If ∆ is an operation such that for integers a and b we have a ∆ b = a × b - 2 × a × b + b × b (-a) × b + b × b then find
i. 4 ∆ (-3)
ii. (-7) ∆ (-1)
Also show that 4 ∆ (-3) ≠ (-3) ∆ 4 and (-7) ∆ (-1) ≠ (-1) ∆ (-7)

Solution:

In the given problem we have a ∆ b = a × b - 2 × a × b + b × b (-a) × b + b × b now to find the value of 4 ∆ (-3) we will substitute the values a = 4 and b = -3 in the given expression

i. 4 ∆ (-3)

= 4 × (-3) - 2 × 4 × (-3)+ (-3) × (-3)(-4) × (-3) + (-3) × (-3)

= -12 + 24 + 108 + 9

= -12 + 141

= 129

Similarly, to solve (-7) Δ (-1) we will substitute = -7 and b = -1

ii. (-7) Δ (-1)

= (-7) × (-1) – 2 × (-7) × (-1)+ (-1) × (-1) (7) × (-1) + (-1) × (-1)

= 7 - 14 - 7 + 1

= 8 - 21

= -13

Now we have,

(-3) Δ 4

= (-3) × 4 – 2 × (-3) ×(4) + 4 × 4(3) × 4 + 4 × 4

= -12 + 24 + 192 + 16

= -12 + 232

= 220

But we know that 4 Δ (-3) = 129. Hence, 4 Δ (-3) ≠ (-3) Δ 4 And,

(-1) Δ (-7)

= (-1) × (-7) – 2 × (-1) × (-7) + (-7) × (-7)(1) × (-7) + (-7) × (-7)

= 7 -14 - 343 + 49

= 56 - 357

= -301

But, the value of (-7) Δ (-1) = -13. Hence, (-7) Δ (-1) ≠ (-1) Δ (-7).

✦ Try This: It is given that a ∆ b = a × b - 2 × a × b + b × b (-a) × b + b × b then find 2 ∆ (-12).

The above given question can be simplified using integer rules and by simply substituting a = 2 and b = -12 in the expression a ∆ b = a × b - 2 × a × b + b × b (-a) × b + b × b

☛ Also Check: NCERT Solutions for Class 7 Maths Chapter 1

NCERT Exemplar Class 7 Maths Chapter 1 Exercise Problem 126

If ∆ is an operation such that for integers a and b we have a ∆ b = a × b - 2 × a × b + b × b (-a) × b + b × b then find i. 4 ∆ (-3) ii. (-7) ∆ (-1). Also show that 4 ∆ (-3) ≠ (-3) ∆ 4 and (-7) ∆ (-1) ≠ (-1) ∆ (-7)

Summary:

If ∆ is an operation such that for integers a and b we have a ∆ b = a × b - 2 × a × b + b × b (-a) × b + b × b. After applying the integer rules we examined that 4 Δ (-3) is equal to 129 and (-7) Δ (-1) is equal to -13

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