For the parabola \(y=x^2-2kx+(k^2-9)\), where \(k\) is any real number, how will it intersect the \(x\)-axis?
Answer and explanation
Correct answer: At two distinct points
Here, \(a=1, b=-2k, c=k^2-9\). Therefore, the discriminant is \(D=b^2-4ac=(-2k)^2-4(1)(k^2-9)=36\), which is positive for every real value of \(k\). Hence, the equation has two distinct real roots, so the parabola intersects the \(x\)-axis at two distinct points. In fact, the roots are \(x=k+3\) and \(x=k-3\). Option B would require \(D=0\), which is not the case here. Exam tip: \(D>0\) indicates two distinct real intersection points.
Frequently asked questions
What is the correct answer to this question?
At two distinct points
Why is this the correct answer?
Here, \(a=1, b=-2k, c=k^2-9\). Therefore, the discriminant is \(D=b^2-4ac=(-2k)^2-4(1)(k^2-9)=36\), which is positive for every real value of \(k\). Hence, the equation has two distinct real roots, so the parabola intersects the \(x\)-axis at two distinct points. In fact, the roots are \(x=k+3\) and \(x=k-3\). Option B would require \(D=0\), which is not the case here. Exam tip: \(D>0\) indicates two distinct real intersection points.
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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