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

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

Question:

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them :
3x2 – 4√3x + 4 = 0.

Given:

The quadratic equation: 3x2 – 4√3x + 4 = 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 3x2 – 4√3x + 4 = 0.
Comparing it with the standard form ax2 + bx + c = 0, we have:

a = 3

b = -4√3

c = 4
Calculate the discriminant (Δ):

⇒ Δ = b2 - 4ac

⇒ Δ = (-4√3)2 - 4(3)(4)

⇒ Δ = (16 × 3) - 48

⇒ Δ = 48 - 48

⇒ Δ = 0
Determine the nature of the roots:

Since Δ = 0, the equation has two equal real roots.
Find the roots (since they are real):

⇒ x = -b ± √Δ 2a

⇒ x = -(-4√3) ± √0 2 × 3

⇒ x = 4√3 6
Simplify the fraction:

⇒ x = 2√3 3

Result:

The discriminant of the equation 3x2 – 4√3x + 4 = 0 is 0.

Since the discriminant is equal to zero (Δ = 0), there are two equal real roots.

The roots are x = 2√3 3 .
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