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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 (–7) ∆ (–1)​

By Jade

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 (–7) ∆ (–1)​

About the author
Jade

Quu- A, B, C teen sajhendar hai,jinki punji kramshah 8 lakh,12 lakh ,aur 15 lakh hai. A kaam bhi krta hai aur kul labh ka 12.5% sa

2. Write the numeral for sixty-two
million eighty-three thousand five
hundred sixty-
one
OCX 000​

1 thought on “If ∆ is an operation such that for integers a and b we have,<br />a ∆ b = a×b–2×a×b+b×b(–a)×b+b×b<br />Then find (–7) ∆ (–1)​”

  1. Answer:

    aΔb=a×b−2×a×b+b×b(−a)×b+b×b

    So,

    i.

    4Δ(−3)=[4×(−3)]−[2×4×(−3)]+[(−3)×(−3)(−4)×(−3)]+[(−3)×(−3)]

    =[−12]−[−24]+[108]+[9]

    =−12+24+108+9

    =−12+141

    =129

    ii.

    (−7)Δ(−1)=[(−7)×(−1)]−[2×(−7)×(−1)]+[(−1)×(−1)(7)×(−1)]+[(−1)×(−1)]

    =7−14+[−7]+1

    =7−14−7+1

    =8−21

    =−13

    Now (−3)Δ4=[(−3)×4]−[2(−3)×4]+[4×4(3)×4]+[4×4]

    =−12+24+192+16

    =−12+232

    =220

    Therefore, [4Δ(−3)=129]

    =[(−3)Δ4=220]

    Now, case II:

    (−1)Δ(−7)=[(−1)×(−7)]−[2×(−1)×(−7)]+[(−7)×(−7)(1)×(−7)]+[(−7)×(−7)]

    =7−14−343+49

    =56−357

    =−301

    Therefore, [−7Δ(−1)=−13]

    =[(−1)Δ(−7)=−301]

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