Question

Roots of equation x2 + 2x + 5 = 0 are:​

Answer 1

Step-by-step explanation:

The equation x^2 +2x+5=0 has to be solved.

The roots of a quadratic equation ax^2 + bx + c = 0 are given by (-b+-sqrt(b^2-4ac))/(2a)

For the given equation, a = 1, b = 2 and c = 5.

The roots of the equation are:

(-2+-sqrt(2^2-4*1*5))/(2*1)

= (-2+-sqrt(4 - 20))/2

= (-2+-sqrt(-16))/2

= (-2+- 4*i)/2

= -1+- 2*i

The roots of the equation are -1+- 2*i

MARIA-VIVANCO | STUDENT

1. subtract -5 to both sides:

x^2 +2x=-5

then you find half of what 2 is and add it's square to both sides

(x+1)^2=-4

Then you square root each side to get rid of the exponent. Since -4 is negative we use i.

So that becomes x+1=2i

x1: -1+2i

x2: -1-2i

To solve the equation x^2+2x+5 = 0. by completing the square.

x^2+2x +5 = 0.

=> x^2+2x = -5.

The left side x^2+2x+becomes a perfect square x^2+2x+1= x^+1^2, if we add 1. So we add 1 to both sides:

=> x^2+2x+1 = -5+1 = -4.

=> (x+1)^2 = -4.

=(x+1)^2 = (2)^2 *(i)^2 , where i^2 = -1.

Therefore x+1 = {(2)^2*i^2}^(1/2) . Or x+1 = - {(2)^2*i^2}^(1/2).

=> x+1= 2i or x+1= -2i.

Therefore x = -1+2i. Or x= -(1+2i).

For the beginning, we'll add -5 both sides, to move the constant on the right side of the equation, so that being more clear what we have to add to the left side to complete the square.

x^2 +2x = -5

We'll complete the square by adding the number 1 to both side, to get a perfect square to the left side.

x^2 +2x + 1 = -5 + 1

We'll write the left side as a perfect square:

(x + 1)^2 = -4

x + 1 = sqrt -4

x + 1 = 2i

x1 = -1 + 2i

x2 = -1 - 2i

The complex solutions of the equation are: { -1 + 2i ; -1 - 2i}.

Answer 2

SOLUTION

TO DETERMINE

The roots of the equation x² + 2x + 5 = 0

EVALUATION

Here the given equation is

x² + 2x + 5 = 0

Comparing with the general equation

ax² + bx + c = 0 we get a = 1 , b = 2 , c = 5

We now use the Sridhar Acharya formula to find the roots as below

$\displaystyle \sf{ x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a} }$

$\displaystyle \sf{ \implies x = \frac{ - 2 \pm \: \sqrt{ {2}^{2} - 4 \times 1 \times 5} }{2 \times 1} }$

$\displaystyle \sf{ \implies x = \frac{ - 2 \pm \: \sqrt{4 - 20} }{2} }$

$\displaystyle \sf{ \implies x = \frac{ - 2 \pm \: \sqrt{ - 16} }{2} }$

$\displaystyle \sf{ \implies x = \frac{ - 2 \pm \: 4i }{2} }$

$\displaystyle \sf{ \implies x = - 1 \pm \: 2i }$

Hence the required roots are - 1 + 2i , - 1 - 2i

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