Question

`2/3`and 1 are the solutions of equation mx² + nx + 6 = 0. Find the values of m and n.

Answer 1

Given equation be mx² + nx + 6 = 0  ...(1)

Since `2/3` be the solution of the equation (1)

∴ It satisfies equation (1)

Putting `x = 2/3` in equation (1)

We have

`m(2/3)^2 = n(2/3) + 6 = 0`

`\implies (4m)/9 + (2n)/3 + 6 = 0`

`\implies` 4m + 6n + 54 = 0

`\implies` 2m + 3n + 27 = 0  ...(2)

Since x = 1 is the solution of equation (1)

Thus, it must satisfy equation (1)

Putting x = 1 in equation (1)

We have

m + n + 6 = 0   ...(3)

Multiply equation (3) with (2)

We have

2m + 2n + 12 = 0

Subtracting equation (4) from equation (2), we get

(2m + 3n + 27) – (2m + 2n + 12) = 0

`\implies` n + 15 = 0

`\implies` n = –15

From (3); m – 15 + 6 = 0

`\implies` m – 9 = 0

`\implies` m = 9

Hence, value of m = 9 and n = –15

Answer 2

mx² + nx + 6 = 0

k(x − α) (x − β) = 0

α = `3/2`​, β = 1

k(x − `3/2​`) (x − 1) = 0

`k (x^2 - 5/3x + 2/3) = kx^2 - (5k)/3 x  + (2k)/3`

Coefficient of x²: m = k

Coefficient of x: n = `−(5k)/3`

Constant term: `(2k)/3 = 6`

`(2k)/3 = 6` ⇒ 2k = 18 ⇒ k = 9

m = k = 9

`n = -5/3 xx 9 = -15`

m = 9, n = −15