If p(x) = -x^2 + 25, what are the real zeros of its graph?
Answer and explanation
Correct answer: 5 and -5
Set p(x)=0 to find zeros:
\(-x^2+25=0\) gives \(x^2=25\), so \(x=\pm5\). Thus the real zeros are 5 and −5. Option B is incorrect because it confuses the value of \(x^2\) (25) with the values of x; the roots are the square roots, \(\pm5\). Option C is wrong since p(0)=25, not 0. Option D is wrong because real zeros do exist. Exam tip: always solve p(x)=0 (or set y=0) and then solve the resulting equation step by step.
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What is the correct answer to this question?
5 and -5
Why is this the correct answer?
Set p(x)=0 to find zeros:
\(-x^2+25=0\) gives \(x^2=25\), so \(x=\pm5\). Thus the real zeros are 5 and −5. Option B is incorrect because it confuses the value of \(x^2\) (25) with the values of x; the roots are the square roots, \(\pm5\). Option C is wrong since p(0)=25, not 0. Option D is wrong because real zeros do exist. Exam tip: always solve p(x)=0 (or set y=0) and then solve the resulting equation step by step.
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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