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Question

Evaluate the following :

(i) 2x3+2x27x+72, when x=35i2(ii) x44x3+4x2+8x+44, when x=3+2i(iii) x4+4x3+6x2+4x+9, when x=1+i2(iv) x6+x4+x2+1, when x=1+i2(v) 2x4+5x3+7x2x+41, when x=23i


Solution

(i) We have,x=35i2 2x=35i 2x3=5i (2x3)2=(5i)2 4x2+912x=25 4x212x+34=0 2(2x26x+17)=0 2x26x+17=0    ...(i) 2x3+2x27x+72  =x(2x26x+17)+6x217x+2x27x+72     (adding and subtracting ) 6x2 and 17x=x×0+8x224x+72  [using (i)]=4(2x26x+17)+4=4×0+4=4        [using (i)]

(ii) We have,x=3+2i x3=2i (x3)2=(2i)2 x2+322×3×x=4 x2+96x×4=0 x26x+13=0        ...(i)Now, x44x3+4x2+8x+44=x2(x26x+13)+6x213x24x3+4x2+8x+44    (adding and subtracting  6x3 and 13x2)=x2×0+2x39x2+8x+44    [using (i)]=2x(x26x+13)+12x226x9x2+8x+44   (adding and subtracting 12x2 and 26x)=2x×0+3x218x+44=3(x26x+13)+5    [using (i)]=3×0+5=5

(iii) we have x=1+i2 x+1=i2 (x+1)2=(i2)2 x2+1+2x=2 x2+2x+3=0              ....(i)Now, x4+4x3+6x2+4x+9=x2(x2+2x+3)+2x3+3x2+4x+9=x2×0+2x(x2+2x+3)x22x+9       [using (i)]=2x×0(x2+2x+3)+3+9    [using (i) and adding and subtracting 3]=0+3+9        [using (i)]=12

(iv) we have,x=1+i2 2x=1+i (2x)2=(1+i)2 (squaring both sides) 2x2=12+(i)2+2×1×i=11+2i 2x2=2i x2=i (x2)2=(i)2     (squaring both sides) x4=1 x4+1=0             ....(i)Nowx6+x4+x2+1=x6+x2+x4+1=x2(x4+1)+1(x4+1)=x2×0+1×0    [using (i)]=0

(v) x+2=3i x2+4x+7=0 2x2+5x3+7x2x+41=(x2+4x+7)(2x23x+5)+6=0×(2x23x+5)+6=6


Mathematics
RD Sharma
Standard XI

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