It seems like you are working on question 5 from the image. The problem asks to:

Show algebraically that the line $y = kx + 5$ intersects the parabola $y = x^2 + 2$ twice for all values of $k$.

Let's go through the solution step-by-step.

1. Setting the two equations equal:

We need to find where the line $y = kx + 5$ intersects the parabola $y = x^2 + 2$. So we set the two equations equal to each other:

$kx + 5 = x^2 + 2$

2. Rearranging the equation:

Rearrange this to form a quadratic equation:

$x^2 - kx + 2 - 5 = 0$ $x^2 - kx - 3 = 0$

3. Solving the quadratic:

The quadratic equation is now $x^2 - kx - 3 = 0$. The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is given by $\Delta = b^2 - 4ac$. For the quadratic to have two real solutions, the discriminant must be positive or zero ($\Delta \geq 0$).

Here, $a = 1$, $b = -k$, and $c = -3$. So, the discriminant is:

$\Delta = (-k)^2 - 4(1)(-3)$ $\Delta = k^2 + 12$

4. Analyzing the discriminant:

Since $\Delta = k^2 + 12$, this value is always greater than zero for all values of $k$ (since $k^2 \geq 0$ and 12 is positive). This means that the quadratic equation has two distinct real solutions for any value of $k$.

Conclusion:

The line $y = kx + 5$ will always intersect the parabola $y = x^2 + 2$ twice, regardless of the value of $k$.

Would you like any further clarification or details on this? Here are some related questions you might want to explore:

1. Can the line ever be tangent to the parabola?
2. What happens to the intersection points when $k = 0$?
3. Can we solve for the exact intersection points for a specific $k$?
4. How does the slope $k$ affect the steepness of the line and its intersections?
5. Could we solve this problem using graphical methods?

Tip: Always analyze the discriminant of a quadratic equation to determine the number of solutions.