Question

If (a - b) = 0.9 and ab = 0.36, find the values of (i) (a + b) (ii) (a² - b²).​

Answer 1

Answer:

1.5 & 1.35

Step-by-step explanation:

→ a - b = 0.9

Square on both sides:

→ (a - b)² = (0.9)²

→ a² + b² - 2ab = 0.81

Add and subtract 2ab on LHS:

→ a² + b² + 2ab - 2ab - 2ab = 0.81

→ (a + b)² - 2ab - 2ab = 0.81

→ (a + b)² - 4ab = 0.81

→ (a + b)² - 4(0.36) = 0.81 {ab=0.36}

→ (a + b)² = 0.81 + 1.44

→ (a + b)² = 2.25 = (1.5)²

→ a + b = 1.5

Therefore,

→ a² - b² {formula, a²-b²=(a+b)(a-b)}

→ (a + b)(a - b)

→ (1.5)(0.9)

→ 1.35

Answer 2

Answer:

(2.25)^0.5,1.35 are the answers respectively for (i) and (ii)

Step-by-step explanation:

As we know,

(a-b)^2=a^2+b^2-2ab

0.81=a^2+b^2-0.72

a^2+b^2=1.53

As we know,

(a+b)^2=a^2+b^2+2ab

(a+b)^2=1.53+0.72

(a+b)-->(2.25)^0.5

As we know,

(a+b)(a-b)=a^2-b^2

((2.25)^0.5)*0.9=1.35

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