Question:
At how many points will the curves \(y=x^2\) and \(y=-x^2-2x-1\) intersect in the real \((x,y)\) plane?
At how many points will the curves \(y=x^2\) and \(y=-x^2-2x-1\) intersect in the real \((x,y)\) plane?
Show Hint
Set the two expressions for y equal and check whether the resulting quadratic has real roots.
Updated On: Jul 20, 2026
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The Correct Option is A
Solution and Explanation
Step 1: Set up the equation for intersection.
Points where the two curves meet must satisfy both equations at the same time, so set the two expressions for \(y\) equal to each other.
\[ x^2=-x^2-2x-1 \]
Step 2: Simplify into a single quadratic equation.
Bring every term to one side.
\[ x^2+x^2+2x+1=0 \] \[ 2x^2+2x+1=0 \]
Step 3: Check the discriminant.
For a quadratic \(ax^2+bx+c=0\), the number of real roots is decided by the discriminant \(D=b^2-4ac\). Here \(a=2\), \(b=2\), \(c=1\).
\[ D=(2)^2-4(2)(1)=4-8=-4 \]
Step 4: Interpret the discriminant.
Since \(D<0\), the quadratic equation \(2x^2+2x+1=0\) has no real solution for \(x\). This means there is no real value of \(x\) at which the two curves give the same \(y\).
Step 5: Final conclusion.
Since no real \(x\) satisfies both equations, the two curves do not meet at any real point.
\[ \boxed{0} \]
Hence, the correct option is (A).
Points where the two curves meet must satisfy both equations at the same time, so set the two expressions for \(y\) equal to each other.
\[ x^2=-x^2-2x-1 \]
Step 2: Simplify into a single quadratic equation.
Bring every term to one side.
\[ x^2+x^2+2x+1=0 \] \[ 2x^2+2x+1=0 \]
Step 3: Check the discriminant.
For a quadratic \(ax^2+bx+c=0\), the number of real roots is decided by the discriminant \(D=b^2-4ac\). Here \(a=2\), \(b=2\), \(c=1\).
\[ D=(2)^2-4(2)(1)=4-8=-4 \]
Step 4: Interpret the discriminant.
Since \(D<0\), the quadratic equation \(2x^2+2x+1=0\) has no real solution for \(x\). This means there is no real value of \(x\) at which the two curves give the same \(y\).
Step 5: Final conclusion.
Since no real \(x\) satisfies both equations, the two curves do not meet at any real point.
\[ \boxed{0} \]
Hence, the correct option is (A).
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