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#1 2013-06-22 01:04:52

niharika_kumar
Member
From: Numeraland
Registered: 2013-02-12
Posts: 1,028

quadratic equations

well i have 3 questions here which are prove based

Q.1.if the roots of the equation (x+a)(x+b)+(x+b)(x+c)+(x+c)(x+a)=0 be equal,prove that 2/b=1/a+1/c .

Q.2. prove that the equation x^2(a^2+b^2)+2x(ac +bd)+(c^2+d^2)=0 has no real roots,if 'ad' is not equal to 'bc'.

Q.3. if p , q, r and s are real numbers such that pr=2(q+s), then show that atleast one of the equations x^2+px+q=0 and x^2+rx+s=0 has real roots.


these questions are given in srijan mathematics for class 10.please help me solve these!


friendship is tan 90°.

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#2 2013-06-22 04:11:55

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,531

Re: quadratic equations

hi niharika_kumar

I think these problems all need the same idea from quadratic equations:

This comes straight from the quadratic formula.  If the square root part exists and is not zero then

D>0 means two real roots

If it is zero then

D=0 means just one real root


And if there is no real square root

D<0 means no real roots.

Q2. Find out D and consider what is needed for real roots.  You should find D can never be > 0 and is equal to 0 for that condition.

Q3.  (p + r)^2 ≥ 0  but pr = 2(q+s)

So consider what happens if p^2 < 4q AND r^2 < 4s .... add these two equations, make use of the above and you should arrive at something that can never be true.  Therefore one or other equation has real roots.

Working on Q1   smile

Several hours later:

Q1 has me beat at the moment.  Here's my analysis; maybe someone can spot an error.

For equal roots

So I plotted the graph

for a variety of values of a and c trying to find some x-axis crossing points to get a 'b' for my chosen 'a' and 'c'

I couldn't get the graph to cross the x-axis for any a and c values  ???

So I thought I'd apply the D = rule to this quadratic.

This will always be negative unless a = c

In that one case

The original certainly has equal roots in this case and the required result also follows.  As far as I can tell there are no other real solutions.

Help sad

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#3 2013-06-22 08:23:45

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,531

Re: quadratic equations

Thought I'd start a fresh post rather than adding to the above.

This question is getting more and more bizarre.

The original equation is 'symmetrical' in a, b and c.

By which I mean, if you replace a by b, b by c, and c by a, you get the same equation back.  Similarly, a by c, c by b, b by a.

So let's pretend we have a way to prove:

Then it must also be possible to prove

Subtracting (ii) from (i)

and by the symmetry that must work for c too, so a = b = c.

So <equal roots> => a = b = c     AND    <expression to prove>  => a = b = c

Only trouble is,  this still gets me no nearer to <equal roots> => <expression to prove>

dizzy dizzy dizzy

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#4 2013-06-22 20:00:26

niharika_kumar
Member
From: Numeraland
Registered: 2013-02-12
Posts: 1,028

Re: quadratic equations

thnx for the solution !
but pls give the solutions for the other ques too ,i m nt able to solve them.


friendship is tan 90°.

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#5 2013-06-22 20:07:54

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,531

Re: quadratic equations

hi niharika_kumar

Did you read my first post?

I have suggested a way for you to solve these.  If you are unclear please post back with what you have tried and where you are stuck.

Q1 has caused a lot of discussion.  Please check the exact wording.  It would help to know if a, b and c are {real numbers} or if they may be {complex}.

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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