Question

7. What is the solution to x2 + 10x + 27 = 0 when written in the form a ± bi? (1 point)

a. x equals negative five plus or minus two i square root of two end square root

b. x equals five plus or minus i square root of two end square root

c. x equals five plus or minus two i square root of two end square root

d. x equals negative five plus or minus i square root of two end square root

Answer 1

The solution is
`-5(+-)1.4i`

Explanation:

• To calculate the roots of a quadratic equation , you should remember the well known formula :
  `(-b(+-)√(b^2-4ac) )/(2a)`, where "a" is the term linked to the squared term, "b" is the term linked to the linear term, and "c" is the constant term.
• In this case, a=1, b=10 and c=27. Then, applying this formula means doing the following calulations:
  `(-10(+-)√(10^2-4*1*27) )/(2*1) =(-10(+-)√(100-108) )/(2) =(-10(+-)√(-8) )/(2)`.
• Because we know that the imaginary unit is
  `i=√(-1)`, or alternatively,
  `i^2=-1`, we can express the previous expression as follows:
  `(-10(+-)√(8*(-1)) )/(2)`.
• We can now apply distributive property of the radical of a product , and obtain the following:
  `(-10(+-)√(8)*√(-1))/(2) =(-10(+-) 2.83i)/(2) =-5(+-)1.41i`, which is our final expression.

Answer 2

The correct answer is D. -5 +/- i√2

Explanation:

Let's solve for x, the equation given to us, this way, using the quadratic formula:

x = (- b +/- √b² - 4ac)/2a

Replacing with the real values, we have:

x = (-10 +/- √10² - 4 * 1 * 27)/2

x = (-10 +/- √100 -108)/2

x = (-10 +/- √-8)/2

x = (-10 +/- √-1 * 4 * 2)/2 Let's recall that √-1 = i

x = (-10 +/- 2i√2)/2

x = 2 (-5 +/- i√2)/2 (Dividing by 2 both the numerator and the denominator)

x₁ = -5 + i√2

x₂ = -5 - i√2

The correct answer is D. -5 +/- i√2