ONLY 1C PLEASE

for a i got: a+b= -3.5

ab= 4

b:17/4

please show me how to do 1c thanks

YEEEEEET Feb 28, 2019

#1**+3 **

(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

CPhill Feb 28, 2019

#3**+2 **

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

tertre Feb 28, 2019

#4**+2 **

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

tertre Feb 28, 2019