Roots of Quad Eqn

I am stumped...can help please ?

If a + b and a - b are the roots of the quad eqn (x^2 + 4x + m = 0) and (a - 1) / a and

(b - 1) / b are the roots of the quad eqn (12x^2 - nx + 15 = 0), find the value of m and n.

Answer ( m = - 32, n = 28). Thank you.

(x-(a+b))(x-(a-b)) = x^2+4x+m
12(x-(a-1)/a)(x-(b-1)/b) = 12x^2-nx+15

solve the two equations

sori jay but dun understand how LHS of eqn are related to RHS... can explain? tks

sori jay but dun understand how LHS of eqn are related to RHS... can explain? tks

You can directly compare the coefficients on both sides.

(x-(a+b))(x-(a-b)) = x^2+4x+m

x^2 - (a+b) x - (a-b) x + (a^2 - b^2) = x^2 + 4x + m

x^2 - 2ax + (a^2-b^2) = x^2 + 4x + m

= > -2a = 4 ; a^2-b^2 = m

In this case, - 2a = 4, hence a = - 2

12 (x-(a-1)/a)(x-(b-1)/b) = 12x^2-nx+15

Since a = - 2, it becomes

12 ( x - 3/2 ) ( x - (b-1)/b ) = 12x^2 -nx + 15

12 ( x^2 - 3/2x - ((b-1)/b)x + 3(b-1)/2b = 12x^2 -nx + 15

12 ( x^2 - [3/2 + (b-1)/b] x + 3(b-1)/2b ) = 12x^2 -nx + 15

= > 12[3/2 + (b-1)/b] = n; 36(b-1)/2b = 15

From 36(b-1)/2b = 15, we get b =6

We sub it into a^2-b^2 = m along with a = - 2, and we will get m = -32

Since 12[3/2 + (b-1)/b] = n, and we know now b = 6, n = 28

Thank you for the answer. :)

Hi,

Here is an alternative approach :)

From the first quadratic equation, consider its sum of roots

(a + b) + (a - b) = -4

i.e. 2a = -4

i.e. a = -2.

From the second quadratic equation, consider its product of roots

{(a - 1)/a} . {(b - 1)/b} = 15/12

i.e. (3/2) . {(b - 1)/b} = 15/12

i.e. (b-1)/b = 5/6

i.e. b = 6.

Now we consider product of roots from the first quadratic equation,

m = (a + b)(a - b) = (4)(-8) = -32.

Finally we consider the sum of roots from the second quadratic equation,

n/12 = (a - 1)/a + (b - 1)/b = 3/2 + 5/6 = 7/3

i.e. n = 28.

Thanks!

Cheers,
Wen Shih