If x + y = 8 and xy = 3 ¾, find the values of

If x + y = 8 and xy = 3 ¾, find the values of

(i) x – y

(ii) 3 (x² + y²)

(iii) 5 (x² + y²) + 4 (x – y).

Answer :

(i) We know that

(x – y)² = x² + y² – 2xy

It can be written as

(x – y)² = x² + y² + 2xy – 4xy

(x – y)² = (x + y)² – 4xy

It is given that

x + y = 8 and xy = 3 ¾ = 15/4

Substituting the values

(x – y)² = 8² – 4 × 15/4

So we get

(x – y)² = 64 – 15 = 49

x – y = ± √49 = ± 7

(ii) We know that

(x + y)² = x² + y² + 2xy

We can write it as

x² + y² = (x + y)² – 2xy

It is given that

x + y = 8 and xy = 3 ¾ = 15/4

Substituting the values

x² + y² = 8² – 2 × 15/4

So we get

x² + y² = 64 – 15/2

Taking LCM

x² + y² = (128 – 15)/ 2 = 113/2

We get

3 (x² + y²) = 3 × 113/2 = 339/2 = 169 ½

(iii) We know that

5 (x² + y²) + 4 (x – y) = 5 × 113/2 + 4 × ± 7

By further calculation

= 565/2 ± 28

We can write it as

= 565/2 + 28 or 565/2 – 28

= 621/2 or 509/2

It can be written as

= 310 ½ or 254 ½

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