Question

# Evaluate the following : (i) 2x3+2x2−7x+72, when x=3−5i2(ii) x4−4x3+4x2+8x+44, when x=3+2i(iii) x4+4x3+6x2+4x+9, when x=−1+i√2(iv) x6+x4+x2+1, when x=1+i√2(v) 2x4+5x3+7x2−x+41, when x=−2−√3i

Open in App
Solution

## (i) We have,x=3−5i2⇒ 2x=3−5i⇒ 2x−3=−5i⇒ (2x−3)2=(−5i)2⇒ 4x2+9−12x=−25⇒ 4x2−12x+34=0⇒ 2(2x2−6x+17)=0⇒ 2x2−6x+17=0 ...(i)∴ 2x3+2x2−7x+72 =x(2x2−6x+17)+6x2−17x+2x2−7x+72 (adding and subtracting ) 6x2 and 17x=x×0+8x2−24x+72 [using (i)]=4(2x2−6x+17)+4=4×0+4=4 [using (i)] (ii) We have,x=3+2i⇒ x−3=2i⇒ (x−3)2=(2i)2⇒ x2+32−2×3×x=−4⇒ x2+9−6x×4=0⇒ x2−6x+13=0 ...(i)Now, x4−4x3+4x2+8x+44=x2(x2−6x+13)+6x2−13x2−4x3+4x2+8x+44 (adding and subtracting 6x3 and 13x2)=x2×0+2x3−9x2+8x+44 [using (i)]=2x(x2−6x+13)+12x2−26x−9x2+8x+44 (adding and subtracting 12x2 and 26x)=2x×0+3x2−18x+44=3(x2−6x+13)+5 [using (i)]=3×0+5=5 (iii) we have x=−1+i√2⇒ x+1=i√2⇒ (x+1)2=(i√2)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)−x2−2x+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+i√2⇒ √2x=1+i⇒ (√2x)2=(1+i)2 (squaring both sides)⇒ 2x2=12+(i)2+2×1×i=1−1+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+7x2−x+41=(x2+4x+7)(2x2−3x+5)+6=0×(2x2−3x+5)+6=6

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
De Moivre's Theorem
MATHEMATICS
Watch in App
Join BYJU'S Learning Program