# NCERT Solution Exercise – 4.3 Class – 10 Maths

x=-b±√b²-4ac/2a

For Finding Nature of roots:

First we find Discriminant (D) : b²- 4ac

There are three different cases

Case I :

b²-4ac<0

Nature of Roots :  Imaginary Roots/ No Real roots

Case II :

b² – 4ac = 0

Nature of Roots : Real and two different roots

Case III :

b²- 4ac>0

Nature of Roots : Real and two different roots.

Q1. find the roots of the following quadratic equations, if they exist, by the method of the quadratic formula.

(I) 2x²-7x+3=0

Nature of nature:

b²- 4ac = (-7)²-4×2×3

⇒ b²- 4ac = 49 – 24

⇒ b²- 4ac = 25>0

Roots are Real and  Different

x = -b±√b²-4ac/2a

⇒ x = -(-7)±√25/2×2

⇒ x = 7±5/4

⇒ x=7+5/4  ;  x= 7-5/4

⇒ x=12/4   ;   x= 2/4

⇒ x = 3     ;  x=1/2

⇒ x =3,1/2

(II) 2x²+x-4=0

Nature of roots :

b²- 4ac = (1)² – 4×2×-4

⇒  b²- 4ac =  1+32

⇒  b²- 4ac =  33>0

⇒ b²-4ac >0

Hence Roots are real & Two Different

x = -b±√(b²-4ac) / 2a

⇒ x= -(1)±√33 / 2×2

⇒ x= -1±√33 / 4

⇒ x = -1+√33 / 4 ;    x = -1-√33/4

(III) 4x² + 4√3x + 3 = 0

Nature of roots:

⇒ b²-4ac  = (4√3)² – 4×4×3

⇒ b²-4ac = 16×3-48

⇒ b²-4ac =  48 – 48

⇒ b²-4ac = 0

Hence Roots are real and two equal roots

x = -b±√(b²-4ac)/2a

⇒ x = -4√3 ± √0/ 2×4

⇒ x= -4√3 ± 0/ 8

⇒ x = -4√3+0/8         ;     x =-4√3-0/8

⇒ x = -4√3/8             ;   x   = – 4√3/8

⇒ x = – √3/2  , -√3/2

(IV) 2x²+x+4=0

Nature of roots:

b²-4ac = (1)²-4(2)(4)

⇒ b²-4ac =  1-32

⇒ b²-4ac = -31<0

b²-4ac<0

Hence Roots are imaginary no solution.

Q3. Find the roots of the following  equations:-

(I) x-1/x=3 ; x≠0

Solution:

x²-1/x = 3

⇒ x²-1 = 3x

⇒ x² – 1 – 3x = 0

⇒ x² – 3x – 1 = 0

x = -b±√(b²-4ac)/2a

⇒ x = -(-3)±√(-3)²-4×1×-1/2×1

⇒ x = 3±√(9+4)/2

⇒ x = 3±√13/2

x = 3+√13/2      ,   3-√13/2

x = 3+√13/2 , 3-√13/2

(II) 1/(x+4)  –  1/(x-7) = 11/30 ; x≠-4,7

⇒ [(x-7)-(x+4)]/(x+4)(x-7)  = 11/30

⇒ [x – 7 – x – 4 ]/ x(x-7) + 4(x-7) = 11/30

⇒  -11/(x²-7x+4x-28) = 11/30

⇒  -1/(x² – 3x – 28 ) = 1/30

⇒ -30 = x² – 3x – 28

⇒ 0 = x² – 3x – 28 + 30

⇒ 0 = x² – 3x + 2

⇒ x² – 3x + 2 = 0

By Formula Method:

x = -b ± √(b²-4ac) / 2a

⇒ x = -(-3)±√[(-3)²+4(1)(2)]/2×1

x = 3±√1/2

x = 3±1/2

x = 3+1/2 , x = 3-1/2

x = 4/2 ; x = 2/2

x = 2 ; x = 1

x = 2,1

Q 4. The sum of the reciprocals of Rehman’s ages,(in years) 3 years ago and 5 years from now is 1/3, Find his present age.

