NCERT solution Class 10 quadratic equations

NCERT Solutions

Class 10 Maths

Quadratic Equations

NCERT Questions

Ex. 4.1 Q1
Ex. 4.1 Q2
Ex. 4.2 Q1
Ex. 4.1 Q2
Ex. 4.2 Q3
Ex. 4.2 Q4
Ex. 4.2 Q5
Ex. 4.2 Q6
Ex. 4.3 Q1
Ex. 4.3 Q2
Ex. 4.3 Q3
Ex. 4.3 Q4
Ex. 4.3 Q5

Ex. 4.3 Q1

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

(i) 2x² – 3x + 5 = 0

(ii) 3x² – 4√3x + 4 = 0

(iii) 2x² – 6x + 3 = 0

(i) 2x² – 3x + 5 = 0

Comparing the given quadratic equation with ax² + bx + c = 0, we have:

a = 2, b= – 3, c = 5

Now discriminant = b² – 4ac

= (-3)² – 4 * 2 * 5

= 9 – 40

= -31 < 0

Since b² – 4ac is negative.

Hence, the given quadratic equation has no real roots.

(ii) 3x² – 4√3x + 4 = 0

Comparing the given quadratic equation with ax² + bx + c = 0, we have:

a = 3, b= -4√3, c = 4

Now discriminant = b² – 4ac

= (-4√3)² – 4* 3 * 4

= 48 – 48

= 0

Thus, the given quadratic equation has two real roots which are equal.

Here, the roots are: -b/2a and –b/2a

i.e -(-4√3)/(2 * 3) and -(-4√3)/(2 * 3)

=> 4√3/6 and 4√3/6

=> 2√3/3 and 2√3/3

=> 2√3/(√3 * √3) and 2√3/(√3 * √3)

=> 2/√3 and 2/√3

(iii) 2x² – 6x + 3 = 0

Comparing the given quadratic equation with ax² + bx + c = 0, we have:

a = 2, b= -6, c = 3

Now discriminant = b² – 4ac

= (-6)² – 4* 2 * 3

= 36 – 24

= 12 > 0

Thus, the given quadratic equation has two real and distinct roots which are given as:

x = {-b ± √(b² – 4 * a * c)}/2a

=> x = [-(-6) ± √{(-6)² – 4 * 2 * 3)}]/(2 * 2)

=> x = {6 ± √(36 – 24)}/4

=> x = (6 ± √12)/4

=> x = (6 ± 2√3)/4

=> x = 2(3 ± √3)/4

=> x = (3 ± √3)/2

Thus, the roots are: (3 + √3)/2 and (3 - √3)/2

Video solution:

Quadratic Equations Class 10 Maths NCERT Chapter 4 NCERT Solutions

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NCERT solutions of Quadratic Equations Class 10 Maths - all questions of all exercises are solved. Q. No. 1, 2, 3, and so on. Complete explanation and detailed solution makes them the best videos for Class 10 Maths Quadratic Equations NCERT solutions.