Question

If a+b=6 and ab = 8, evaluate (1) a² + b2 and (2) a - b​

Answer 1

Answer:

$(a + b)^{2} = {a}^{2} + {b}^{2} + 2ab \\ \therefore \: (6)^{2} = {a}^{2} + {b}^{2} + 2 \times 8 \\ \therefore \: 36= {a}^{2} + {b}^{2} +16 \\ \therefore \: 36 - 16= {a}^{2} + {b}^{2} \\ \therefore \: 20= {a}^{2} + {b}^{2} \\ \implies \: \huge \fbox{ {a}^{2} + {b}^{2} = 20} \\Now\\</p><p>(a - b)^{2} = (a + b)^{2} - 4ab\\</p><p> \implies (a - b)^{2} =6^2 -4\times 8\\</p><p> \implies (a - b)^{2} =36-32\\</p><p> \implies (a - b)^{2} =4\\</p><p> \implies \huge \fbox {a - b = 2}$

Answer 2

Step-by-step explanation:

A)

1. Squaring a +b

2. Using formula 'a² + b²'

a² + b² = a + b+2ab

a²+ b² = 6 + (2×8)

a² + b² = 6 + 16

a² + b² = 22

B)

1. Squaring 'a-b'
2. Using formula 'a² - b²'

a² - b² = a + b - 2ab

a² - b² = 6 - (2×8)

a² - b² = 6 - 16

a² - b² = (-10)

a - b = √10