The Quadratic Function and Sum/Product of roots: (1 Viewer)

[ADrew]

Used to know how to do these but I forgot as it was ages ago when I first learnt it. Was wondering if someone would kindly point me in the right direction:

"Given that A and B are the roots of (2m-1)x^2+(m+1)x+1=0, find m if:"
a) A=2
b) A+B=2AB

AND

"If one zero of y=x^2+mx+n is the square of the other, prove (without finding the roots) that m^3=n(3m-n-1)"

Thanks for any help.

 

[SpiralFlex]

I feel like purple today.

a) Sum of the roots:



Since we already know one root,





Product of roots,



Since we already know one root,





Now,



I got .

b) Changing it around,





I got

c) Let one of the roots be

Therefore the other root is

With this result,



















Note: Someone should check my answer as I am tired today.

 

[funnytomato]

Note: Someone should check my answer as I am tired today.

for part a of question 1 , sub in x=A=2 would make you less tired

 

[funnytomato]

SpiralFlex is correct

 

[Drongoski]

SpiralFlex is correct

So m = 1/10 and the quadratic eqn becomes:





A + B =/= 2 AB ???


Did I make a serious error somewhere??

 

[SpiralFlex]

You have the wrong coefficient of .

 

[funnytomato]

So m = 1/10 and the quadratic eqn becomes:





A + B =/= 2 AB ???


Did I make a serious error somewhere??

i see what u r doing, it's not if a) AND b)

 

[Drongoski]

You have the wrong coefficient of .

What is the correct resulting quadratic eqn then - based on m = 1/10 ??

 

[funnytomato]

So m = 1/10 and the quadratic eqn becomes:





A + B =/= 2 AB ???


Did I make a serious error somewhere??

i think your eqn was right

except for that it didn't need to(AND cannot) satisfy A + B = 2 AB

 

[SpiralFlex]

What is the correct resulting quadratic eqn then - based on m = 1/10 ??

Wait sorry, your one is right.

 

[ADrew]

Thanks so much guys. I understand what I was doing wrong now.