Differentiation (Part-38) | S N De

 

5. Solve the following quadratic equations using Sridhar Acharya's formula :

$\,(i)\, 2x^2+(4i-5)x+8+i=0 \\ \,(ii)\, y^2-(1-2i)y+1+5i=0$

Sol.$\,(i)\, \,\,2x^2+(4i-5)x+8+i=0 \cdots(1)$

The discriminant of (1) is :

$\,D= (4i-5)^2-4 \times 2 \times (8+i)\\~~~~=-16-40i+25-64-8i\\~~~~=-55-48i\\~~~~=3^2+(8i)^2- 2\times 3 \times 8i\\~~~~=(3-8i)^2 \cdots(2)$

From (1), we get $\,\,x=\frac{-(4i-5) \pm \sqrt{D}}{2 \times 2} \\=\frac{-4i+5 \pm (3-8i)}{4}\,\,[\text{By (2)}]\\ \Rightarrow x=\frac{-4i+5+3-8i}{4}\\=\frac{-12i+8}{4}\\=2-3i\,\,\, \\ \text{or} \,\,\,x=\frac{-4i+5-(3-8i)}{4}\\ =\frac{4i+2}{4}\\=\frac{1}{2}+i $

Sol. $\,\,(ii)\, y^2-(1-2i)y+1+5i=0 \cdots(1)$

The discriminant of (1) is :

$\,D=[-(1-2i)]^2-4 \times 1 \times (1+5i)\\=(1-2i)^2-4-20i \\=1-4i-4-4-20i\\=-7-24i\\=3^2+(4i)^2- 2\times 3 \times 4i\\=(3-4i)^2 \cdots(2)$

Now the solution of (1) is given by : $\,\,x=\frac{-[-(1-2i)] \pm\sqrt{D}}{2 \times 1}\\=\frac{1-2i \pm(3-4i)}{2}\,\,\,[\text{By (2)}]\\ \Rightarrow x=\frac{1-2i+3-4i}{2}=(4-6i)/2=2-3i \\ \text{or}\,\,x=\frac{1-2i-(3-4i)}{2}=(-2+2i)/2=-1+i$

Read More :  QUADRATIC EQUATIONS (Part-19) | S.N. Dey Math Solution Series

$\,(iii)\, 2y^2+3y+8-6i=0 \, \\ \,(iv)\,\, 6x^2-(5+3i)x+11i-3=0$

Sol. $(iii)\, 2y^2+3y+8-6i=0 \cdots (1)$

The discriminant of (1) is : 

$\,D=3^2-4 \times 2 \times (8-6i)\\=9-8(8-6i)\\=9-64+48i\\=3^2+(8i)^2 + 2 \times 3 \times 8i \\=(3+8i)^2 \cdots (2)$

Hence , the solution of (1) is :

$\,y=\frac{-3\pm \sqrt{D}}{2 \times 2}\\=\frac{-3 \pm (3+8i)}{4}\\ \Rightarrow x=\frac{-3+3+8i}{4}=2i \\ \text{or}\,\,x=\frac{-3-(3+8i)}{2 \times 2}\\=(-6-8i)/4\\=-\frac 12(3+4i)$  

Sol. $\,(iv)\,6x^2-(5+3i)x+11i-3=0 \cdots(1)$

The discriminant of (1) is : 

$\,D=[-(5+3i)]^2-4 \times 6\times (11i-3)\\=(5+3i)^2-24(11i-3)\\=25+30i-9-264i+72\\=88-234i\\=13^2+(9i)^2 - 2\times 13\times 9i\\=(13-9i)^2 \cdots(2)$

Hence , the solution of (1) is :

$\,x=\frac{(5+3i) \pm \sqrt{D}}{2 \times 6}\\=\frac{5+3i \pm(13-9i)}{12} \\ \Rightarrow x=\frac{18-6i}{12}=\frac 12(3-i) \\ \text{or}\,\,x=\frac{5+3i-13+9i}{12}=\frac 13(-2+3i)$

$5(v)\,\, ix^2-6x-9i=0 \\ 5(vi)\,\, 3x^2+(11i-2)x+4-8i=0$

Sol. $5(v)\,\, ix^2-6x-9i=0 \\ \Rightarrow i^2x^2-6ix-9i^2=0 \\ \Rightarrow -x^2-6ix+9=0 \\ \Rightarrow  x^2+6ix-9=0 \cdots(1)$

The discriminant of (1) is : 

$D=(6i)^2 -4 \times 1 \times (-9)\\~~~=-36+36\\~~~=0 \cdots(2)$ 

Hence , the solution of (1) is : $\,\,x=\frac{-6i \pm \sqrt{D}}{2 \times 1}\\=\frac{-6i \pm 0}{2}\,\,\,[\text{By (2)}]\\=-3i,-3i$

Sol. $5(vi)\,\, 3x^2+(11i-2)x+4-8i=0  \cdots(1)$

The discriminant of (1) is : 

$D= (11i-2)^2-4 \times 3 \times (4-8i) \\=-121-44i+4-48+96i\\=-165+52i\\=2^2+(13i)^2+ 2 \times 2 \times 13i\\=(2+13i)^2 \cdots(2)$ 

Hence , the solution of (1) is : $\,\,x=\frac{-(11i-2) \pm \sqrt{D}}{2 \times 1}\\=\frac{-11i+2 \pm (2+13i)}{2}\,\,\,[\text{By (2)}]\\=\frac{1}{3}(2+i),\,\,-4i$

To get full PDF of S.N. Dey  Math Solutions on Quadratic Equations , click here.