If \(p(x)=2x^2-8\), at which \(x\)-values will the graph cut the \(x\)-axis?
Answer and explanation
Correct answer: -2 and 2
The graph meets the x-axis where \(p(x)=0\). Solve \(2x^2-8=0\) to get \(x^2=4\), so \(x=\pm 2\). Equivalently factor as \(2(x-2)(x+2)\), giving zeros \(-2\) and \(2\). Choice B would follow from the wrong step \(x^2=16\); C is wrong because \(p(0)=-8\) (not zero); D is wrong because there are two zeros, both \(2\) and \(-2\), not only \(4\). Exam tip: set \(p(x)=0\) first and factor out common constants to simplify solving.
Frequently asked questions
What is the correct answer to this question?
-2 and 2
Why is this the correct answer?
The graph meets the x-axis where \(p(x)=0\). Solve \(2x^2-8=0\) to get \(x^2=4\), so \(x=\pm 2\). Equivalently factor as \(2(x-2)(x+2)\), giving zeros \(-2\) and \(2\). Choice B would follow from the wrong step \(x^2=16\); C is wrong because \(p(0)=-8\) (not zero); D is wrong because there are two zeros, both \(2\) and \(-2\), not only \(4\). Exam tip: set \(p(x)=0\) first and factor out common constants to simplify solving.
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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