roots of polynomial integer coefficient

(c)  If  1/a^2 and 1/b^2   are roots...we must have that

(x - 1/a^2 ) ( x - 1/b^2)  = 0

x^2   - [ 1/ a^2 + 1/b^2] x  +   1 /(ab)^2   = 0

x^2 - [  1/a^2 +1/ b^2] / (ab^2) + 1/4^2  =  0

x^2 -  [  ( a^2 + b^2) / (ab)^2 ] x  + 1/16  = 0

x^2  - [ (17/4) / 16 ] x  + 1/16 = 0

x^2 - (17/64)x + 1/16 = 0       multiply through by 64

64x^2 - 17x + 4 =  0

So....the quadratic is

64x^2 - 17x + 4

Part (a) is just Vieta's.

The sum of the roots is -b/a, or in the quadratic equation, it is -7/2.

On the other hand, the product of the roots is c/a, so the answer is 8/2=4.

Hope that helped,

tertre 

Part (b) is using \((x+b)^2.\)

Remember that \((x+b)^2=x^2+b^2+2xb.\)

Therefore, plugging in the values, we get \((-\frac{7}{2})^2=x^2+b^2+2(4).\)

Simplifying, \(\frac{49}{4}-8=x^2+b^2\), and that is equal to \(\frac{49}{4}-\frac{32}{4}=\boxed{\frac{17}{4}}.\)

Hope that helped,

tertre