If a - b = 4 and a + b = 6; find :

(i) a² + b²

(ii) ab

(i) We know that,

(a - b)² = a² + b² - 2ab ………….(1)

(a + b)² = a² + b² + 2ab …………..(2)

Adding equation (1) and (2), we get :

(a - b)² + (a + b)² = 2(a² + b²)

(a² + b²) = $\dfrac{(a - b)^2 + (a + b)^2}{2}$

Substituting values we get :

$\Rightarrow a^2 + b^2 = \dfrac{4^2 + 6^2}{2} \\[1em] \Rightarrow a^2 + b^2 = \dfrac{16 + 36}{2} \\[1em] \Rightarrow a^2 + b^2 = \dfrac{52}{2} \\[1em] \Rightarrow a^2 + b^2 = 26.$

Hence, a² + b² = 26.

(ii) We know that,

(a - b)² = a² + b² - 2ab ………….(1)

(a + b)² = a² + b² + 2ab …………..(2)

Subtracting equation (1) from (2), we get :

⇒ (a + b)² - (a - b)² = a² + b² + 2ab - (a² + b² - 2ab)

⇒ (a + b)² - (a - b)² = 4ab

⇒ ab = $\dfrac{(a + b)^2 - (a - b)^2}{4}$

Substituting values we get :

$\Rightarrow ab = \dfrac{6^2 - 4^2}{4} \\[1em] = \dfrac{36 - 16}{4} \\[1em] =\dfrac{20}{4} \\[1em] = 5.$

Hence, ab = 5.