### Solve the following quadratic equations: (iii) (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...

### Solve the following quadratic equations: (i) (ii)

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

### Solving the following quadratic equations by factorization method: (iii) (iv)

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

### Solve the following quadratic equations by factorization method only:

$x^{2}+x+1=0$ $x^{2}+x+1 / 4+3 / 4=0$ $x^{2}+2(x)(1 / 2)+(1 / 2)^{2}+3 / 4=0$ $(x+1 / 2)^{2}+3 / 4=0\left[\right.$ Since, $\left.(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$ $(x+1 / 2)^{2}+3 / 4 \times 1=0$...

### Solve the following quadratic equations by factorization method only:

$x^{2}-x+1=0$ $x^{2}-x+1 / 4+3 / 4=0$ $x^{2}-2(x)(1 / 2)+(1 / 2)^{2}+3 / 4=0$ $(x-1 / 2)^{2}+3 / 4=0\left[\right.$ Since, $\left.(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$ $(x-1 / 2)^{2}+3 / 4 \times 1=0$...

### Solve the following quadratic equations by factorization method only:

$x^{2}+2 x+2=0$ $x^{2}+2 x+1+1=0$ $x^{2}+2(x)(1)+1^{2}+1=0$ $(x+1)^{2}+1=0\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$ $\mathrm{since}, \mathrm{i}^{2}=-1 \Rightarrow 1=-\mathrm{i}^{2}$...

### Solve the following quadratic equations by factorization method only:

$4 x^{2}-12 x+25=0$ $4 x^{2}-12 x+9+16=0$ $(2 x)^{2}-2(2 x)(3)+3^{2}+16=0$ $(2 x-3)^{2}+16=0\left[\right.$ Since, $\left.(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$ $(2 x-3)^{2}+16 \times 1=0$...