If $a^2 + b^2 = 13$ and ab = 6, find :

(i) a + b

(ii) a - b

(iii) $a^2 - b^2$

(iv) $3(a+b)^2-2(a-b)^2$

(i) Given, $a^2 + b^2 = 13$ and ab = 6

We need to find the value of (a + b),

$⇒ (a + b)^2 = a^2 + b^2 + 2ab$

Substituting the value of $a^2 + b^2$ and ab,

$⇒ (a + b)^2 = 13 + 2 \times 6\\[1em] ⇒ (a + b)^2 = 13 + 12\\[1em] ⇒ (a + b)^2 = 25\\[1em] ⇒ a + b = \sqrt{25}\\[1em] ⇒ a + b = 5 \text{ or } -5$

Hence, a + b = 5 or -5.

(ii) We need to find the value of (a - b),

$⇒ (a - b)^2 = a^2 + b^2 - 2ab$

Substituting the value of $a^2 + b^2$ and ab,

$⇒ (a - b)^2 = 13 - 2 \times 6\\[1em] ⇒ (a - b)^2 = 13 - 12\\[1em] ⇒ (a - b)^2 = 1\\[1em] ⇒ a - b = \sqrt{1}\\[1em] ⇒ a - b = 1 \text{ or } -1$

Hence, a - b = 1 or -1.

(iii) $a^2 - b^2$ = (a - b)(a + b)

Using (i) and (ii),

When (a - b) = 1 and (a + b) = 5

⇒ $a^2 - b^2$ = 1 x 5

= 5

When (a - b) = -1 and (a + b) = 5

⇒ $a^2 - b^2$ = -1 x 5

= -5

When (a - b) = 1 and (a + b) = -5

⇒ $a^2 - b^2$ = 1 x (-5)

= -5

When (a - b) = -1 and (a + b) = -5

⇒ $a^2 - b^2$ = (-1) x (-5)

= 5

Hence, $a^2 - b^2$ = 5 or -5.

(iv) $3(a+b)^2-2(a-b)^2$

From (i) and (ii),

a - b = 1 or -1

⇒ (a - b)² = 1² or (-1)²

⇒ (a - b)² = 1

And, a + b = 5 or -5

⇒ (a + b)² = 5² or (-5)²

⇒ (a + b)² = 25

So,

$3(a + b)^2 - 2(a - b)^2\\[1em] = 3 \times 25 - 2 \times 1\\[1em] = 75 - 2\\[1em] = 73$

Hence, $3(a + b)^2 - 2(a - b)^2$ = 73.