Differentiation (Part-38) | S N De

 

3. Solve by factorization: 

$\,(i)\, 3x^2+8ix+3=0 \\ \Rightarrow 3x^2+8ix-3i^2=0 \\ \Rightarrow 3x^2+9ix-ix-3i^2=0 \\ \Rightarrow 3x(x+3i)-i(x+3i)=0 \\ \Rightarrow (x+3i)(3x-i)=0 \\ \therefore x=\frac i3\,\,\text{or}\,\,-3i.$

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

$\,(iii)\, x^2+(i-3\sqrt2)x-3\sqrt2i=0 \\ \Rightarrow x^2+ix-3\sqrt2x-3\sqrt2i=0 \\ \Rightarrow x(x+i)-3\sqrt2(x+i)=0 \\ \Rightarrow (x+i)(x-3\sqrt2)=0 \\ \therefore x= -i,\,\, 3\sqrt2.$

$\,(iv)\,2x^2-ix+6=0 \\ \Rightarrow 2x^2-ix-6i^2=0\,\,\,[\text{Using}\,\,i^2=-1]\\ \Rightarrow 2x^2-4xi+3xi-6i^2=0 \\ \Rightarrow 2x(x-2i)+3i(x-2i)=0 \\ \Rightarrow (x-2i)(2x+3i)=0 \\ \Rightarrow x=2i,\,\,-\frac{3i}{2}.$

$\,(v)\, x^2+3\sqrt2xi+8=0 \\ \Rightarrow x^2+3\sqrt2xi-8i^2=0 \\ \Rightarrow x^2+4\sqrt2xi-\sqrt2xi-8i^2=0 \\ \Rightarrow x(x+4\sqrt2i)-\sqrt2i(x+4\sqrt2i)=0\\ \Rightarrow (x+4\sqrt2i)(x-\sqrt2i)=0 \\ \therefore x=-4\sqrt2i,\,\, \sqrt2i.$

$\,(vi)\,12ix^2-x+6i=0 \\ \Rightarrow 12i^2x^2-ix+6i^2=0 \\ \Rightarrow -12x^2-ix+6i^2=0 \\ \Rightarrow 12x^2+ix-6i^2=0 \\ \Rightarrow 12x^2+9ix-8ix-6i^2=0 \\ \Rightarrow 3x(4x+3i)-2i(4x+3i)=0 \\ \Rightarrow (4x+3i)(3x-2i)=0 \\ \therefore \,\,x=-\frac{3i}{4},\,\,\,\frac{2i}{3}$

4. Solving the equation $\,\,x^2+(i-7)x-(i-18)=0,\,\,$ prove that, the roots of the equation are not complex conjugate numbers.

Sol. Comparing the given equation $\,\,x^2+(i-7)x-(i-18)=0 \cdots(1)$ with $\,\,ax^2+bx+c=0\,\,$ we get $\,\,a=1,b=(i-7),c=-(i-18)$ and so the discriminant of the given equation is : 

$\,\, D=b^2-4ac\\=(i-7)^2+4.1.(i-18)\\=i^2-2.i.7+7^2+4i-72\\=-1-14i+49+4i-72\\=-24-10i\\=1^2+(5i)^2-2.1.5i\\=(1-5i)^2 \cdots(2) $

So, the solution of (1) is : $\,\,x=\frac{-(i-7)\pm \sqrt{D}}{2}\\~~~~=\frac{-i+7 \pm(1-5i)}{2}\,\,\,[\text{By (2)}]\\~~~~=\frac{-i+7+1-5i}{2},\,\,\frac{-i+7-1+5i}{2}\\~~~~=(4-3i),\,\,(3+2i) \cdots(3)$

From (3), it is evident that the roots of the equation are not complex conjugate numbers. (proved)

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