# Algberic question

• Jun 16th 2013, 04:12 AM
sscgeek
Algberic question
If a^4+a^2 b^2 + b^4 = 8 and a^2 + ab + b^2 = 4, then the value of ab is

(A) -1 (B) 0 (C) 2 (D) 1
• Jun 16th 2013, 05:18 AM
Trefoil2727
Re: Algberic question
Quote:

Originally Posted by sscgeek
If a^4+a^2 b^2 + b^4 = 8 and a^2 + ab + b^2 = 4, then the value of ab is

(A) -1 (B) 0 (C) 2 (D) 1

a^2 + b^2 =4-ab
(a+b)^2 = 4+ ab...(1)

a^4 + b^4 = 8- (ab)^2
(a+b)^2 = 2ab+ √ (8+ (ab)^2)...(2)

4+ab= 2ab +√ (8+ (ab)^2)
4-ab= √ (8+ (ab)^2)
16-8ab + (ab)^2 = 8+ (ab)^2
8ab + 8
ab=1
• Jun 16th 2013, 05:38 AM
Prove It
Re: Algberic question
From the first equation:

\displaystyle \begin{align*} a^4 + a^2b^2 + b^4 &= 8 \\ \left( a^2 \right) ^2 + 2a^2b^2 + \left( b^2 \right) ^2 - a^2b^2 &= 8 \\ \left( a^2 + b^2 \right) ^2 - a^2b^2 &= 8 \end{align*}

and from the second equation:

\displaystyle \begin{align*} a^2 + a\,b + b^2 &= 4 \\ a^2 + b^2 &= 4 - a\,b \\ \left( a^2 + b^2 \right) ^2 &= \left( 4 - a\,b \right) ^2 \\ \left( a^2 + b^2 \right) ^2 &= 16 - 8\,a\,b + a^2b^2 \end{align*}

Substituting into the first equation gives

\displaystyle \begin{align*} 16 - 8\,a\,b + a^2b^2 - a^2b^2 &= 8 \\ 16 - 8\,a\,b &= 8 \\ -8\,a\,b &= -8 \\ a\,b &= 1 \end{align*}
• Jun 16th 2013, 08:39 AM
Soroban
Re: Algberic question
Hello, sscgeek!

Quote:

$\text{Given: }\:\begin{Bmatrix}a^4+a^2 b^2 + b^4 &=& 8 & [1] \\ a^2 + ab + b^2 &=& 4 & [2]\end{Bmatrix}$

$\text{Find the value of }ab.$

. . $(A)\;-1 \qquad (B)\;0 \qquad (C)\;2 \qquad (D)\;1$

$\begin{array}{ccccccc}\text{Square [2]:} & a^4 + 2a^3b + 3a^2b^2 + 2ab^3 + b^4 &=& 16 \\ \text{Subtract [1]:} & a^4\quad\;\;+ \quad\;\; a^2b^2 \quad\;\; + \quad\;\; b^4 &=& 8 \end{array}$

$\text{We have: }\;\; 2a^3b + 2a^2b^2 + 2ab^2 \;=\;8$

. . . . . . . . . . $ab\underbrace{(a^2 + ab + b^2)}_{\text{This is 4}} \;=\; 4$

. . . . . . . . . . . . . . . . . $ab(4) \;=\;4$

. . . . . . . . . . . . . . . . . . . $ab \;=\;1 \;\;\text{ answer (D)}$