NCERT Class X Chapter 4: Quadratic Equation Exercise 4.3 Question 1 (iii)

NCERT Class X Chapter 4: Quadratic Equation Exercise 4.3 1 (iii)

Question:

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them : 2x2 – 6x + 3 = 0

Given:

The quadratic equation: 2x2 – 6x + 3 = 0

To Find:

The nature of the roots. If real roots exist, find them.

Formula:

WKT, for a quadratic equation of the form ax2 + bx + c = 0, the discriminant (Δ) is given by:

Δ = b2 - 4ac


WKT, the nature of roots is determined by the value of the discriminant (Δ):

• If Δ > 0, there are two distinct real roots.

• If Δ = 0, there are two equal real roots.


• If Δ < 0, there are no real roots.


WKT, if the roots are real, they can be found using the quadratic formula:

x = -b ± √Δ 2a

Solution:

The given quadratic equation is 2x2 – 6x + 3 = 0.
Comparing it with the standard form ax2 + bx + c = 0, we have:

a = 2,  b = -6 and c = 3
Calculate the discriminant (Δ):

⇒ Δ = b2 - 4ac

⇒ Δ = (-6)2 - 4(2)(3)

⇒ Δ = 36 - 24

⇒ Δ = 12
Determine the nature of the roots:

Since Δ = 12, which is greater than 0 (Δ > 0), the equation has two distinct real roots.
Find the roots (since they are real):

⇒ x = -b ± √Δ 2a

⇒ x = -(-6) ± √12 2 × 2

⇒ x = 6 ± √(4 × 3) 4

⇒ x = 6 ± 2√3 4
Factor out 2 from the numerator and simplify:

⇒ x = 2(3 ± √3) 4
⇒ x = 3 ± √3 2
The two distinct real roots are:

x1 = 3 + √3 2

x2 = 3 - √3 2

Result:

The discriminant of the equation 2x2 – 6x + 3 = 0 is 12.

Since the discriminant is greater than zero (Δ > 0), there are two distinct real roots.

The roots are x = 3 + √3 2 and x = 3 - √3 2 .
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