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 the two expressions for y equal and check the discriminant of the resulting quadratic.

Answer 1

The Correct Option is A

Step 1: Set the two curves equal to find intersection points.
Points where the two curves meet must satisfy both equations at once, so we set the two expressions for \(y\) equal to each other.
\[ x^2 = -x^2 - 2x - 1 \]

Step 2: Collect all terms on one side.
Add \(x^2 + 2x + 1\) to both sides to bring everything to the left.
\[ x^2 + x^2 + 2x + 1 = 0 \]
\[ 2x^2 + 2x + 1 = 0 \]

Step 3: Check the discriminant of this quadratic.
A quadratic \(ax^2+bx+c=0\) has real roots only when its discriminant \(D = b^2 - 4ac\) is zero or positive. Here \(a=2\), \(b=2\), \(c=1\).
\[ D = (2)^2 - 4(2)(1) = 4 - 8 = -4 \]

Step 4: Interpret the result.
Since \(D = -4\) is negative, the equation \(2x^2+2x+1=0\) has no real roots, only a pair of complex conjugate roots. That means there is no real value of \(x\) for which the two curves have the same \(y\)-value, so the two curves never meet anywhere in the real \((x,y)\) plane.

Final Answer:
The curves do not intersect at any real point. \[ \boxed{0} \]