Help Kaplan 800

[kwhite]

What is the value of a - b?

1) a^2 - b^2 = 9

2) a^2 - 2ab + b^2 = 1



I am really stumped and the book explanation isn't helping me at all. Going off of #1 a^2 - b^2 = 9 --> (a-b)(a+b)=9 Therefore:

can't we take (a-b) = 9?? or is my factoring just really off tonight??

thank you for your help.

OA = E

[sanju09]

What is the value of a - b?

1) a^2 - b^2 = 9

2) a^2 - 2ab + b^2 = 1



I am really stumped and the book explanation isn't helping me at all. Going off of #1 a^2 - b^2 = 9 --> (a-b)(a+b)=9 Therefore:

can't we take (a-b) = 9?? or is my factoring just really off tonight??

thank you for your help.

OA = E

I. Since a^2 - b^2 = (a - b) (a + b) = 9, therefore a - b = 9/ (a + b); and we don't know the value of a + b. Insufficient.

II. Since a^2 - 2ab + b^2 = (a - b) ^2 = 1, therefore a - b = ±1; two values! Insufficient

Taken together, we still need to know the value of a + b to answer the question. Insufficient


Take E

Note: You can't take a - b = 9 when statement II translates in a - b = ±1.

[kwhite]

Can you go into further detail about (a+b)(a-b)=9?

I'm under the impression that when we factor an equation we can set them to zero to find the answer... Ex:

x (x^2 + 7x + 6) = 0

-> x (x+1) (x+6) = 0

--> Therefore we can set each variable factor = 0 x=0 ; x+1=0 ; x+6=0

Why does method not work for the original problem (a+b)(a-b)=9 ?

(a-b) = 9 ; (a+b) = 9 ?

Thank you in advance for your help. I just can't seem to wrap my head around this.

[sanju09]

Can you go into further detail about (a+b)(a-b)=9?

I'm under the impression that when we factor an equation we can set them to zero to find the answer... Ex:

x (x^2 + 7x + 6) = 0

-> x (x+1) (x+6) = 0

--> Therefore we can set each variable factor = 0 x=0 ; x+1=0 ; x+6=0

Why does method not work for the original problem (a+b)(a-b)=9 ?

(a-b) = 9 ; (a+b) = 9 ?

Thank you in advance for your help. I just can't seem to wrap my head around this.

Yes, we can set each variable factor = 0 only if we have the expression to the left hand side equal to 0. If a b c... = 0, then a, b, c, ... can be set equal to zero, but if a b c... = 9, we cannot set each of the factor equal to 9. No...no

[kwhite]

that answer cleared up a world of confusion for me. Thank you!

[sanju09]

that answer cleared up a world of confusion for me. Thank you!

I am happy if it served a purpose to you, all the best!!