Question

1. x2 + 4x + 3 = 0
2. x2 - 5x + 4 = 0
3. x2 + 7 = 0​

Answer 1

$(i) \: {x}^{2} + 4x + 3 = 0 \\ {x}^{2} + x + 3x + 3 = 0 \\ x(x + 1) + 3(x + 1) = 0 \\ (x + 1)(x + 3) = 0 \\ x = - 1, - 3$

$(ii) \: {x}^{2} - 5x + 4 \\ {x}^{2} - x - 4x + 4 = 0 \\ x(x - 1) - 4(x - 1) = 0 \\ (x - 1)(x - 4) = 0 \\ x = 1,4$

(iii) Solve : x2+7 = 0

Subtract 7 from both sides of the equation :

x2 = -7

When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:

x = ± √ -7

In Math, i is called the imaginary unit. It satisfies i2 =-1. Both i and -i are the square roots of -1

Accordingly, √ -7 =

√ -1• 7 =

√ -1 •√ 7 =

i • √ 7

The equation has no real solutions. It has 2 imaginary, or complex solutions.

x= 0.0000 + 2.6458 i

x= 0.0000 - 2.6458 i