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
Originally posted by Jt061952: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