(xii) 
(xi) $x^{2}-(\sqrt{2}+i) x+\sqrt{2 i}=0$ $x^{2}-(\sqrt{2 x}+i x)+\sqrt{2 i}=0$ $x^{2}-\sqrt{2 x}-i x+\sqrt{2 i}=0$ $x(x-\sqrt{2})-i(x-\sqrt{2})=0$ $(x-\sqrt{2})(x-i)=0$ $(x-\sqrt{2})=0$ or $(x-i)=0$...
(x) 
(ix) $i x^{2}-x+12 i=0$ $i x^{2}+x(-1)+12 i=0\left[\right.$ We know, $\left.i^{2}=-1\right]$ substituting $-1=\mathrm{i}^{2}$ $\mathrm{ix}^{2}+\mathrm{xi}^{2}+12 \mathrm{i}=0$...
(viii) 
(vii) 2x2 + √15ix – i = 0 applying discriminant rule, x = (-b ±√(b2 – 4ac))/2a a = 2, b = √15i, c = -i =>15 – 8i = 16 – 1 – 8i 15 – 8i = 16 + (–1) – 8i = 16 + i2 – 8i [∵ i2 = –1] = 42 + (i)2 –...
(vi) 
(v) $i x^{2}-4 x-4 i=0$ $i x^{2}+4 x(-1)-4 i=0\left[W\right.$ e know, $\left.i^{2}=-1\right]$ substituting $-1=\mathrm{i}^{2}$ $\mathrm{ix}^{2}+4 \mathrm{xi}^{2}-4 \mathrm{i}=0$ $i\left(x^{2}+4 i...
(iv) 
(iii) (2 + i)x2 – (5- i)x + 2 (1 – i) = 0 applying discriminant rule, x = (-b ±√(b2 – 4ac))/2a a = (2+i), b = -(5-i), c = 2(1-i) since, i2 = –1 substituting –1 = i2 x = (1 – i) or 4/5 – 2i/5 ∴ The...
(ii) 
(i) $x^{2}-(3 \sqrt{2}+2 i) x+6 \sqrt{2} i=0$ $x^{2}-(3 \sqrt{2} x+2 i x)+6 \sqrt{2 i}=0$ $x^{2}-3 \sqrt{2 x}-2 i x+6 \sqrt{2 i}=0$ $x(x-3 \sqrt{2})-2 i(x-3 \sqrt{2})=0$ $(x-3 \sqrt{2})(x-2 i)=0$...
(iv) 
(iii) $x^{2}-(2 \sqrt{3}+3 i) x+6 \sqrt{3} i=0$ $x^{2}-(2 \sqrt{3} x+3 i x)+6 \sqrt{3} i=0$ $x^{2}-2 \sqrt{3} x-3 i x+6 \sqrt{3} i=0$ $x(x-2 \sqrt{3})-3 i(x-2 \sqrt{3})=0$ $(x-2 \sqrt{3})(x-3 i)=0$...
(ii) 
(i) $x^{2}+10 i x-21=0$ $x^{2}+10 i x-21 \times 1=0$ since, $i^{2}=-1 \Rightarrow 1=-\mathrm{i}^{2}$ substituting $1=-\mathrm{i}^{2}$ $x^{2}+10 i x-21\left(-i^{2}\right)=0$ $x^{2}+10 i x+21...