Which are the solutions of x^2=19x+1?

Which are the solutions of x^2=19x+1?

The Correct Answer and Explanation is:

To solve the quadratic equation (x^2 = 19x + 1), we first rearrange it into standard form. Standard form for a quadratic equation is given by:

[
ax^2 + bx + c = 0
]

In our case, we can rewrite the equation as:

[
x^2 – 19x – 1 = 0
]

Here, (a = 1), (b = -19), and (c = -1). To find the solutions to this quadratic equation, we will use the quadratic formula:

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

Substituting the values of (a), (b), and (c) into the formula, we first calculate the discriminant ((b^2 – 4ac)):

1. Calculate (b^2):

[
b^2 = (-19)^2 = 361
]

2. Calculate (4ac):

[
4ac = 4 \cdot 1 \cdot (-1) = -4
]

3. Find the discriminant:

[
b^2 – 4ac = 361 – (-4) = 361 + 4 = 365
]

Now that we have the discriminant, we can substitute back into the quadratic formula:

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

Next, we calculate (\sqrt{365}). Approximating this value:

[
\sqrt{365} \approx 19.1
]

Now we substitute this approximation back into our formula:

[
x = \frac{{19 \pm 19.1}}{2}
]

This yields two possible solutions:

1. (x_1 = \frac{{19 + 19.1}}{2} \approx \frac{38.1}{2} \approx 19.05)
2. (x_2 = \frac{{19 – 19.1}}{2} \approx \frac{-0.1}{2} \approx -0.05)

Therefore, the solutions to the equation (x^2 = 19x + 1) are approximately:

[
x_1 \approx 19.05 \quad \text{and} \quad x_2 \approx -0.05
]

In conclusion, the solutions to the quadratic equation are (x \approx 19.05) and (x \approx -0.05).