Given the equation kx2 +(k-1)x + 2k + 3 = 0, where k is a non-zero integer, find the value of k for which one root is the negative of the other.
Firstly I divide the equation by k and you get.
x2 + (k-1/k)x + (2k+3/k) = 0
Let the roots be alfa (a) and -alfa (-a)
a + (-a) = -(k-1)/k
a-a = -(k-1)/k
a x a (ALSO SIMILAR TO ALFA x BETA because -a should be taken as beta but the roots is negative of the other so we can't use bet)
-a2 = (2k+3)/k
refer back to first part where there's (a-a)
0 = -(k-1)/k
0 = (-k+1)/k
0=-k+1
k = 1 (answer)
Notice that I do not have to make use of the -a2 part? is my method of doing correct? I think the part where I multiply -a and a should be used, but I didn't make use of that to derive the answer. Is anything missing?
Hi,
Your approach is correct. The roots (if you have learnt complex ones) are 5i and -5i, which correspond nicely to alpha and negative alpha. Thanks!
Cheers,
wee_ws
Originally posted by wee_ws:Hi,
Your approach is correct. The roots (if you have learnt complex ones) are 5i and -5i, which correspond nicely to alpha and negative alpha. Thanks!
Cheers,
wee_ws
O, much thanks.
There is this question which goes something like :
Obtain the quadratic equation whose roots are the square of the roots of the equation 3x2-x+2.
so i changed the equation to :
x2 - 1/3 x + 2/3 =0
alfa (a) + beta (b) = 1/3
(a)(b) = 2/3
(1/3)2 = 1/9
(2/3)2 = 4/9
x2 - 1/9 x + 4/9 =0
so
9x2 - x + 4 =0 (answer)
but the "supposedly" correct answer is 9x2 - 11x + 4 =0
can anyone spot any mistakes in my statements? thanks.
is ur answer wrong? i got 9x^2 + 11x + 4 = 0 instead =(
oh i found ur error!! =D
(a+b)^2 = a^2 +2ab + b^2, not a^2+b^2.
does that help? =D
Originally posted by SBS261P:oh i found ur error!! =D
(a+b)^2 = a^2 +2ab + b^2, not a^2+b^2.
does that help? =D
If you take a look carefully, I didn't make use of the (a+b)2 formula. I actually attempted to find the roots of the equation before squaring them and sbustituting them into another equation.
3x^2 - x + 2 =0
the original equation for alfa and beta is x^2 - (SUM OF ROOT)x + (PRODUCT OF ROOTS) = 0
so when i divide it by 3,
x2- 1/3 x +2/3 = 0
This means that the sum of roots is 1/3 and the product of roots is 2/3.
square 1/3 and you'll get 1/9.
square 2/3 and you'll get 4/9.
so now, we can form the new equation.
x2 - 1/9 x + 4/9 =0
9x^2 - 1x + 4 = 0
thanks SBS261p! but i'm sorry to say i still can't manage to spot my error. maybe you can try to elaborate further so i can catch what you're trying to say. thanks!
a + b = 1/3 (sum of roots)
but in ur answer u want a^2+b^2 (sum of root in de quadratic equation u want to get)
but by squaring 1/3 u get a^2 + 2ab + b^2, which is 2ab too much
so... u need to minus away 2ab
1/9 - 4/3 = -11/9