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September 16, 2014

September 16, 2014

Posted by **Jayden Haddy** on Thursday, August 15, 2013 at 4:09am.

a^2-(1/b) and b^2-(1/a).

- maths -
**Jai**, Thursday, August 15, 2013 at 7:43amWe can further transform the given equation in such a way that the numerical coefficient of x^2 is 1. That is, by dividing all terms by 2. In this form, recall that the sum of roots is equal to negative of the numerical coeff of x, and the product of roots is equal to the constant:

2x^2 - 4x + 3 = 0

x^2 - 2x + 3/2 = 0

Thus, if a and b are the roots,

a + b = 2

ab = 3/2

Given the roots of the quadratic equation we're looking for, we can also get their sum & products:

sum: a^2-(1/b) + b^2-(1/a) = [(ab)(a^2 + b^2) - (a+b)]/ab

product: (a^2-(1/b))*(b^2-(1/a)) = (a^2)(b^2) + 1/ab - (a+b)

From the values of a + b and ab, we can actually use them in order to get the values of the long expressions above.

Squaring the sum equation:

a + b = 2

a^2 + 2ab + b^2 = 4

Since we only need a^2 + b^2, we subtract 2ab to both sides:

(a^2 + 2ab + b^2) - 2ab = 4 - 2ab

a^2 + b^2 = 4 - 2(3/2)

a^2 + b^2 = 1

Substituting,

sum:

[(ab)(a^2 + b^2) - (a+b)]/ab

[(3/2)(1) - 2]/(3/2)

(3/2 - 2)*2/3

= -1/3

product:

(a^2)(b^2) + 1/ab - (a+b)

(3/2)^2 + 1/3/2 - (2)

9/4 + 2/3 - 2

= 11/12

Therefore, we can now write the equation:

x^2 - (sum)x + (product) = 0

x^2 + 1/3 x + 11/12 = 0

The other way to do this is to find the exact roots of the first equation then substitute it to the long expressions. But for me I think that is harder if the roots contain imaginary terms.

Hope this helps~ :)

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