FORMULA FOR A SQUARE PLUS B SQUARE

Problem 1 :

If p + q = 12 and pq = 22, then find p² + q².

Solution :

p + q = 12 and pq = 22

Write the formula for a² + b² = (a + b)² - 2ab

p² + q² = (p + q)² - 2pq

= (12)² - 2(22)

= 144 – 44

p² + q² = 100

Problem 2 :

If a + b = 25 and a² + b² = 225, then find ab.

Solution :

a + b = 25 and a² + b² = 225

Write the formula for a² + b² = (a + b)² - 2ab

225 = (25)² - 2ab

225 = 625 – 2ab

2ab = 625 – 225

2ab = 400

ab = 400 / 2

ab = 200

Problem 3 :

If x – y = 13 and xy = 28, then find x² + y².

Solution :

x – y = 13 and xy = 28

Write the formula for a² + b² = (a - b)² + 2ab

x² + y² = (x - y)² + 2xy

= (13)² + 2(28)

= 169 + 56

x² + y² = 225

Problem 4 :

If m - n = 16 and m² + n² = 400, then find mn.

Solution :

m - n = 16 and m² + n² = 400

Write the formula for a² + b² = (a - b)² + 2ab

m² + n² = (m - n)² + 2mn

400 = (16)² - 2mn

400 = 256 – 2mn

2mn = 400 – 256

2mn = 144

mn = 144 / 2

mn = 72

Problem 5 :

If a² + b² = 74 and ab = 35, then find a + b.

Solution :

a² + b² = 74 and ab = 35

Write the formula for a² + b² = (a + b)² - 2ab

74 = (a + b)² - 2(35)

74 = (a + b)² - 70

(a + b)² = 74 – 70

(a + b)² = 4

a + b = 2

Problem 6 :

Factorise y² + 4y + 4

Solution :

= y² + 4y + 4

= y² + 2(y)(2) + 2²

It is in the form of a² + 2ab + b²

= (y + 2)²

= (y + 2)(y + 2)

Problem 7 :

Factorise 9x²y - 24xy² + 16y³

Solution :

= 9x²y - 24xy² + 16y³

= y(9x² - 24xy + 16y²)

= y(3²x² - 24xy + 4²y²)

= y[(3x)² - 2(3x)(4y) + (4y)²]

It is in the form of a² - 2ab + b²

= y(3x + 4y)²

= y (3x + 4y)(3x + 4y)

Problem 8 :

If x² + 1/x² = 6, find the value of x⁴ + 1/x⁴ 

Solution :

x² + 1/x² = 6

a² + b² = (a + b)² - 2ab

 x⁴ + 1/x⁴ =  (x²)² + (1/x²)²

= (x² + 1/x²)² - 2x²(1/x²)

= (x² + 1/x²)² - 2

= 6² - 2

= 36 - 2

 x⁴ + 1/x⁴ = 34

Problem 9 :

Find the value of k if

xy²k = (4xy + 3y)² - (4xy - 3y)²

Solution :

xy²k = (4xy + 3y)² - (4xy - 3y)²

= (4xy)² + 2(4xy)(3y) + (3y)² - [(4xy)² + 2(4xy)(3y) + (3y)²]

= 16x²y² + 24xy² + 9y² - [16x²y² - 24xy² + 9y²]

= 16x²y² + 24xy² + 9y² - 16x²y² + 24xy² - 9y²

= 48xy²

This can be compared with xy²k

k = 48

Problem 10 :

If a² + b² = 9 and ab = 4

Find the value of 3(a + b)² - 2(a - b)²

Solution :

a² + b² = 9 and ab = 4

= 3(a + b)² - 2(a - b)²

= 3(a² + 2ab + b²) - 2(a² - 2ab + b²)

= 3a² + 6ab + 3b² - 2a² + 4ab - 2b²

= 3a² - 2a² + 6ab + 4ab + 3b²  - 2b²

= a² + 10ab + b²

= 9 + 10(4)

= 9 + 40

= 49

Problem 11 :

If x + 1/x = 8, find the value of x² + 1/x²

Solution :

x + 1/x = 8

x² + 1/x² = (x + 1/x)² - 2 x(1/x)

= (x + 1/x)² - 2

Applying the value of x + 1/x, we get 

= 8² - 2

= 64 - 2 

= 62

Problem 12 :

If a - 1/a = 5, find the value of a² + 1/a²

Solution :

a - 1/a = 5

a² + 1/a² = (a - 1/a)² + 2 a(1/a)

= (a - 1/a)² - 2

Applying the value of a + 1/a, we get 

= 5² - 2

= 25 - 2 

= 23

Problem 13 :

If a² + 1/a² = 23, find the value of (a + 1/a)

Solution :

a² + 1/a² = 23

a² + 1/a² = (a + 1/a)² - 2 a(1/a)

23 = (a + 1/a)² - 2

(a + 1/a)² = 23 + 2

(a + 1/a)² = 25

(a + 1/a) = √25

(a + 1/a) = -5 and 5

Problem 14 :

If x² + 1/x² = 51, find the value of (x - 1/x)

Solution :

x² + 1/x² = 51

x² + 1/x² = (x - 1/x)² + 2 x(1/x)

Applying the given value, we get

51 = (x - 1/x)² + 2

(x - 1/x)² = 51 - 2

(x - 1/x)² = 49

(x - 1/x) = √49

x - 1/x = -7 and 7