If the equation \(x^2-2\sqrt{n}x+4=0\) has real and equal roots, what is the value of \(n\)?
Answer and explanation
Correct answer: 4
For a quadratic equation to have real and equal roots, its discriminant must be zero. Here, \(a=1\), \(b=-2\sqrt{n}\), and \(c=4\), so \(D=b^2-4ac=(-2\sqrt{n})^2-4(1)(4)=4n-16\). Setting \(D=0\) gives \(4n-16=0\), hence \(n=4\). As an exam tip, use \(D=0\) immediately whenever the roots are stated to be equal and real.
Frequently asked questions
What is the correct answer to this question?
4
Why is this the correct answer?
For a quadratic equation to have real and equal roots, its discriminant must be zero. Here, \(a=1\), \(b=-2\sqrt{n}\), and \(c=4\), so \(D=b^2-4ac=(-2\sqrt{n})^2-4(1)(4)=4n-16\). Setting \(D=0\) gives \(4n-16=0\), hence \(n=4\). As an exam tip, use \(D=0\) immediately whenever the roots are stated to be equal and real.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Nature of Roots.
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