If \(p(x)=25x^2-36\), what are the x-axis intersections of its graph?
Answer and explanation
Correct answer: \(\left(\frac{6}{5},0\right),\ \left(-\frac{6}{5},0\right)\)
x-axis intersections occur where \(p(x)=0\). So solve \(25x^2-36=0\). Recognize a difference of squares: \((5x)^2-6^2=0\), hence \((5x-6)(5x+6)=0\). Solving gives \(x=\pm\tfrac{6}{5}\). Thus the intersections are \(\left(\tfrac{6}{5},0\right)\) and \(\left(-\tfrac{6}{5},0\right)\). Distractor B errs by effectively taking \(\sqrt{36}=6\) without accounting for the factor 25 on \(x^2\). Exam tip: either factor as a difference of squares or divide the equation by 25 first to simplify.
Frequently asked questions
What is the correct answer to this question?
\(\left(\frac{6}{5},0\right),\ \left(-\frac{6}{5},0\right)\)
Why is this the correct answer?
x-axis intersections occur where \(p(x)=0\). So solve \(25x^2-36=0\). Recognize a difference of squares: \((5x)^2-6^2=0\), hence \((5x-6)(5x+6)=0\). Solving gives \(x=\pm\tfrac{6}{5}\). Thus the intersections are \(\left(\tfrac{6}{5},0\right)\) and \(\left(-\tfrac{6}{5},0\right)\). Distractor B errs by effectively taking \(\sqrt{36}=6\) without accounting for the factor 25 on \(x^2\). Exam tip: either factor as a difference of squares or divide the equation by 25 first to simplify.
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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