Solve [tex][tex]$2x^2 + 26 = 0$[/tex][/tex] to identify the roots.

A. [tex]x = -\sqrt{13}, \, x = \sqrt{13}[/tex]
B. [tex]x = -i\sqrt{13}, \, x = i\sqrt{13}[/tex]
C. [tex]x = -2\sqrt{28}, \, x = 2\sqrt{28}[/tex]
D. [tex]x = -2i\sqrt{26}, \, x = 2i\sqrt{26}[/tex]



Answer :

To solve the quadratic equation [tex]\(2x^2 + 26 = 0\)[/tex], we will follow these steps:

1. Rewrite the equation in standard form:
The given equation is already in standard form [tex]\(ax^2 + bx + c = 0\)[/tex]:
[tex]\[2x^2 + 26 = 0.\][/tex]

2. Identify the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[a = 2,\][/tex]
[tex]\[b = 0,\][/tex]
[tex]\[c = 26.\][/tex]

3. Calculate the discriminant [tex]\(\Delta\)[/tex]:
The discriminant [tex]\(\Delta\)[/tex] is given by:
[tex]\[\Delta = b^2 - 4ac.\][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[\Delta = 0^2 - 4 \cdot 2 \cdot 26 = 0 - 208 = -208.\][/tex]

4. Analyze the discriminant:
The discriminant [tex]\(\Delta = -208\)[/tex] is negative, which indicates that the roots of the quadratic equation are complex numbers (non-real roots).

5. Use the quadratic formula:
The general formula to solve the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is:
[tex]\[x = \frac{-b \pm \sqrt{\Delta}}{2a}.\][/tex]
Since [tex]\(\Delta = -208\)[/tex], let's substitute [tex]\(\Delta\)[/tex] and the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[x = \frac{0 \pm \sqrt{-208}}{2 \cdot 2} = \frac{\pm \sqrt{-208}}{4}.\][/tex]

6. Simplify the root:
We know that [tex]\(\sqrt{-208} = \sqrt{-1 \cdot 208} = i \sqrt{208},\)[/tex]
where [tex]\(i\)[/tex] is the imaginary unit ([tex]\(i = \sqrt{-1}\)[/tex]).

Therefore:
[tex]\[x = \frac{\pm i \sqrt{208}}{4}.\][/tex]

7. Simplify further:
We can simplify [tex]\(\sqrt{208}\)[/tex]:
[tex]\[208 = 16 \times 13 \implies \sqrt{208} = \sqrt{16 \times 13} = 4\sqrt{13}.\][/tex]

Thus:
[tex]\[x = \frac{\pm i \cdot 4 \sqrt{13}}{4} = \pm i \sqrt{13}.\][/tex]

Therefore, the roots of the equation [tex]\(2x^2 + 26 = 0\)[/tex] are:
[tex]\[x = -i \sqrt{13} \quad \text{and} \quad x = i \sqrt{13}.\][/tex]

So, the correct answer is:

[tex]\[x = -i \sqrt{13}, \quad x = i \sqrt{13}.\][/tex]