Question

Let p(x) = 4x^2+6x+4 and q(x)= 4y^2-12y+25. find the unique pair of real nos (x, y) that satisfy p(x).p(y) = 28

Answer 1

Given: Two equations p(x) = 4x^2+6x+4 and q(x)= 4y^2-12y+25

To find: The unique pair of real nos (x, y) that satisfy p(x).p(y) = 28

Solution:

• Now we have given two equations: p(x) = 4x^2+6x+4 and q(x)= 4y^2-12y+25

• Let p(x) = K

• So according to the condition:

            p(x).p(y) = 28

            K.(4y^2 - 12y + 25) = 28

            4y^2 - 12y + 25 - 28/K = 0

• Now discriminant will be equal to zero as (x,y) is unique pair.

           12^2   −4 x 4 x ( 25−  28 /K ) = 0

           144 - 400 + 448/K  = 0

           448/K = 256

           K = 448/256

           K = 7/4

           4x^2 + 6x + 4 = 7/4

           4x^2 + 6x + 4 - 7/4 = 0

           4x^2 + 6x + 9/4 = 0

           x = -6 ±√(36 - 4x4x9/4) / 2(4)

           x = -3/4

• Now,

           K.(4y^2 - 12y + 25) = 28

           7/4 . (4y^2 - 12y + 25) = 28

           4y^2 - 12y + 9 = 0

           (2y - 3)² = 0

           y = 3/2

• So, (x,y) = (-3/4,3/2)

Answer:

                The unique pair of real nos (x, y) that satisfy p(x).p(y) = 28 is  (-3/4, 3/2)