Which are the solutions of [tex]$x^2 = 19x + 1$[/tex]?

A. [tex]\frac{19 - \sqrt{19}}{2}, \frac{19 + \sqrt{19}}{2}[/tex]
B. [tex]\frac{19 - \sqrt{365}}{2}, \frac{19 + \sqrt{365}}{2}[/tex]
C. [tex]\frac{-19 - \sqrt{19}}{2}, \frac{-19 + \sqrt{19}}{2}[/tex]
D. [tex]\frac{-19 - \sqrt{365}}{2}, \frac{-19 + \sqrt{365}}{2}[/tex]



Answer :

To solve the quadratic equation \( x^2 - 19x - 1 = 0 \), let's use the quadratic formula:

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

Here, the coefficients are:
[tex]\[ a = 1, \; b = -19, \; c = -1 \][/tex]

Firstly, calculate the discriminant:

[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = (-19)^2 - 4(1)(-1) \][/tex]
[tex]\[ \Delta = 361 + 4 \][/tex]
[tex]\[ \Delta = 365 \][/tex]

Now, using the quadratic formula:

[tex]\[ x = \frac{-(-19) \pm \sqrt{365}}{2 \times 1} \][/tex]
[tex]\[ x = \frac{19 \pm \sqrt{365}}{2} \][/tex]

So, we have two solutions:

[tex]\[ x_1 = \frac{19 - \sqrt{365}}{2} \][/tex]
[tex]\[ x_2 = \frac{19 + \sqrt{365}}{2} \][/tex]

Therefore, the solutions to the equation \( x^2 - 19x - 1 = 0 \) are:

[tex]\[ \frac{19 - \sqrt{365}}{2} \;\; \text{and} \;\; \frac{19 + \sqrt{365}}{2} \][/tex]

Examining the given options, the correct solutions are:

[tex]\[ \frac{19 - \sqrt{365}}{2}, \frac{19 + \sqrt{365}}{2} \][/tex]

These match the second option given in the problem.

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