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What does the Fundamental Theorem of Algebra state about the equation [tex]$2x^2 - 6x + 10 = 0$[/tex]?

A. The Fundamental Theorem of Algebra tells you that the equation will have two complex roots since the leading coefficient of the equation is 2. The roots are [tex]$x = \frac{3 \pm i \sqrt{11}}{2}$[/tex].

B. The Fundamental Theorem of Algebra tells you that the equation will have two complex roots since the degree of the polynomial is 2. The roots are [tex][tex]$x = 3 \pm i \sqrt{11}$[/tex][/tex].

C. The Fundamental Theorem of Algebra tells you that the equation will have two complex roots since the degree of the polynomial is 2. The roots are [tex]$x = \frac{3 \pm i \sqrt{11}}{2}$[/tex].

D. The Fundamental Theorem of Algebra tells you that the equation will have two complex roots since the leading coefficient of the equation is 2. The roots are [tex]$x = 3 \pm i \sqrt{11}$[/tex].



Answer :

GinnyAnswer
To thoroughly understand the solution to the quadratic equation [tex]\( 2x^2 - 6x + 10 = 0 \)[/tex] using the Fundamental Theorem of Algebra, follow these steps:

Step 1: Understand the Quadratic Equation
The quadratic equation provided is:
[tex]\[ 2x^2 - 6x + 10 = 0 \][/tex]

Step 2: Identify the Coefficients
Identify the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] from the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex]:
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = -6 \)[/tex]
- [tex]\( c = 10 \)[/tex]

Step 3: Use the Quadratic Formula
To find the roots of the quadratic equation, you can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here:
- [tex]\(a = 2\)[/tex]
- [tex]\(b = -6\)[/tex]
- [tex]\(c = 10\)[/tex]

Step 4: Compute the Discriminant
The discriminant [tex]\(\Delta\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substitute in the values:
[tex]\[ \Delta = (-6)^2 - 4(2)(10) = 36 - 80 = -44 \][/tex]

Step 5: Interpret the Discriminant
Since the discriminant [tex]\(\Delta\)[/tex] is negative ([tex]\(\Delta = -44\)[/tex]), the quadratic equation will have two complex (non-real) roots.

Step 6: Calculate the Real and Imaginary Parts of the Roots
Given the discriminant, the roots can be computed as:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
With [tex]\(\Delta = -44\)[/tex], the square root [tex]\(\sqrt{-44}\)[/tex] is imaginary:
[tex]\[ \sqrt{-44} = \sqrt{44}i \][/tex]

[tex]\[ x = \frac{-(-6) \pm \sqrt{-44}}{2(2)} = \frac{6 \pm \sqrt{44}i}{4} \][/tex]

Simplify the expression:
[tex]\[ x = \frac{6 \pm 2\sqrt{11}i}{4} = \frac{3 \pm \sqrt{11}i}{2} \][/tex]

Therefore, the roots are:
[tex]\[ x = \frac{3 + \sqrt{11}i}{2} \][/tex]
[tex]\[ x = \frac{3 - \sqrt{11}i}{2} \][/tex]

Step 7: Verify the Answer
To confirm, compare these results to the given complex formats:
Given result breakdown:
- Discriminant: [tex]\(\Delta = -44\)[/tex]
- Real part of the roots: [tex]\( \frac{-b}{2a} = \frac{6}{4} = 1.5 \)[/tex]
- Imaginary part: [tex]\( \frac{\sqrt{|\Delta|}}{2a} = \frac{\sqrt{44}}{4} \approx 1.658 \)[/tex]

Les the roots match:
[tex]\[ 1.5 \pm 1.658i \text{ or } \frac{3 \pm i\sqrt{11}}{2} \][/tex]

So, the roots found match [tex]\( \left(1.5 + 1.658i\right) \)[/tex] and [tex]\( \left(1.5 - 1.658i\right) \)[/tex], which confirm:
[tex]\[ x = \frac{3 \pm \sqrt{11}i}{2} \][/tex]

Summary
The Fundamental Theorem of Algebra states that a polynomial of degree [tex]\(n\)[/tex] (where [tex]\(n \geq 1\)[/tex]) has exactly [tex]\(n\)[/tex] roots in the complex number system (including real and non-real roots). For the quadratic equation [tex]\(2x^2 - 6x + 10 = 0\)[/tex], it has two complex, non-real roots:
[tex]\[ x = \frac{3 \pm \sqrt{11}i}{2} \][/tex]

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