Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that: Class 9

Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that: Class 9

Question 1.

Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that: Class 9

i. p(0) = 5.

ii. The polynomial p(x) - q(x) cuts the x-axis at (3,0).

iii. The sum p(x) + q(x) is equal to 6x + 4 for all real x.

Find the polynomials p(x) and q(x).

Solution:

p(x) = ax + b ...(i)

q(x) = cx + d ...(ii)

Given p( 0) = 5

∴ p(0) = a(0) + b

∴ b = 5

According to the given condition,

p(x) - q(x) = 0 at x = 3

∴ p(3) - q(3) = 0

∴ (3a + b) - (3c + d) = 0 ....[From (i) and (ii)]

∴ 3a + b - 3c - d = 0

∴ (3a - 3c) + (b - d) = 0

∴ 3(a - c) + (5 - d) = 0 ...(iii) ...[∵ b = 5]

According to the second condition,

p(x) + q(x) = 6x + 4, for all real x

∴ (ax + b) + (cx + d) = 6x + 4

∴ (ax + cx) + (b + d) = 6x + 4

∴ (a + c)x + (b + d) = 6x + 4

Equating coefficients on both sides, we get

a + c = 6 ...(iv)

b + d = 4

∴ 5 + d = 4 ...[∵ b = 5]

∴ d = -1

Substituting d = -1 in (iii), we get

3(a - c) + (5 - (-1)) = 0

∴ 3(a - c) + 6 = 0

∴ 3(a - c) = - 6

∴ a - c = -2

∴ a = -2 + c ...(v)

Now, substituting a = -2 + c in (iv), we get

(-2 + c) + c = 6

∴ -2 + 2c = 6

∴ 2c = 6 + 2

∴ 2c = 8

∴ c = $\frac{8}{2}$ = 4

Now, Substituting c = 4 in equation (v), we get

a = -2 + 4

∴ a = 2

∴ The two polynomials are,

p(x) = ax + b = 2x + 5

q(x) = cx + d = 4x + (- 1) = 4x -1