Advertisements
Advertisements
प्रश्न
Solve the following quadratic equation:
x2 – 4x + 13 = 0
Advertisements
उत्तर
The 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 xx 1)`
= `(4 +- 6 i)/2`
= 2 ± 6i
∴ The roots of the given equation are 2 + 6i and 2 – 6i.
APPEARS IN
संबंधित प्रश्न
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:
8x2 + 2x + 1 = 0
Solve the following quadratic equation:
`2x^2 - sqrt(3) x + 1` = 0
Solve the following quadratic equation:
3x2 – 7x + 5 = 0
Solve the following quadratic equation:
x2 + 3ix + 10 = 0
Solve the following quadratic equation:
x2 + 4ix – 4 = 0
Solve the following quadratic equation:
ix2 – 4x – 4i = 0
Solve the following quadratic equation:
x2 – (2 + i) x – (1 – 7i) = 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 equation for x, y ∈ R:
`(x + "i"y)/(2 + 3"i")` = 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`
Solve the following quadratic equation.
`8"x"^2 + 2"x" + 1 = 0`
Solve the following quadratic equation.
8x2 + 2x + 1 = 0
Solve the following quadratic equation.
8x2 + 2x + 1 = 0
Solve the following quadratic equation.
8x2 + 2x + 1 = 0
Solve the following quadratic equation.
8x2 + 2x + 1 = 0
Solve the following quadratic equation.
8x2 + 2x + 1 = 0
