Question

If a and b are roots of q.P 2x2+7x+5 =0 then write a q.E whose roots are 2a+3 and 2b+3

Answer 1

We are given that the equation x^2 + 3x + 1 = 0 has roots a, b.

By using Sum and Product of roots formulae, we'll get


a + b = -3               (1)

a.b = 1                   (2)


Also, we know that any Quadratic equation can be written as

x^2 - (sum of its roots)x + (product of its roots) = 0

Keeping these in mind, we will try finding equations for the pairs of roots given in the question.


(i) We have to find the equation whose roots are a/b and b/a. We can easily get the equation if we calculate the sum and product of its roots.


Sum of roots

= a/b + b/a

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

= [(a + b)^2 - 2ab] / ab



Using (1) and (2), we'll get


= [(-3)^2 - 2(1)] / 1

= 9 - 2

= 7


Product of roots

= a/b * b/a

= 1


We've found that the equation has got sum of its roots equal to 7 and product of its roots equal to 1.

Hence, the equation is x^2 - (sum of roots)x + product of roots = 0

=> x^2 - 7x + 1 = 0


Similarly, we'll solve the other bits of question


(ii) The roots are a^3, b^3


Sum of roots

= a^3 + b^3


Using the identity x^3 + y^3 = (x + y)(x^2 - xy + y^2)


= (a + b)(a^2 - ab + b^2)

= (a + b)[(a + b)^2 - 2ab - ab)]

= (a + b)[(a + b)^2 - 3ab]


Now, using (1) and (2)


= (-3)[(-3)^2 - 3(1)]

= (-3)(9 - 3)

= (-3)(6)

= -18


Product of roots

= a^3 . b^3

= (ab)^3

= (1)^3               (because ab = 1)

= 1


The sum of roots has come out to be -18 and product of the roots is found to be 1. The equation is, thus,

=> x^2 -(-18)x + 1 = 0

=> x^2 + 18x + 1 = 0


(iii) The roots are a/(2b+3) and b/(2a+3)


Sum of roots

= a/(2b+3) + b/(2a+3)

= [a(2b+3) + b(2a+3)] / (2a+3)(2b+3)

= (2ab + 3a + 2ab + 3b) / (4ab + 6a + 6b + 9)

= (4ab + 3(a + b)) / (4ab + 6(a + b) + 9)


Using (1) and (2)


= (4(1) + 3(-3)) / (4(1) + 6(-3) + 9)

= (4 - 9) / (4 - 18 + 9)

= -5/-5

= 1


Product of roots

= a/(2b+3) * b/(2a+3)

= ab / (2a+3)(2b+3)


We've already calculated (2a+3)(2b+3) in the denominator while calculating the sum of roots. It came out to be -5.

Again, using (2), we'll get


= (1)/(-5)

= -1/5


The sum of roots is 1 and product of roots is -1/5. So, the equation is

x^2 + (-1)x + (-1/5) = 0

=> 5x^2 - 5x - 1 = 0