) + (A+1)/((A+1)(B+1))
= (A+B+2)/((A+1)(B+1))
= (A+B+2)/(AB + A + B +1)
= (1 + 2)/(-3 + 1 + 1)
= 3/-1
= -3


b) 1-64d^6f^6 = (1^2)^3 - (4d^2f^2)^3
= 1^3 - (4d^2f^2)^3
= (1-4d^2f^2)(1 +4d^2f^2 +16d^4f^4)
= (1+2df)(1-2df)(1+4d^2f^2 + 16d^4f^4)

c) 2tanx +1 = 0
2tanx = -1
tanx = -1/2
x = -26.5650511... degrees
= -26 degrees 34 minutes (to nearest minutes).
If there is a domain, you should follow the domain by adjusting this angle to the positive.

d) First equation: y = 2x -6
Therefore, m1 = 2
Second equation: x - 3y +7 = 0
3y = x +7
y = x/3 + 7/3
Threfore, m2 = 1/3

Now, use the formula: tan @ = (absolute value of (m1 -m2))/(1+ m1m2)
tan @ = (5/3)/(5/3)
tan @ = 1

Therefore, @ = 45 degrees

 

[tommykins]

Got some questions I need help with:

A) If A and B are roots of 2x^2-2x-6=0, find the value of (A+1)^-1 + (B+1)^-1.

B) Factorise fully 1-64d^6f^6.

C) Solve 2tan x+1 =0 for 0_<x_<360 degrees.

D) Find the acute angle between y=2x-6 and x-3y+7=0. << are you ment to graph it to work it out?



thanks in advance

a seems to be done by evyerone already.

b) 1-64d^6f^6. let m = d³ and n = f³

eqn becomes 1-64m²n² = difference of two squares

1-64m²n² = (1-8mn)(1+8mn) = (1-8d³f³)(1+8d³f³) = (1-2df)(1+2df+d²f²)(1+8d³f³)

2tan x + 1 = 0
2tanx = -1
tanx = -1/2

x = tan^-1 (-1/2) there should be 2 solutions as it is 0 < x < 360.

D) Find the acute angle between y=2x-6 and x-3y+7=0. << are you ment to graph it to work it out?

Solve simulatneously to find point of interesection.

(1) y = 2x-6
(2) x - 3y + 7 = 0 -> 3y = x + 7 -> y = (x+7)/3

1 into 2.

x-3(2x-6) + 7 =0
x - 6x + 18 + 7 = 0
-5x +25 = 0
-5x = -25
x = 5.

y = 2(5) - 6 = 4.

P(5,4).

Gradient of (1) is 2, gradient of (2) is 1/3.

The formula is tan @ = | m1 - m2 / 1+m1m2 | = | 2-(1/3) / 1+(2*1/3) | = 1.

Thus @ = tan^-1 1 = 45 degrees.

EDIT : no idea why i found the point of intersection.

PS. lyounamu, check my solution for b), yours seems realyl messy.

 

[lyounamu]

a seems to be done by evyerone already.

b) 1-64d^6f^6. let m = d³ and n = f³

eqn becomes 1-64m²n² = difference of two squares

1-64m²n² = (1-8mn)(1+8mn) = (1-8d³f³)(1+8d³f³) = (1-2df)(1+2df+d²f²)(1+8d³f³)

2tan x + 1 = 0
2tanx = -1
tanx = -1/2

x = tan^-1 (-1/2) there should be 2 solutions as it is 0 < x < 360.

D) Find the acute angle between y=2x-6 and x-3y+7=0. << are you ment to graph it to work it out?

Solve simulatneously to find point of interesection.

(1) y = 2x-6
(2) x - 3y + 7 = 0 -> 3y = x + 7 -> y = (x+7)/3

1 into 2.

x-3(2x-6) + 7 =0
x - 6x + 18 + 7 = 0
-5x +25 = 0
-5x = -25
x = 5.

y = 2(5) - 6 = 4.

P(5,4).

Gradient of (1) is 2, gradient of (2) is 1/3.

The formula is tan @ = | m1 - m2 / 1+m1m2 | = | 2-(1/3) / 1+(2*1/3) | = 1.

Thus @ = tan^-1 1 = 45 degrees.

EDIT : no idea why i found the point of intersection.

PS. lyounamu, check my solution for b), yours seems realyl messy.

You made a mistake in the question (b).

