# Wipro Sample Questions On Roots Of An Equation

Mangesh Shinde
13:57:00
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**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

(p

^{2}- 1) (p

^{2}- 9) = 9

p

^{4}- 10p

^{2}+ 9 = 9

p

^{2}(p

^{2}- 10) = 0

p

^{2}=0 or p

^{2}-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

(p

^{2}- 4) (p

^{2}- 16) = 64

p

^{4}- 20p

^{2}+ 64 = 64

p

^{4}- 20p

^{2}= 0

p

^{2}(p

^{2}- 20) = 0

p

^{2}= 0 or p

^{2}-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[(2

^{2})-(sqrt(20))

^{2}] / 27 = -32/27

Hence the answer is -32/27.

**Question 3**

Find the product of the roots of x

^{3}-7x

^{2}+13x-7 = 0

a)1 b)3 c)5 d)7

**Answer :**d)7

Solution:

Let f(x)= x

^{3}- 7x

^{2}+ 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 x

^{3}-7x

^{2}+13x-7 can be written as(x-1)Q(x). We shall find Q(x) by dividing x

^{3}-7x

^{2}+13x-7 by (x - 1) as follows :

x-1) xSo Q(x) = x^{3}-7x^{2}+13x-7 (x^{2}-6x+7 is Q(x) by division. x^{3}-x^{2}------------- -6x^{2}+13x -6x^{2}+ 6x ------------ 7x-7 7x-7 ------------ 0

^{2}- 6x + 7.

Therefore f(x) = (x-1) (x

^{2}- 6x + 7) .

So the other two roots can be determined from the quadratic x

^{2}- 6x + 7= 0.

(Note : Roots x1 and x2 of any equation of the form ax

^{2}+bx+c = 0 are given by the formula. x1 = (-b + sqrt(b

^{2}- 4ac))/2a and x2 = (-b - sqrt(b

^{2}- 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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