How many real roots does the equation \(x^2+16=0\) have?
Answer and explanation
Correct answer: 0
For a quadratic \(ax^2+bx+c=0\), the number of real roots depends on the discriminant \(D=b^2-4ac\). Here \(a=1,\;b=0,\;c=16\), so \(D=0^2-4\times1\times16=-64<0\). A negative discriminant means no real roots. Alternatively, since for any real x, \(x^2\ge0\), \(x^2+16\) cannot be zero. Exam tip: check the discriminant first — \(D>0\) gives 2 roots, \(D=0\) gives 1, \(D<0\) gives 0 real roots. The option “cannot be determined” is wrong because all coefficients are given.
Frequently asked questions
What is the correct answer to this question?
0
Why is this the correct answer?
For a quadratic \(ax^2+bx+c=0\), the number of real roots depends on the discriminant \(D=b^2-4ac\). Here \(a=1,\;b=0,\;c=16\), so \(D=0^2-4\times1\times16=-64<0\). A negative discriminant means no real roots. Alternatively, since for any real x, \(x^2\ge0\), \(x^2+16\) cannot be zero. Exam tip: check the discriminant first — \(D>0\) gives 2 roots, \(D=0\) gives 1, \(D<0\) gives 0 real roots. The option “cannot be determined” is wrong because all coefficients are given.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Roots of a Quadratic Equation.
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