NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 4: Quadatric Equation - Exercise 4.1
Question 8

Question.  8

The quadratic \(2x^2-\sqrt{5}\,x+1=0\) has

(A)

two distinct real roots

(B)

two equal real roots

(C)

no real roots

(D)

more than two real roots

Detailed Answer with Explanation:

We are solving the quadratic equation \(2x^2 - \sqrt{5}x + 1 = 0\).

To check how many real roots it has, we use the discriminant formula:

\(D = b^2 - 4ac\), where \(a=2\), \(b=-\sqrt{5}\), and \(c=1\).

Now, calculate step by step:

\(b^2 = (-\sqrt{5})^2 = 5\).

\(4ac = 4 \times 2 \times 1 = 8\).

So, \(D = 5 - 8 = -3\).

Because \(D < 0\), the quadratic has no real roots. Instead, it will have complex roots.

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 4: Quadatric Equation – Exercise 4.1 | Detailed Answers