To find the points of intersection between the curve $y = x^2 + 10x + 20$ and the straight line $y = 2x + 5$, we need to set the two equations equal to each other and solve for $x$.

Step 1: Set the equations equal to each other

$x^2 + 10x + 20 = 2x + 5$

Step 2: Rearrange the equation to form a quadratic equation

Subtract $2x + 5$ from both sides:

$x^2 + 10x + 20 - 2x - 5 = 0$

Simplify:

$x^2 + 8x + 15 = 0$

Step 3: Solve the quadratic equation

This quadratic equation can be solved using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the equation $x^2 + 8x + 15 = 0$, the coefficients are:

• $a = 1$
• $b = 8$
• $c = 15$

Substitute these values into the quadratic formula:

$x = \frac{-8 \pm \sqrt{8^2 - 4(1)(15)}}{2(1)}$

$x = \frac{-8 \pm \sqrt{64 - 60}}{2}$

$x = \frac{-8 \pm \sqrt{4}}{2}$

$x = \frac{-8 \pm 2}{2}$

This gives two solutions for $x$:

$x = \frac{-8 + 2}{2} = \frac{-6}{2} = -3$ $x = \frac{-8 - 2}{2} = \frac{-10}{2} = -5$

Step 4: Find the corresponding $y$ values

Now substitute these $x$ values back into the equation for the straight line $y = 2x + 5$ to find the corresponding $y$ coordinates:

1. For $x = -3$: $y = 2(-3) + 5 = -6 + 5 = -1$

2. For $x = -5$: $y = 2(-5) + 5 = -10 + 5 = -5$

Final Answer:

The points of intersection are:

• $(-3, -1)$
• $(-5, -5)$

Would you like more details or have any questions about the steps?

Here are 5 related questions:

1. How can we confirm that the points found are indeed points of intersection?
2. What method can be used to graphically determine the points of intersection?
3. How does the discriminant of a quadratic equation help in predicting the number of intersection points?
4. How would the solution change if the quadratic curve had no real intersection with the line?
5. Can we solve the intersection problem using a system of equations approach?

Tip: Always double-check your solutions by substituting the values back into both original equations to ensure they satisfy both.