If the equation of a parabola is \(y=x^2-2kx+(k^2-4)\), how will it intersect the \(x\)-axis?
Answer and explanation
Correct answer: At two distinct points
On the \(x\)-axis, \(y=0\). Therefore, we solve \(x^2-2kx+k^2-4=0\). Its discriminant is \(\Delta=(-2k)^2-4(1)(k^2-4)=16>0\), so for every real value of \(k\), the equation has two distinct real roots. In fact, the roots are \(x=k-2\) and \(x=k+2\), so the parabola intersects the \(x\)-axis at \((k-2,0)\) and \((k+2,0)\). Remember: \(\Delta>0\) indicates two distinct points of intersection.
Frequently asked questions
What is the correct answer to this question?
At two distinct points
Why is this the correct answer?
On the \(x\)-axis, \(y=0\). Therefore, we solve \(x^2-2kx+k^2-4=0\). Its discriminant is \(\Delta=(-2k)^2-4(1)(k^2-4)=16>0\), so for every real value of \(k\), the equation has two distinct real roots. In fact, the roots are \(x=k-2\) and \(x=k+2\), so the parabola intersects the \(x\)-axis at \((k-2,0)\) and \((k+2,0)\). Remember: \(\Delta>0\) indicates two distinct points of intersection.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Nature of Roots.
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