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If ∆is an operation such that for integers a and b we have a∆b=a×a-2×a×b+b×b then find the value of 4∆(-3) |
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Answer» Step-by-step explanation: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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