If \(p(x)=16x^2-9\), what are the x-axis intersections of the graph?
Answer and explanation
Correct answer: \(\left(\tfrac{3}{4},0\right)\) और \(\left(-\tfrac{3}{4},0\right)\)
x-axis intersections are the points where \(p(x)=0\). Solve \(16x^2-9=0\). Factor as \((4x-3)(4x+3)=0\), giving \(4x-3=0\) or \(4x+3=0\), hence \(x=\pm\tfrac{3}{4}\). Thus the intersections are \(\left(\tfrac{3}{4},0\right)\) and \(\left(-\tfrac{3}{4},0\right)\). Why other options fail: (C) \(\tfrac{4}{3}\) is the reciprocal error from mishandling \(4x\); (B) \(\pm3\) ignores the coefficient 16. Exam tip: view \(16x^2\) as \((4x)^2\) and use difference of squares to factor quickly.
Frequently asked questions
What is the correct answer to this question?
\(\left(\tfrac{3}{4},0\right)\) और \(\left(-\tfrac{3}{4},0\right)\)
Why is this the correct answer?
x-axis intersections are the points where \(p(x)=0\). Solve \(16x^2-9=0\). Factor as \((4x-3)(4x+3)=0\), giving \(4x-3=0\) or \(4x+3=0\), hence \(x=\pm\tfrac{3}{4}\). Thus the intersections are \(\left(\tfrac{3}{4},0\right)\) and \(\left(-\tfrac{3}{4},0\right)\). Why other options fail: (C) \(\tfrac{4}{3}\) is the reciprocal error from mishandling \(4x\); (B) \(\pm3\) ignores the coefficient 16. Exam tip: view \(16x^2\) as \((4x)^2\) and use difference of squares to factor 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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