Maths Question - Quadratic Equation and Inequalities

bonkysleuth SG

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?

wee_ws SG

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

bonkysleuth SG

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

9x2 - x + 4 =0 (answer)

but the "supposedly" correct answer is 9x2 - 11x + 4 =0

can anyone spot any mistakes in my statements? thanks.

SBS261P SG

is ur answer wrong? i got 9x^2 + 11x + 4 = 0 instead =(

SBS261P SG

oh i found ur error!! =D

(a+b)^2 = a^2 +2ab + b^2, not a^2+b^2.

does that help? =D

bonkysleuth SG

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!

SBS261P SG

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