Solve the following quadratic equation: x2 – 4x + 13 = 0 - Mathematics and Statistics

Solve the following quadratic equation:

x² – 4x + 13 = 0

बेरीज

Given equation is x² – 4x + 13 = 0

Comparing with ax² + bx + c = 0, we get

a = 1, b = –4, c = 13

Discriminant = b² – 4ac

= (– 4)² – 4 × 1 × 13

= 16 – 52

= –36 < 0

So, the given equation has complex roots.

These roots are given by

x = `(-"b" ± sqrt("b"^2 - 4"ac"))/(2"a")`

= `(-(- 4) ± sqrt(- 36))/(2 1)`

= `(4 ± 6"i")/2`

= 2 ± 6i

∴ The roots of the given equation are 2 + 6i and 2 – 6i.

shaalaa.com

Solution of a Quadratic Equation in Complex Number System

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

पाठ 3: Complex Numbers - EXERCISE 3.2 [पृष्ठ ४०]

पाठ 3 Complex Numbers
EXERCISE 3.2 | Q 2) iv) | पृष्ठ ४०

Find the values of x and y which satisfy the following equations (x, y ∈ R):

`(x + 1)/(1 + "i") + (y - 1)/(1 - "i")` = i

Solve the following quadratic equation:

`2x^2 - sqrt(3) x + 1` = 0

Solve the following quadratic equation:

x² + 3ix + 10 = 0

Solve the following quadratic equation:

`x^2 - (3 sqrt(2) + 2"i") x + 6 sqrt(2)"i"` = 0

Solve the following quadratic equation:

x² – (5 – i)x + (18 + i) = 0

Solve the following quadratic equation:

(2 + i) x² – (5 – i) x + 2(1 – i) = 0

Solve the following equation for x, y ∈ R:

(1 – 3i) x + (2 + 5i) y = 7 + i

Solve the following equation for x, y ∈ R:

(x + iy)(5 + 6i) = 2 + 3i

Solve the following quadratic equations.

`8x^2+2x+1=0`