 

[tommykins]

Where? highlight it please

EDIT - yes i did, i apoligise. did it on paper.

full solution is

1-64d^6f^6

= (1- 2df)(1+2df+4d²f²)(1+2df)(1-2df+4d²f²)

If it's wrong again, damn I need to review yr 11 stuff

 

[lyounamu]

a seems to be done by evyerone already.

b) 1-64d^6f^6. let m = d³ and n = f³

eqn becomes 1-64m²n² = difference of two squares

1-64m²n² = (1-8mn)(1+8mn) = (1-8d³f³)(1+8d³f³) = (1-2df)(1+2df+d²f²)(1+8d³f³)

2tan x + 1 = 0
2tanx = -1
tanx = -1/2

x = tan^-1 (-1/2) there should be 2 solutions as it is 0 < x < 360.

D) Find the acute angle between y=2x-6 and x-3y+7=0. << are you ment to graph it to work it out?

Solve simulatneously to find point of interesection.

(1) y = 2x-6
(2) x - 3y + 7 = 0 -> 3y = x + 7 -> y = (x+7)/3

1 into 2.

x-3(2x-6) + 7 =0
x - 6x + 18 + 7 = 0
-5x +25 = 0
-5x = -25
x = 5.

y = 2(5) - 6 = 4.

P(5,4).

Gradient of (1) is 2, gradient of (2) is 1/3.

The formula is tan @ = | m1 - m2 / 1+m1m2 | = | 2-(1/3) / 1+(2*1/3) | = 1.

Thus @ = tan^-1 1 = 45 degrees.

EDIT : no idea why i found the point of intersection.

PS. lyounamu, check my solution for b), yours seems realyl messy.

Red line & answer

 

[tommykins]

I've just gotten out a simpler solution -

1-64f^6d^6= (1- 2df)(1+2df+4d²f²)(1+2df)(1-2df+4d²f²)

part of me feels like its wrong however, haven't facotirsed cubics in ages.

 

[lyounamu]

I've just gotten out a simpler solution -

1-64f^6d^6= (1- 2df)(1+2df+4d²f²)(1+2df)(1-2df+4d²f²)

part of me feels like its wrong however, haven't facotirsed cubics in ages.

First & third ones are right. 2nd and 4th ones are wrongly factorised.
Just leave it as 1+4d^2f^2+16d^4f^4 (sorry for messy answer)

 

[tommykins]

if it's (1+8d³f³)(1-8d³f³) you can factorise both of them out, which is what the question is asking.
triple checked- my answer is not wrong.

 

[lyounamu]

if it's (1+8d³f³)(1-8d³f³) you can factorise both of them out, which is what the question is asking.
triple checked- my answer is not wrong.

Ok. I can see what you did.

 

[tommykins]

thanks for pointing out my first error however =)

 

[Aerath]

Lyounamu, for question A, his equation was 2x^2 - 2x - 6 = 0. Not 2x^2 - 3x - 6 =0

 

[lyounamu]

Lyounamu, for question A, his equation was 2x^2 - 2x - 6 = 0. Not 2x^2 - 3x - 6 =0

......

I better get my glasses back from the rubbish bin.

Thanks by the way, I fixed it.

 

[Aerath]

Still wrong. Alpha*Beta = -3, not positive three.

 

[adidasboy]

For B is = -26 degrees 34 minutes (to nearest minutes). the final answer?
Do i have to do the A, S, T, C quadrant thing?

 

[lyounamu]

For B is = -26 degrees 34 minutes (to nearest minutes). the final answer?
Do i have to do the A, S, T, C quadrant thing?

Were you given a domain? Then you should follow it (by doing A, S, T C quadrant thing). If not, just leave the answer like that.

 

[lyounamu]

Still wrong. Alpha*Beta = -3, not positive three.

Double failure.

What goes around, comes around (I pointed out tommykins' mistake and you pointed out my mistake, twice!).

 

[adidasboy]

Prove the Identities:
A) (a+b+c)(ab+bc+ca)-abc=(a+b)(b+c)(c+a)
B) (ax+by)^2+(ay-bx)^2+c^2(x^2+y^2)=(x^2+y^2)(a^2+b^2+c^2)

How do I work these ones out?

Am I meant to just simplify LHS & RHS. And they will equal?



C) If 2x =a+b+c, show that (x-a)^2+(x-b)^2+(x-c)^2+x^2=a^2+b^2+c^2

Thanks in advance.

 

[Aerath]

For B is = -26 degrees 34 minutes (to nearest minutes). the final answer?
Do i have to do the A, S, T, C quadrant thing?

Your domain is 0 < x < 360.

Therefore: your answer will be 153*26* and 333*26*.