# Fp1/p4 - Q

WatchPage 1 of 1

Go to first unread

Skip to page:

This discussion is closed.

My exams finished later than anyone else in my school so i missed a few maths lessons. i need help on this question which i know nothing about, just to get me going.

By using substitution, find the equation whose roots are given by adding 2 to the roots of the equation

3x^3 - 13x^2 - 51x - 36 = 0

thanx

By using substitution, find the equation whose roots are given by adding 2 to the roots of the equation

3x^3 - 13x^2 - 51x - 36 = 0

thanx

0

Report

#2

Shifting the graph

y = 3x^3 - 13x^2 - 51x - 36

rightwards by 2 units gives

y = 3(x - 2)^3 - 13(x - 2)^2 - 51(x - 2) - 36

So the required equation is

3(x - 2)^3 - 13(x - 2)^2 - 51(x - 2) - 36 = 0

y = 3x^3 - 13x^2 - 51x - 36

rightwards by 2 units gives

y = 3(x - 2)^3 - 13(x - 2)^2 - 51(x - 2) - 36

So the required equation is

3(x - 2)^3 - 13(x - 2)^2 - 51(x - 2) - 36 = 0

0

what about this one? this is more simple.

the roots of x^2 - 3x + 1 = 0 are a and b. find the equation whose roots are a+1 and b+1, without finding the values of a and b.

the roots of x^2 - 3x + 1 = 0 are a and b. find the equation whose roots are a+1 and b+1, without finding the values of a and b.

0

Report

#4

(Original post by

what about this one? this is more simple.

the roots of x^2 - 3x + 1 = 0 are a and b. find the equation whose roots are a+1 and b+1, without finding the values of a and b.

**confused?**)what about this one? this is more simple.

the roots of x^2 - 3x + 1 = 0 are a and b. find the equation whose roots are a+1 and b+1, without finding the values of a and b.

if roots of new eqn are a+1,b+1

sum of roots = a+b+2=5

product of roots=(a+1)(b+1)=ab+1+a+b=1+3+1= 5

so eqn is

x^2-5x+5=0

0

Report

#5

**confused?**)

what about this one? this is more simple.

the roots of x^2 - 3x + 1 = 0 are a and b. find the equation whose roots are a+1 and b+1, without finding the values of a and b.

f(x) = 0 has 2 roots a, b

So f(x-1) = 0 will have 2 roots a+1 and b +1

0

thanx, im getting the hang of it.

solve this equation 32x^3 - 14x +3 = 0 given that one root is twice another.

how would you do this cubic

solve this equation 32x^3 - 14x +3 = 0 given that one root is twice another.

how would you do this cubic

0

Report

#7

(Original post by

thanx, im getting the hang of it.

solve this equation 32x^3 - 14x +3 = 0 given that one root is twice another.

how would you do this cubic

**confused?**)thanx, im getting the hang of it.

solve this equation 32x^3 - 14x +3 = 0 given that one root is twice another.

how would you do this cubic

f(x) = 0 has 3 roots a, b , c

So f(x-1) = 0 has 3 roots a+1, b+1, c+1

and f(x-1) = 32(x-1)^3 - 14(x-1) + 3

-> that gives you the equation.

0

(Original post by

Still like that.

f(x) = 0 has 3 roots a, b , c

So f(x-1) = 0 has 3 roots a+1, b+1, c+1

and f(x-1) = 32(x-1)^3 - 14(x-1) + 3

-> that gives you the equation.

**BCHL85**)Still like that.

f(x) = 0 has 3 roots a, b , c

So f(x-1) = 0 has 3 roots a+1, b+1, c+1

and f(x-1) = 32(x-1)^3 - 14(x-1) + 3

-> that gives you the equation.

0

Report

#9

(Original post by

am i supposed to substitute.

**confused?**)am i supposed to substitute.

0

(Original post by

Hmm, it's a long way. What if they give you a question like x^5 + 4x^4 +7x^3 +...???

**BCHL85**)Hmm, it's a long way. What if they give you a question like x^5 + 4x^4 +7x^3 +...???

__p4__. it only goes up to cubic equations. do you know the method p4 students should be using

0

Report

#11

(Original post by

im only doing

**confused?**)im only doing

__p4__. it only goes up to cubic equations. do you know the method p4 students should be using
0

**BCHL85**)

Still like that.

f(x) = 0 has 3 roots a, b , c

So f(x-1) = 0 has 3 roots a+1, b+1, c+1

and f(x-1) = 32(x-1)^3 - 14(x-1) + 3

-> that gives you the equation.

i still dont get it mate, sorry. How wold you solve this??

solve this equation 32x^3 - 14x +3 = 0 given that one root is twice another.

0

(Original post by

i still dont get it mate, sorry. How wold you solve this??

solve this equation 32x^3 - 14x +3 = 0 given that one root is twice another.

**confused?**)i still dont get it mate, sorry. How wold you solve this??

solve this equation 32x^3 - 14x +3 = 0 given that one root is twice another.

why are the roots a+1 when it says twice another root?

0

Report

#14

(Original post by

why are the roots a+1 when it says twice another root?

**confused?**)why are the roots a+1 when it says twice another root?

So if f(x) = 0 has 3 roots a, b, c, f(x/2) = 0 will have 3 roots 2a, 2b, 2c.

0

Report

#15

**confused?**)

thanx, im getting the hang of it.

solve this equation 32x^3 - 14x +3 = 0 given that one root is twice another.

how would you do this cubic

32(x-a)(x-2a)(x-b) = 32(x^3 - (3a+b)x^2 + (2a^2+3ab)x - 2a^2b).

Comparing the x^2 coefficients gives

3a+b = 0.

Comparing constant coefficients gives

-64a^2 b = 3

Putting b = - 3a gives

a^3 = 1/64

So a= 1/4 and b = -3/4.

Just to be on the safe side we can check

2a^2+3ab = 2/16 - 9/16 = -7/16 = -14/32 as required.

So the roots are 1/4, 1/2, -3/4 and the original cubic factors as

(4x-1)(2x-1)(4x+3)

0

X

Page 1 of 1

Go to first unread

Skip to page:

new posts

Back

to top

to top