If \(p(x)=x^2-6kx+9k^2\), at which x-value will its graph touch the x-axis?
Answer and explanation
Correct answer: \(x=3k\)
Factorizing gives \(p(x)=x^2-6kx+9k^2=(x-3k)^2\), so there is a repeated root at \(x=3k\) and the parabola touches the x-axis at that x-value. Alternatively, the discriminant \(D=(-6k)^2-4\cdot1\cdot9k^2=0\) shows a double root. Note: when \(k=0\) all choices collapse to 0, but for a general (nonzero) k the unique touching point is \(x=3k\). Exam tip: spot a perfect square trinomial or check discriminant = 0 to identify a tangent to the x-axis quickly.
Frequently asked questions
What is the correct answer to this question?
\(x=3k\)
Why is this the correct answer?
Factorizing gives \(p(x)=x^2-6kx+9k^2=(x-3k)^2\), so there is a repeated root at \(x=3k\) and the parabola touches the x-axis at that x-value. Alternatively, the discriminant \(D=(-6k)^2-4\cdot1\cdot9k^2=0\) shows a double root. Note: when \(k=0\) all choices collapse to 0, but for a general (nonzero) k the unique touching point is \(x=3k\). Exam tip: spot a perfect square trinomial or check discriminant = 0 to identify a tangent to the x-axis quickly.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Polynomials. Topic: Geometrical meaning of the zeroes of a polynomial..
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