Question

At how many points will the curves \( y = x^2 \) and \( y = -x^2 - 2x - 1 \) intersect in the real \( (x, y) \) plane?

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

Set both equations equal, form a quadratic in x, and check its discriminant.

Answer 1

The Correct Option is A

Step 1: Understand the question:
We need to find how many points in the real \( (x, y) \) plane satisfy both curve equations at the same time.
An intersection point is a point where both curves give the same \( y \) for the same \( x \).

Step 2: Set the two equations equal:
At an intersection, both expressions for \( y \) must be equal, so we write:
\[ x^2 = -x^2 - 2x - 1 \]
This gives one equation in \( x \) alone, and every real root of this equation is an intersection point.

Step 3: Simplify into standard quadratic form:
Move every term to one side:
\[ x^2 + x^2 + 2x + 1 = 0 \]
\[ 2x^2 + 2x + 1 = 0 \]
This is now a standard quadratic equation \( ax^2 + bx + c = 0 \), with \( a = 2 \), \( b = 2 \), \( c = 1 \).

Step 4: Use the discriminant to count real roots:
The number of real roots of a quadratic depends on its discriminant \( \Delta = b^2 - 4ac \).
\[ \Delta = (2)^2 - 4(2)(1) = 4 - 8 = -4 \]
Since \( \Delta < 0 \), the equation has no real roots, only a pair of complex roots.

Final Answer:
No real value of \( x \) satisfies the equation, so the two curves never meet in the real plane. \[ \boxed{0} \]