To show that the [tex]\( x \)[/tex]-coordinates of points [tex]\( A \)[/tex] and [tex]\( B \)[/tex] satisfy the equation [tex]\( 5x^2 + 4xp + p^2 - 34 = 0 \)[/tex], we need to follow a sequence of algebraic steps to derive this quadratic equation from the given equations of the line and the circle.
### Step-by-Step Solution:
1. Write the equations:
- The equation of the line is: [tex]\( y = 2x + p \)[/tex].
- The equation of the circle is: [tex]\( x^2 + y^2 = 34 \)[/tex].
2. Substitute the equation of the line into the equation of the circle:
- Since [tex]\( y = 2x + p \)[/tex], substitute [tex]\( y \)[/tex] in the circle's equation:
[tex]\[
x^2 + (2x + p)^2 = 34
\][/tex]
3. Expand the substituted equation:
- Expand [tex]\( (2x + p)^2 \)[/tex]:
[tex]\[
(2x + p)^2 = 4x^2 + 4xp + p^2
\][/tex]
- Now substitute back into the circle's equation:
[tex]\[
x^2 + 4x^2 + 4xp + p^2 = 34
\][/tex]
4. Combine like terms:
- Combine all the terms on one side of the equation to set it to 0:
[tex]\[
5x^2 + 4xp + p^2 - 34 = 0
\][/tex]
Thus, we have successfully shown that the [tex]\( x \)[/tex]-coordinates of the points [tex]\( A \)[/tex] and [tex]\( B \)[/tex] satisfy the equation:
[tex]\[
5x^2 + 4xp + p^2 - 34 = 0
\][/tex]
This completes the demonstration.
