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

Answer:

The correct answer is D. -5 +/- i√2

Step-by-step 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

Answer 2

Answer:

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

Step-by-step explanation:

• To calculate the roots of  a quadratic equation , you should remember the well known formula : $\frac{-b(+-)\sqrt{b^2-4ac} }{2a}$, 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: $\frac{-10(+-)\sqrt{10^2-4\times1\times27} }{2\times1} =\frac{-10(+-)\sqrt{100-108} }{2} =\frac{-10(+-)\sqrt{-8} }{2}$.
• Because we know that the imaginary unit is $i=\sqrt{-1}$, or alternatively, $i^2=-1$, we can express the previous expression as follows: $\frac{-10(+-)\sqrt{8\times(-1)} }{2}$.
• We can now apply distributive property of the radical of a product , and obtain the following:$\frac{-10(+-)\sqrt{8}\times\sqrt{-1}}{2} =\frac{-10(+-) 2.83i}{2} =-5(+-)1.41i$, which is our final expression.