Question

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)​

Answer 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]