Quadratic formula / Shreedharacharya sutra:
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
By quadratic formula
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
By Quadratic formula:
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
By Quadratic Formula:
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
By Formula (Quadratic Formula) Method:
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