If x + y = 8 and xy = $3\dfrac{3}{4}$, find the values of

(i) x - y

(ii) 3(x² + y²)

(iii) 5(x² + y²) + 4(x - y).

(i) We know that,

(x + y)² = x² + y² + 2xy …..(i)

(x - y)² = x² + y² - 2xy …..(ii)

Subtracting eqn. (ii) from (i) we get,

(x + y)² - (x - y)² = x² - x² + y² - y² + 2xy - (-2xy) = 4xy.

⇒ (x + y)² - (x - y)² = 4xy.

∴ (x - y) = $\sqrt{(x + y)^2 - 4xy}$.

Substituting values we get,

$x - y = \sqrt{8^2 - 4 \times 3\dfrac{3}{4}} \\[1em] = \sqrt{64 - 4 \times \dfrac{15}{4}} \\[1em] = \sqrt{64 - 15} \\[1em] = \sqrt{49} \\[1em] = \pm 7.$

Hence, x - y = ±7.

(ii) We know that,

3(x² + y²) = 3[(x + y)² - 2xy].

Substituting values we get,

$3(x^2 + y^2) = 3\Big[(8)^2 - 2 \times 3\dfrac{3}{4}\Big] \\[1em] = 3\Big(64 - 2 \times \dfrac{15}{4}\Big) \\[1em] = 3\Big(64 - \dfrac{15}{2}\Big) \\[1em] = 3\Big(\dfrac{128 - 15}{2}\Big) \\[1em] = 3 \times \dfrac{113}{2} \\[1em] = \dfrac{339}{2} \\[1em] = 169\dfrac{1}{2}.$

Hence, 3(x² + y²) = $169\dfrac{1}{2}.$

(iii) From parts (i) and (ii) we get,

(x - y) = ±7 and x² + y² = $\dfrac{113}{2}$.

When (x - y) = 7,

Substituting values we get,

$5(x^2 + y^2) + 4(x - y) = 5 \times \dfrac{113}{2} + 4 \times 7 \\[1em] = \dfrac{565}{2} + 28 \\[1em] = \dfrac{565 + 56}{2} \\[1em] = \dfrac{621}{2} \\[1em] = 310\dfrac{1}{2}.$

When (x - y) = -7,

Substituting values we get,

$5(x^2 + y^2) + 4(x - y) = 5 \times \dfrac{113}{2} + 4 \times -7 \\[1em] = \dfrac{565}{2} - 28 \\[1em] = \dfrac{565 - 56}{2} \\[1em] = \dfrac{509}{2} \\[1em] = 254\dfrac{1}{2}.$

Hence, 5(x² + y²) + 4(x - y) = $310\dfrac{1}{2} \text{ or } 254\dfrac{1}{2}$.