Differentiation (Part-38) | S N De

1. Show that  the roots of each of the following quadratic equations are complex numbers. Find the solutions in each case :

$\,(i)\,x^2+2x+2=0$

Sol. (i) $\,\,x^2+2x+2=0\\ \Rightarrow x^2+2x+1+1=0 \\ \Rightarrow (x+1)^2=-1 \\ \Rightarrow x+1=\pm\sqrt{i^2}\,\,[\,\,\text{Using}\,\,i^2=-1]\\ \Rightarrow x+1=\pm i \\ \Rightarrow x=-1\pm i.$ 

$(ii)\,\,x^2+4x+8=0$

Sol. $\,\,(ii)\,\,x^2+4x+8=0 \\ \Rightarrow (x+2)^2=-4 \\ \Rightarrow x+2 =\pm 2i \,\,[\,\,\text{Using}\,\,i^2=-1]\\ \Rightarrow x=-2 \pm 2i.$

$\,(iii)\,2x^2-3x+4=0$

Sol. $\,(iii)\,2x^2-3x+4=0 \\ \Rightarrow x^2-\frac 32 x+2=0 \\ \Rightarrow x^2-2.x.\frac 34+(\frac 34)^2=-2+\frac{9}{16} \\ \Rightarrow (x-\frac 34)^2=-\frac{23}{16} \\ \Rightarrow x-\frac 34=\pm \sqrt{i^2 \times \frac{23}{16}} \\ \Rightarrow x=\frac 34 \pm i\left(\frac{\sqrt{23}}{4}\right) \\ \Rightarrow x=\frac 14 (3 \pm i\sqrt{23})$

$\,(iv)\,3x^2-7x+5=0$

Sol. $\,(iv)\,3x^2-7x+5=0 \\ \Rightarrow x^2-\frac 73 x+\frac 53=0 \\ \Rightarrow x^2-2.x.\frac 76+(\frac 76)^2=-\frac 53+\frac{49}{36} \\ \Rightarrow (x-\frac 76)^2=-\frac{11}{36}\\~~~~~~~~~~~~~~~~~~~~=\left(\pm i\frac{\sqrt{11}}{6}\right)^2\\ \Rightarrow x-\frac 76=\pm i \frac{\sqrt{11}}{6} \\ \Rightarrow x=\frac 76 \pm i \frac{\sqrt{11}}{6} \\ \Rightarrow x=\frac 16 (7 \pm i\sqrt{11})$

$\,(v)\,\,\frac{2x-1}{x-2}=\frac{x}{x-1}$

Sol. $\,(v)\,\,\frac{2x-1}{x-2}=\frac{x}{x-1}\\ \Rightarrow (2x-1)(x-1)=x(x-2) \\ \Rightarrow 2x^2-3x+1=x^2-2x \\ \Rightarrow x^2-x+1=0 \\ \Rightarrow x^2-2.x.\frac 12+(\frac 12)^2=-1+(\frac 12)^2\\ \Rightarrow(x-\frac 12)^2=-\frac 34 \\ \Rightarrow x-\frac 12=\pm i\frac{\sqrt3}{2}\,\,\,[\text{Using}\,\,i^2=-1]\\ \Rightarrow x=\frac 12(1 \pm i\sqrt3)$

$\,(vi)\, \sqrt3 x^2+x+\sqrt3=0$

Sol. $\,(vi)\, \sqrt3 x^2+x+\sqrt3=0 \\ \Rightarrow x^2+\frac{1}{\sqrt3}+1=0 \\ \Rightarrow x^2+2.x.\frac{1}{2\sqrt3}+\left(\frac{1}{2\sqrt3}\right)^2=-1+\frac{1}{12} \\ \Rightarrow (x+\frac{1}{2\sqrt3})^2=-\frac{11}{12} \\ \Rightarrow x+\frac{1}{2\sqrt3}=\pm i \frac{\sqrt{11}}{2\sqrt3} \\ \Rightarrow x= \frac{1}{2\sqrt3}(-1\pm i\sqrt{11})$

2. Show that the roots of each of the following quadratic equations are conjugate complex numbers: 

$\,(i)\,\,\,3x^2-4x+3=0 \quad (ii)\,\,\frac{1}{x-3}+\frac x5=0 \\ (iii)\,\frac{1}{x-1}+\frac{1}{x-2}=\frac{2}{2x-3}$

Sol. $\,(i)\,\,\,3x^2-4x+3=0 \cdots(1)$

The discriminant of (1) : 

$\,D =(-4)^2-4.3.3=16-36=-20<0$

Hence, the roots of (1), are imaginary.

Again, $\,\,3x^2-4x+3=0 \\ \Rightarrow x^2-\frac 43 x+1=0 \\ \Rightarrow x^2-2.x.\frac 23+(\frac 23)^2=-1+\frac 49 \\ \Rightarrow(x-\frac 23)^2=-\frac 59 \\ \Rightarrow x-\frac 23=\pm i\frac{\sqrt5}{3} \\ \Rightarrow x=\frac 23 \pm i \frac{\sqrt5}{3}=\frac 13(2 \pm i\sqrt5) \cdots(2)$

Clearly, the roots of (1) are conjugate complex numbers which is evident from (2). 

Sol. $\,(ii)\,\,\,\frac{1}{x-3}+\frac x5=0 \\ \Rightarrow \frac{5+x(x-3)}{5(x-3)}=0 \\ \Rightarrow x^2-3x+5=0\cdots(1)$

The discriminant of (1) : 

$\,D =(-3)^2-4.1.5=16-36=-20<0$

Hence, the roots of (1), are imaginary.

Again, $\,\,3x^2-4x+3=0 \\ \Rightarrow x^2-\frac 43 x+1=0 \\ \Rightarrow x^2-2.x.\frac 23+(\frac 23)^2=-1+\frac 49 \\ \Rightarrow(x-\frac 23)^2=-\frac 59 \\ \Rightarrow x-\frac 23=\pm i\frac{\sqrt5}{3} \\ \Rightarrow x=\frac 23 \pm i \frac{\sqrt5}{3}=\frac 13(2 \pm i\sqrt5) \cdots(2)$

Clearly, the roots of (1) are conjugate complex numbers which is evident from (2). 

Sol. $\,(iii)\,\frac{1}{x-1}+\frac{1}{x-2}=\frac{2}{2x-3} \\ \Rightarrow \frac{x-2+x-1}{(x-1)(x-2)}=\frac{2}{2x-3}  \\ \Rightarrow \frac{2x-3}{x^2-3x+2}=\frac{2}{2x-3}\\ \Rightarrow4x^2-12x+9=2x^2-6x+4 \\ \Rightarrow 2x^2-6x+5=0 \\ \Rightarrow x^2-3x+\frac 52=0 \cdots(1)$

Now, the discriminant of (1) :

$\,D=(-3)^2-4 \times 1 \times \frac 52=-1<0$

Hence, the roots of (1) are imaginary. 

Again, from (1) we get, $\,x^2-3x+\frac 52=0 \\ \Rightarrow x^2-2.x.\frac 32+(\frac 32)^2=-\frac 52+\frac 94 \\ \Rightarrow (x-\frac 32)^2=-\frac 14 \\ \Rightarrow x-\frac 32=\pm i \frac 12\,\,\,[\text{Using}\,\,i^2=-1]\\ \Rightarrow x=\frac 12(3 \pm i) \cdots(2)$

Clearly, the roots of (1) are conjugate complex numbers which is evident from (2). 

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