IIT JEE , AIEEE Forum - Mathematics , Algebra

serioussam303

Dear Sir/Madam,

I have some doubts in The Theory of Quadratic Equations.

A)) In an equation(more than 2 degree)....what is the sum and product of the roots.
Eg.,
in a 3 degree equation...if the roots are p,q,r.....
is it correct that
p+q+r=0....if yes..then why
also..what is p.q.r

B)) Please provide me with the solutions of the following questions in detail.

1)) if m and n are the roots of x²?p(x+1) ? c = 0,show that

(m+1)(n+1)=1-c . Hence prove that

[(m²+2m+1) / (m²+2m+c)]+ [(n²+2n+1) / (n²+2n+c)]=1

2)) Ramesh and Mahesh solve an equation. In solving Mahesh commits a mistake in the constant term and finds the roots to be 8 and 2. Mahesh commits a mistake in the coefficient of x and finds the roots to be -9 and -1. Find the correct roots.

3)) The coefficient of x in the quadratic equation x²+ px + q =0 was taken as 17 in place of 13,its roots were found to be -2 and -15.Find the roots of the original equation.

4)) The roots of 8x²? 10x + 3 = 0 are p and q²where q>1/2 then the equation whose roots are (p+iq) ¹⁰⁰ and (p-iq) ¹⁰⁰ is

(a)x²-x+1=0

(b)x²+x+1=0

(c)x²-x-1=0

(d)x²+x-1=0

5)) If m and n are the roots of the equation 6x² ? 6x +1 = 0, then prove that

½(p + qm + rm² + sm³) + ½( p + qn + rn² + sn³) = (p/1)+(q/2)+(r/3)+(s/4)

6)) If p and q are the roots of the equation x + 1 = rx(1 ? rx) and s , t be the two values of m determined from the equation (p/q)+(q/p) = ? ? 2 , show that

(s²/t²) + (t²/s²) + 2 = 4[(?+1)/( ?-1)]²

7)) Solve x3 ? 13x2 + 15x = 189 = 0 , if one root exceeds the other by 2.

8)) Solve x4 ? 2x2 + 8x ? 3 = 0

Dhanraj

catch_arnnie

for part A))

if you have a 3rd degree equation like ax³+bx²+cx+d=0 then, let the roots be p,q,r. so, sum of roots(p+q+r) will be the minus of coefficient of x² divided by coefficient x³ i.e. p+q+r= -b/a. which means sum of roots is not always zero(it's zero only if b=0)

moreover, the product of roots i.e. p.q.r=-d/a .

iitkgp_bipin

For a cubic equation :

ax³ + bx² + cx + d = 0 having roots p,q,r

p+q+r = -b/a

pq+qr+rp = c/a

pqr = -d/a

Please post your other queries on a new page.

aysh

B)
2.)Let the original equation be ax^2+bx+c=0.
Then,
-b/a = 8+2 = 10= sum of correct roots.
c/a = (-9).(-1) = 9=product of correct roots.
Clearly,
correct roots = 9,1.

PLEZZZZZ RATE...

sunayana_kushwaha

i also have same doubts as part A above

I wish to know that for an equation ax⁴+bx³+cx²+dx+e=0
if the roots are m₁,m_2.m₃m₄
what is m1+m2+m3+m4
and
m1m2+ m2m3+m3m4 + m4m1

m1m2m3 + m2m3m4 +m3m4m1

and

m1m2m3m4=?

if possible can anybody provide with general theory

aysh

3.)Given equation :x^2+px+q = 0.
The value of p was taken wrongly.
But, q remains the same.
Thus,
q = (-2).(-15) = 30 = product of correct roots.
Correct value of p = 13.
Sum of correct roots = -13.
Clearly,
correct roots = -10,-3.

PLEZZZZZ RATE...

aysh

Hi sunayana,
it's easy...
For a genral eq. ax^4+bx^3+cx^2+dx+e=0,
sum of roots

= m1+m2+m3+m4 = -b/a.
product of roots taken 2 at a time

= m1m2+m2m3+m3m4+m4m1+m2m4+m1m3 = c/a.
product of roots taken 3 at a time

=m1m2m3+m2m3m4+m3m4m1+m4m1m2 = -d/a.

product of roots = e/a.

Pleazzzzzzz rate....

Avinash_Bhat

I believe you got part A.

Now to part B :

1)

The equation given in the question must be : x² - p(x + 1) - c = 0

ie, x² - px - (p + c) = 0

If m and n are the roots, then (m + n) = p and mn = (- p - c)

So : (m + 1)(n + 1) = mn + (m + n) + 1 = (- p - c) + p + 1 = 1 - c.

2)

Let the equation be : x² + px + q = 0

One solves the equation : x² + px + s = 0 to get the roots 8 and 2.

So : p = -(8 + 2) = -10.

Other one solves the equation : x² + mx + q = 0 to get the roots -1 and -9

So : q = (-1) * (-9) = 9.

SO : The required equation is : x² - 10x + 9 = 0 whose roots are : 1 and 9

3)

As per the question : x² + 17x + q = 0 has the roots -2 and -15. So, here q = (-2) * (-15) = 30.

So, the required equation is : x² + 13x + 30 = 0 whose roots are -10 and -3.

4)

I think the equation given is : 8x² - 10x + 3 = 0 whose roots will be :

p = 1/2

q² = 3/4 so that q =

3 / 2 (q > 1/2)

ie, (p + iq) = ( 1/2 +

3i/2 ) = -

² -

and -

² are the complex

(p - iq) = ( 1/2 -

3i/2 ) = -

cube roots of -1

SO : (p + iq)¹⁰⁰ = (-

²)¹⁰⁰ =

²

(p - iq)¹⁰⁰ = (-

)¹⁰⁰ =

ie, The equation whose roots are

and

² is : x² + x + 1 = 0

" Thathwamasi "

Avinash_Bhat

Hey,
I think there is some mistake in the 5th question. Someone please try to answer it kya??.

parag14

B) 3)

Solutions obtained are -2 and -15
Hence by factor theorem, the equation is (x+2)(x+15)=0
=> x² + 17x + 30 = 0
But coefficient of x is 13
therefore the equation is x² + 13x + 30 = 0
=> (x + 10)(x + 3) = 0
=> x = -10 , -3

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