If \(w\) is a real number, what is the nature of the roots of the equation \(4x^2-4wx+w^2=0\)?
Answer and explanation
Correct answer: Real and equal for every real \(w\)
Here, \(a=4\), \(b=-4w\), and \(c=w^2\). Therefore, the discriminant is \(D=b^2-4ac=(-4w)^2-4(4)(w^2)=0\). Hence, for every real \(w\), the roots are real and equal; in fact, the equation is \((2x-w)^2=0\), giving the repeated root \(x=\frac{w}{2}\). Even when \(w=0\), the coefficient of \(x^2\) remains 4, so the equation is still quadratic. Exam tip: When \(D=0\), the roots are real and equal.
Frequently asked questions
What is the correct answer to this question?
Real and equal for every real \(w\)
Why is this the correct answer?
Here, \(a=4\), \(b=-4w\), and \(c=w^2\). Therefore, the discriminant is \(D=b^2-4ac=(-4w)^2-4(4)(w^2)=0\). Hence, for every real \(w\), the roots are real and equal; in fact, the equation is \((2x-w)^2=0\), giving the repeated root \(x=\frac{w}{2}\). Even when \(w=0\), the coefficient of \(x^2\) remains 4, so the equation is still quadratic. Exam tip: When \(D=0\), the roots are real and equal.
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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