Solution:

Let Rehman’s present age be = x years

3 years ago:

Reman’s age was = ( x – 3 ) years

5 years from now:

Rehman’s age = x + 5 years

According to question

1/(x-3) + 1/(x+5) = 1/3

Taking LCM

[ x+5 + x-3 ]/(x-3)(x+5) = 1/3

⇒ (2x + 2)/[x(x+5)-3(x+5)] = 1/3

⇒ (2x+2)/[ x² + 5x – 3x – 15] = 1/3

⇒ (2x+2)/[x² + 2x -15] = 1/3

⇒ 3(2x + 2) = x² + 2x -15

⇒ 6x + 6 = x² + 2x -15

⇒ x² + 2x -15 – 6x – 6 = 0

⇒ x² – 4x – 21 = 0

⇒ x² – (7-3)x – 21 = 0

⇒ x² – 7x + 3x – 21 = 0

⇒ x (x – 7) + 3(x – 7) =0

⇒ (x – 7)(x + 3) = 0

x – 7 = 0 ; x + 3 = 0

⇒ x = 7 ; x = – 3

Hence Rehman’s Present age = 7 Years

Q 5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product if their marks would have been 210, Find her marks in the two subjects.

Solution:

Let Mathematics marks = x

English marks = 30 – x

According to condition

( Mathematics marks + 2 ) ( English Marks -3 ) = 210

⇒ (x + 2) (30 – x – 3) = 210

⇒ (x + 2) (27 – x ) = 210

⇒ x( 27 – x ) + 2(27 – x) = 210

⇒ 27x – x² + 54 – 2x = 210

⇒ – x² + 25x + 54 – 210 = 0

⇒ – x² + 25x – 156 = 0

⇒ x² – 25x  + 156 = 0

⇒ x² – (13+12)x  + 156 = 0

⇒ x² – 13x – 12x  + 156 = 0

⇒ x(x – 13) – 12(x – 13) = 0

⇒ (x – 13)(x – 12) = 0

x – 13 = 0 ; x – 12 = 0

⇒ x = 13 ; x = 12

Hence

• If Shefali’s marks in Mathematics = 12 then in English 18
• If Shefali’s marks in Mathematics = 13 then in English 17

Q 6. The diagonal of a rectangular field is 60 meters more than the shorter side . If the longer side is 30 meters more than the shorter side, Find the sides of the field.

Solution:

Let the shorter side be = x m

Longer side = ( x + 30 ) m

Diagonal = ( x + 60 ) m

By P.G.T.

x² + (x+30)² = (x + 60)²

⇒ x² + x² + 60x + 900 = x² + 120x + 3600

⇒ x² + 60x + 900 – 3600 = 0

⇒ x² + 60x – 2700 = 0

⇒ x² + (90 – 30)x – 2700 = 0

⇒ x² + 90x – 30x – 2700 = 0

⇒ x(x + 90) – 30(x + 90) = 0

⇒ (x + 90)(x – 30) = 0

⇒ x+ 90 = 0 ; x – 30 = 0

⇒ x = – 90 ; x = 30

Hence

• Shorter side = 30 m
• Longer side = 60 m

Q 7. The difference of squares of two numbers is 180, The square of the smaller number is 8 times the larger number. Find the two numbers.

Solution:

Let the larger number be = x

Square of smaller number = eight times the larger number

⇒ (Smaller number)² = 8x

According to condition

(Larger Number)² – (Smaller number)² = 180

⇒ x² – 8x = 180

⇒ x² – (18 – 10)x = 180

⇒ x² – 18x + 10x – 180 = 0

⇒ x(x – 18) + 10(x – 18) = 0

⇒ (x – 18)(x + 10) = 0

⇒ x-18 = 0 ; x + 10 = 0

⇒ x = 18; x = -10

larger number = 18

(Smaller number)² = 8×18

⇒ (Smaller number)² = 144

⇒ (Smaller number) = √144

⇒ Smaller number = 12

Hence two numbers are 12 and 18