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Wipro Sample Questions On Roots Of An Equation

Question 1
What is the sum of the irrational roots of the equation (x-1)(x-3)(x-5)(x-7)=9 ?
a)10 b)8 c)6 d)4
Answer : b)8
Solution:
Given that
(x - 1) (x - 3) (x - 5) (x - 7) = 9
Let x - 4 = p
Then the given eqn becomes
(p + 3) (p + 1) (p - 1) (p - 3) = 9
(p2 - 1) (p2 - 9) = 9
p4 - 10p2 + 9 = 9
p2 (p2 - 10) = 0
p2 =0 or p2-10 =0
p = 0 or p = sqrt(10) or p = - sqrt(10)
then x - 4 = 0, x - 4 = sqrt(10) or x - 4 = - sqrt(10)
Now the roots of the given eqn are 4,4 + sqrt(10) and 4 - sqrt(10)
The irrational roots are 4+sqrt(10) and 4 - sqrt(10)
The sum of the irrational roots = 4 + sqrt(10) + 4 - sqrt(10) = 8.
Hence the answer is 8.
Question 2
The product of the distinct roots of the equation (3x)(3x+2)(3x-4)(3x-6)= 64 is:
a)-32/27 b)-61/5 c)63/16 d)69/12
Answer : a)-32/27
Solution:
Given that
(3x)(3x+2)(3x-4)(3x-6)= 64
Let 3x - 2 = p
Then the given eqn becomes
(p + 2) (p + 4) (p - 2) (p - 4) = 64
(p2 - 4) (p2 - 16) = 64
p4 - 20p2 + 64 = 64
p4 - 20p2 = 0
p2(p2 - 20) = 0
p2 = 0 or p2-20 = 0
p = 0 or p = sqrt(20) or p = - sqrt(20)
then 3x - 2 = 0, 3x - 2 = sqrt(20) or 3x - 2 = - sqrt(20)
and x=2/3, x=[2 + sqrt(20)] / 3 or x = [2 - sqrt(20)] / 3
Now the distinct roots of the given eqn are 2/3, [2 + sqrt(20)] / 3 and [2 - sqrt(20)] / 3
The product of the distinct roots = 2/3 x [2 + sqrt(20)] / 3 x [2 - sqrt(20)] / 3 = 2[(22)-(sqrt(20))2] / 27 = -32/27
Hence the answer is -32/27.
Question 3
Find the product of the roots of x3-7x2+13x-7 = 0
a)1 b)3 c)5 d)7
Answer : d)7
Solution:
Let f(x)= x3 - 7x2 + 13x - 7
Obviously a rational root is 1 , as f(1) = 1-7+13-7= 0. (i.e x = 1 satisfies the equation perfectly)
So x3-7x2+13x-7 can be written as(x-1)Q(x). We shall find Q(x) by dividing x3-7x2+13x-7 by (x - 1) as follows :
x-1) x3-7x2+13x-7 (x2-6x+7 is Q(x) by division.
     x3-x2
     -------------
    -6x2+13x
     -6x2+ 6x
     ------------
           7x-7 
           7x-7
     ------------
           0
So Q(x) = x2 - 6x + 7.
Therefore f(x) = (x-1) (x2 - 6x + 7) .
So the other two roots can be determined from the quadratic x2 - 6x + 7= 0.
(Note : Roots x1 and x2 of any equation of the form ax2+bx+c = 0 are given by the formula. x1 = (-b + sqrt(b2 - 4ac))/2a and x2 = (-b - sqrt(b2 - 4ac))/2a. Refer http://www.teacherschoice.com.au/maths_library/algebra/alg_6.htm for more information.)
So, for our equation, x1 = {-(-6)+sqrt((-6)2-4*7)} / 2 = (3+sqrt(2))is the positive irrational root.
x2 = (3-sqrt(2)) is another positive irrational root.
x = 1 is the positive rational root.
Now the product of the roots = 1(3+sqrt(2))(3-sqrt(2))= (9 - 2)= 7
Hence the answer is 7.

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