Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    4
    ReputationRep:
    Given that A and B are roots of the equation x^2+3x-6=0, find a quadratic equation with integer coefficients whose roots are 2/A and 2/B.

    This is what I done so far:

    A= \frac{-3+\sqrt(33)}{2}   , B=\frac{-3-\sqrt(33)}{2}

    ax^2+bx+c=0

    x_1=\frac{4}{\sqrt(33)-3}   , x_2=\frac{4}{-\sqrt(33)-3}

    \frac{4}(\sqrt(33)-3}+{4}{-\sqrt(33)-3}=\frac{-b}{a}
    \frac{4}{\sqrt(33)-3} \times \frac{4}{-\sqrt(33)-3}=\frac{c}{a}

    Forgot what to do now
    Offline

    16
    ReputationRep:
    The question use sum and product surely

    2/a +2/b = 2(a+b)/ab

    2/a * 2/b = 4/ab

    From the original

    a+b = -3
    ab = -6
    • Study Helper
    Offline

    13
    Study Helper
    (Original post by Prokaryotic_crap)
    ...
    Your method of solving it is harder than it needs to be.

    Recall a quadratic can be written as x^2-("sum of roots") x + "product of roots"=0

    So, A+B = ..., AB=...

    Now do the sum and product of roots for your new quadratic.
    Offline

    0
    ReputationRep:
    How about you just expand out  (x - 2/A)(x - 2/B) ?
    Offline

    2
    ReputationRep:
    Try supposing that your expression is equal to (X-A)(X-B) and see what you can discover.
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    You don't need to solve A and B.

    You should know facts about the roots of polynomials.

    A+B=-\frac{b}{a}

    AB=\frac{c}{a}

    So write down A+B and AB.

    Now consider \frac{2}{A}+\frac{2}{B}

    Can you rewrite this using A+B and AB?

    Finally consider \frac{2}{A} \times \frac{2}{B}

    Rewrite this using AB.

    Now for your final answer. The quadratic is:

    x^2 -(\frac{2}{A}+\frac{2}{B})x+ (\frac{2}{A} \times \frac{2}{B})=0

    You will need to multiply the whole equation by 3 to obtain integer coefficients.
    • Thread Starter
    Offline

    4
    ReputationRep:
    (Original post by Mr M)
    You don't need to solve A and B.

    You should know facts about the roots of polynomials.

    A+B=-\frac{b}{a}

    AB=\frac{c}{a}

    So write down A+B and AB.

    Now consider \frac{2}{A}+\frac{2}{B}

    Can you rewrite this using A+B and AB?

    Finally consider \frac{2}{A} \times \frac{2}{B}

    Rewrite this using AB.

    Now for your final answer. The quadratic is:

    x^2 -(\frac{2}{A}+\frac{2}{B})x+ (\frac{2}{A} \times \frac{2}{B})

    You will need to multiply the whole equation by 3 to obtain integer coefficients.
    ok, this is what I got:

    A+B= -3,AB=-6

    \frac{2}{A}+\frac{2}{B}=\frac{2B  +2A}{AB}=\frac{2(A+B)}{AB}=\frac  {2(-3)}{(-6)}=1

    \frac{2}{A} \times \frac{2}{B}=\frac{4}{AB}=\frac{4  }{(-6)}=\frac{2}{(-3)}

    x^2+x-\frac{2}{3}=0

    3x^2+3x-2=0

    correct?
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by Prokaryotic_crap)
    ok, this is what I got:

    A+B= -3,AB=-6

    \frac{2}{A}+\frac{2}{B}=\frac{2B  +2A}{AB}=\frac{2(A+B)}{AB}=\frac  {2(-3)}{(-6)}=1

    \frac{2}{A} \times \frac{2}{B}=\frac{4}{AB}=\frac{4  }{(-6)}=\frac{2}{(-3)}

    x^2+x-\frac{2}{3}=0

    3x^2+3x-2=0

    correct?
    Look at the coefficient of x.

    Remember the sum of the roots is =-\frac{b}{a}
    Offline

    16
    ReputationRep:
    (Original post by Mr M)
    Look at the coefficient of x.

    Remember the sum of the roots is =- \frac{b}{a}
    Ftfy too late
    • Study Helper
    Offline

    0
    ReputationRep:
    Study Helper
    Did anyone mention the substitution method?

    If x^2+3x-6=0 has roots A and B then (2/u)^2+3(2/u)-6=0 has roots 2/A and 2/B.

    Multiply through by u^2 to finish.
    • Study Helper
    Offline

    13
    Study Helper
    (Original post by BabyMaths)
    ...
    Very nice - PRSOM.
    • Thread Starter
    Offline

    4
    ReputationRep:
    (Original post by Mr M)
    Look at the coefficient of x.

    Remember the sum of the roots is =-\frac{b}{a}
    what do you mean? b = 3 a = 1 so -b/a = -3?
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by Prokaryotic_crap)
    what do you mean? b = 3 a = 1 so -b/a = -3?
    Do you remember doing this in A Level Further Maths?

    I'm talking about the second equation (the one you are forming). Your sign is wrong for the coefficient of x.
    • Thread Starter
    Offline

    4
    ReputationRep:
    (Original post by Mr M)
    Do you remember doing this in A Level Further Maths?

    I'm talking about the second equation (the one you are forming). Your sign is wrong for the coefficient of x.
    yh, i remember?

    so, 3x^2-3x-2=0?

    I though the general form was ax^2+bx+c=0?

    edit: Just spotted it!
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.