Factor over the set of COMPLEX numbers?

Question

n^2 + 4n +7

No matter what I seem to do, the answer always comes out wrong! Somebody help me!!!

Answer 1

Completing the square

n² + 4x + 7 = 0

n² + 4x + 7 - 7 = 0 - 7

n² + 4x = - 7

n² + 4x +_______ = - 7 +______

n² + 4x + 4 = - 7 + 4

(n + 2 )(n + 2) = - 3

(n + 2)² = - 3

(√n + 2)² = ± √- 3

The solution cannot be by real numbers but by imaginary numbers.

n + 2 = ± i√- 3

n + 2 - 2 = - 2 ± i1.732050808

n = - 2 ± i1.732050808

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Solving for +

n = - i0.267949192

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Solving -

n = - 2 - 1.732050808

n = - i3.732050808

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You can solve the above problem by using the quadratid formula.

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Answer 2

The easiest way to factor a complicated quadratic expression like this is to set it equal to zero and use the quadratic formula.

That is, n^2 + 4n + 7 = 0 =>
n = [ -4 +- rt(4^2 - 4*1*7) ] / ( 2*1 )
= [ -4 +- rt(-12) ] / 2
= [ -4 +- 2*rt(3)*i ] / 2
= -2 +- rt(3)*i

So it factors as:.
n^2 + 4n + 7
= [ n - (-2 + rt(3)*i) ] * [ n - (-2 - rt(3)*i) ]
= [ n + 2 - rt(3)*i ] * [ n + 2 + rt(3)*i ]

Answer 3

n^2 + 4n +7=(n+2)^2+3 and in complex numbers
a^2+b^2=(a+bi)(a-bi) so
n^2 + 4n +7=(n+2)^2+3=(n+2+[3^1/2]i)(n+2-[3^1/2]i)

Answer 4

Well, the roots would be, -2 + i*sqrt 3 or -2 - i*sqrt 3,
where i is square root of -1.

Answer 5

set equal to zero & use quadratic formula
n = (-4 +/- sqrt(16-28))/2
n = -2 +/- 1/2sqrt(-12)
n = -2 +/- i*sqrt(3)
factors = -2 +i*sqrt(3) and -2 - i*sqrt(3)

Answer 6

use the formula for ax^2+bx+c=0
x1,x2 (roots of the equation)=( -b +- sqrt(b^2-4ac))/2a
in ur case b=-4, a=1, c=7
n1,n2 = (-4 +-sqrt(16-28))/2*1 = -4+-i*sqrt(12) / 2
= -2+-i*sqrt(3) /2
n1 = (-2+i sqrt(3) )
n2 = (-2-i sqrt(3))

Answer 7

Well show us what you did then. That way we can guide you better.

Just apply the standard formula for solving quadratic equations. This will give you two values for n n1 and n2 when this equatiion is satisfied . The two factors will then be (n-n1) and (n-n2)