Question

Find the values of a and b for which `x = 3/4` and x = –2 are the roots of the equation ax² + bx – 6 = 0.

Answer 1

It is given that `3/4` is a root of ax² + bx – 6 = 0; therefore, we have:

`a xx (3/4)^2 + b xx 3/4 - 6 = 0` 

⇒ `(9a)/16 + (3b)/4 = 6` 

⇒ `(9a + 12b)/16 = 6` 

⇒ 9a + 12b – 96 = 0 

⇒ 3a + 4b = 32   ...(i)  

Again, (–2) is a root of ax² + bx – 6 = 0; therefore, we have: 

a × (–2)² + b × (–2) – 6 = 0

⇒ 4a – 2b = 6 

⇒ 2a – b = 3   ...(ii) 

On multiplying (ii) by 4 and adding the result with (i), we get: 

⇒ 3a + 4b + 8a – 4b = 32 + 12 

⇒ 11a = 44 

⇒ a = 4

Putting the value of a in (ii), we get: 

2 × 4 – b = 3 

⇒ 8 – b = 3

⇒ b = 5 

Hence, the required values of a and b are 4 and 5, respectively.