Question

Solve the following quadratic equations by factorization method :

(i) $x^2+10ix-21=0$

(ii) $x^2+(1-2i)x-2i=0$

(iii) $x^2-(2\sqrt{3}+3i)x+6\sqrt{3}i=0$

(iv)$6x^2-17ix-12=0$

Answer 1

(i) $x^2+10ix-21=0\Rightarrow x^2+10ix+21i^2=0[\because i^2=-1]\Rightarrow x^2+7ix+3ix+21i^2=0\Rightarrow x(x+7i)+3i(x+7i)=0\Rightarrow (x+3i)(x+7i)=0\therefore x=-3i,-7i$

(ii) $x^2+(1-2i)x-2i=0\Rightarrow x^2+x-2i-2i=0\Rightarrow x(x+1)-2i(x+1)=0\Rightarrow (x-2i)(x+1)=0\Rightarrow x=2i,-1$ (iii) $x^2-(2\sqrt{3}+3i)x+6\sqrt{3}i=0\Rightarrow x^2-2\sqrt{3}x-3ix+6\sqrt{3}i=0\Rightarrow x(x-2\sqrt{3})-3i(x-2\sqrt{3})=0\Rightarrow (x-3i)(x-2\sqrt{3})=0\Rightarrow x=3i,2\sqrt{3}$

(iv)$6x^2-17ix-12=0\Rightarrow 6x^2-17ix-12=0[\because i^2=-1]\Rightarrow 6x^2-9ix-8ix+12i^2=0\Rightarrow 3x(2x-3i)-4i(2i-3i)=0\Rightarrow (3x-4i)(2x-3i)=0\Rightarrow x=\frac{4}{3}iOr\frac{3}{2}i$