I need the helps for Maths Ext.1 exam revision. (1 Viewer)

[Forbidden.]

I gave up on these revision questions.

1.

Find the value of k in x²+3x+k-1=0 if the quadratic equation has reciprocal roots.

2.

The polynomial f(x)=ax²+bx+c has zeros 4 and 5, and f(-1)=60. Evaluate a, b and c.

3.

For 1. xy=7 and 3x-5y-1 & 2. y=x² and y=4x-4,
find the number of points of intersection: (i) 2 points (ii) 1 point (iii) no points

4.

Evaluate

(If you can't see, it's k=4 below sigma and 77 above sigma)

 

[~shinigami~]

I gave up on these revision questions.

1.

Find the value of k in x²+3x+k-1=0 if the quadratic equation has reciprocal roots.

reciprocal roots means,

αβ = 1

c/a = 1

k-1 = 1

k = 2

Is that right, I haven't done polynomials in a while.

 

[jyu]

I gave up on these revision questions.

4.

Evaluate

(If you can't see, it's k=4 below sigma and 77 above sigma)

It is an arithm. series of n = 74 terms, first term (k = 4) is a = 19, common difference is d = 6.

Sum of the 74 terms = (n/2)[2a + (n-1)d] = 17612

 

[bos1234]

how did u copy that sigma notation onto this site?

 

[Shrikar]

"The polynomial f(x)=ax<SUP>2</SUP>+bx+c has zeros 4 and 5, and f(-1)=60. Evaluate a, b and c."

Okay zeros mean where the polynomial crosess the x-axis, so sub in f(4) and f(5) and you get two equations. Then you use the information that f(-1)=60 to form a third equation. This is the tricky bit, you have to simultaneously solve the equations. Hope that helps.

 

[~shinigami~]

how did u copy that sigma notation onto this site?

He used an image.

 

[bos1234]

and how did he uplaod it?

 

[pLuvia]

You can use various image uploading sites one such as photobucket and then there is an IMG tag which you can just paste onto a post

 

[pyrodude1031]

1.

product of roots = c/a = k-1 = 1 Hence, k=2.

2.

f(4)=0 ... 16a+4b+c=0 (1)
f(5)=0 ... 25a+5b+c=0 (2)
f(-1)= 60=a-b+c ... (3)

(1)-(3) ... 15a+5b=-60 i.e. b=-12-3a ... (4)
(2)-(3) ... 24a+6b=-60 ... (5)

Put (4) into (5)
24a + 6(-12-3a)=-60
6a=12 ... a=2 ... b=-12-6 = -18 c=60-18-2=40

3.

xy=7 3x-5y-1=0
Put y=7/x into 3x-5y-1=0
3x-5(7/x)-1=0
3x^2-x-35=0
discriminant=(-1)^2-4(3)(-35)=421>0 Hence, two PoI.

y=x^2 y=4x-4
Equate y
x^2=4x-4
x^2-4x+4=0
discrimanant=(-4)2-4(4)=0 Hence, curves touch i.e. 1 PoI . (Strictly speaking, when we say curves touch, they DO NOT intersect. So, there 'really isnt' such thing called "one point of intersection." ..

4.

Sum (k=4 to 77) 6k-5
=6[Sum (k=4 to 77) k] - Sum(k=4 to 77) 5
=6{(74/2)[4+(74-1)(1)]} -5(74)
=16724

Step one to Step two ... relation is simple to prove... and its similar to those relations involving limits and integrals...


BTW... You should not have given up with these revision questions.. they are quite simple .... esp the second one ...

 

[Forbidden.]

It is an arithm. series of n = 74 terms, first term (k = 4) is a = 19, common difference is d = 6.

Sum of the 74 terms = (n/2)[2a + (n-1)d] = 17612

I can't trust the formula, it got me a wrong answer last time.

 

[Forbidden.]

Thanks for the stuff guys, I managed to work it out myself.
By the way pyrodude1031, your answer is wrong in Q4, the answer in the solutions is 17612 which I shall demonstrate later.
You are correct they are a little easy, I just need to spend more time.

1. shinigami, did you mean:

(1/α) x (1/β) = c/a

(βα/αβ) = (k-1/1)

1 = k - 1

.: k = 2

2. This one is something I only remembered for a short time only.

a(4)² + b(4) + c = 0
a(5)² + b(5) + c = 0
a(-1) + b(-1)+c = 60

16a + 4b + c = 0 (1)
25a + 5b + c = 0 (2)
a - b + c = 60 (3)

(1) - (3)

15a + 5b = - 60
3a + b = -12
b = -12 - 3a (4)

(2) - (3)

24a + 6b = -60 (5)

(4) into (5)

24a + 6(-12 - 3a) = -60
24a - 72 -18a = -60
24 - 18a =12
6a = 12

.: a = 2

.: b = -12-3(2)
= -18

.: c = 60 - a + b
= 60 (2) + (-18)
=40

3.

xy = 7
3x - 5y - 1 =0
3x - 5(7/x) - 1 = 0
3x - 35/x - 1 = 0
3x² - 35 - x = 0
3x² - x - 35 = 0

Δ = b² - 4ac
=(-1)² - 4(3)(-35)
=421>0

y = x²
y = 4x - 4

But y = x²

(x²) = 4x - 4

x² - 4x + 4 = 0

Δ = b² - 4ac
=(-4)² - 4(1)(4)
=0 = 0

4.

Let S_i be initial sum.
Let S_o be omitted sum.
Let S_f be final sum.

S_f = S_i - S_o

S_n = (n/2)(a+l)

a_i = 6(77) - 5 = 457
l_i = 6(1) - 5 = 1

S_i = (77/2)[(1)+(457)]
= 17633

We only start from four, calculate the omitted sum from the beginning.

a_o = 6(3) - 5 = 15
l_o = 6(1) - 5 = 1

S_o = (3/2)[(1)+(13)]
= 21

But S_f = S_i - S_o

17633 - 21 = 17612

 

[~shinigami~]

Thanks for the stuff guys, I managed to work it out myself.
By the way pyrodude1031, your answer is wrong in Q4, the answer in the solutions is 17612 which I shall demonstrate later.
You are correct they are a little easy, I just need to spend more time.

1. shinigami, did you mean:

(1/α) x (1/β) = c/a

(βα/αβ) = (k-1/1)

1 = k - 1

.: k = 2

Nah I meant what I posted which is:

αβ = 1

c/a = 1

k-1 = 1

k = 2

Although I'm not sure what you mean.

 

[Forbidden.]

Nah I meant what I posted which is:

αβ = 1

c/a = 1

k-1 = 1

k = 2

Although I'm not sure what you mean.

I didn't know βα/αβ cancel out to form 1, I must have fell asleep during lessons in junior school on those more basic stuff. It's the "αβ = 1" that confused me the most.