If a² + b² = 13 and ab = 6, find

(i) a + b

(ii) a - b

(i) We know that,

(a + b)² = a² + b² + 2ab

∴ (a + b) = $\sqrt{a^2 + b^2 + 2ab}$

Substituting values we get,

$\Rightarrow (a + b) = \sqrt{13 + 2 \times 6} \\[1em] \Rightarrow (a + b) = \sqrt{13 + 12} \\[1em] \Rightarrow (a + b) = \sqrt{25} \\[1em] \Rightarrow (a + b) = \pm 5.$

Hence, a + b = ±5.

(ii) We know that,

(a - b)² = a² + b² - 2ab

∴ (a - b) = $\sqrt{a^2 + b^2 - 2ab}$

Substituting values we get,

$\Rightarrow (a - b) = \sqrt{13 - 2 \times 6} \\[1em] \Rightarrow (a - b) = \sqrt{13 - 12} \\[1em] \Rightarrow (a - b) = \sqrt{1} \\[1em] \Rightarrow (a - b) = \pm 1.$

Hence, a - b = ±1.