For the quadratic equation \(2x^2-5x+q=0\) to have two real and equal roots, what should be the value of \(q\)?
Answer and explanation
Correct answer: \(\frac{25}{8}\)
A quadratic equation \(ax^2+bx+c=0\) has two real and equal roots when its discriminant, \(D=b^2-4ac\), is zero. Here, \(a=2\), \(b=-5\), and \(c=q\). Thus, \((-5)^2-4(2)(q)=0\), giving \(25-8q=0\) and hence \(q=\frac{25}{8}\). Exam tip: For equal roots, immediately apply the condition \(D=0\).
Frequently asked questions
What is the correct answer to this question?
\(\frac{25}{8}\)
Why is this the correct answer?
A quadratic equation \(ax^2+bx+c=0\) has two real and equal roots when its discriminant, \(D=b^2-4ac\), is zero. Here, \(a=2\), \(b=-5\), and \(c=q\). Thus, \((-5)^2-4(2)(q)=0\), giving \(25-8q=0\) and hence \(q=\frac{25}{8}\). Exam tip: For equal roots, immediately apply the condition \(D=0\).
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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