Selina Concise Class 10 Math Chapter 5 Quadratic Equations Exercise 5D Solutions
Quadratic Equations Exercise 5D
(Q1) Solve, each of the following equations, using the formula
(i) x² – 6x = 27
= Solution:
The given equation is x² – 6x = 27
It transform general form of quadratic equation.
x² – 6x – 27 = 0
Comparing equation with general form of quadratic equation
ax2 + bx + c = 0
Here, a = 1, b = -6 and c = -27
By using formula,
x = b ± √b² – 4ac/2a
First we have to solve b² – 4ac,
b² – 4ac = (-6)² – 4×1× (-27)
= 36 + 108
= 144
x = – (-6) ± √144/2×1
= 6 ± 12/2 (√144 =12)
x = 6+12/2 or x = 6-12/2
x = 18/2 or x = -6/2
x = 9 or x = -3
(ii) x² – 10x + 21 = 0
= Solution:
The given equation is x² – 10x + 21 = 0
Comparing equation with general quadratic equation ax² + bx + c = 0
Here, a = 1, b = -10 and c = 21
By using formula,
x = -b ±/√b²- 4ac /2a
First we have to solve b² – 4ac,
b² – 4ac = (-10)² – 4 × 1 × 21
= 100 – 84
= 16
x = – (-10) ± √16/2×1
= 10 ± 4/2 (√16 = 4)
x = 10 + 4/2 or x = 10-4/2
x = 14/2 or x = 6/2
x = 7 or x = 3
(iii) x² – 6x – 10 = 0
= Solution:
The given equation is x² – 6x – 10 = 0
Comparing given equation with ax² + bx + c = 0
Here, a = 1, b = -6 and c = -10
By using formula,
x = – b ± √b² – 4ac/2a
First we have to solve b² – 4ac
b² – 4ac = (-6)² – 4×1× (-10)
= 36 + 40
= 76
x = – (-6) ± √76/2×1
x = 6 ± √76/2
x = 6 ± √19×4/2
x = 6 ± 2√19/2
x = 2 (3 ± √19)/2
x = 3 ± √19
x = 3+√19 or x = 3-√19
(iv) x² + 2x – 6 = 0
= Solution:
The given equation is x² + 2x – 6 = 0
Comparing given equation with ax² + bx + c = 0
Here, a = 1, b = 2 and c = -6
By using formula,
x = -b ± √b² – 4ac/2a
First we have to solve b² – 4ac
b² – 4ac = 2² – 4 × 1 × (-6)
= 4 + 24
= 28
x = -2 ± √28/2×1
x = -2 ± √14×2/2
x = -2 + 7√2/2 or x = -2-7√2/2
(v) 3x² + 2x – 1 = 0
= Solution:
The given equation is 3x² + 2x – 1 = 0
Comparing given equation with ax² + bx + c = 0
Here, a = 3, b = 2 and c = -1
By using formula,
x = b ± √b² – 4ac/2a
First we have to solve b² – 4ac
b² – 4ac = 2² – 4 × 3 × (-1)
= 4 + 12
= 16
x = -2 ± √16/2×3
x = -2 ± 4/6
x = -2+4/6 or x = -2-4/6
x = 2/6 or x = -6/6
x = 1/3 or x = -1
(vi) 2x² + 7x + 5 = 0
= Solution:
The given equation is 2x² + 7x + 5 = 0
Comparing given equation with ax² + bx + xc = 0
Here, a = 2, b = 7 and c = 5
By using formula,
x = – b ± √b² – 4ac/2a
First we have to solve b² – 4ac,
b² – 4ac = 7² – 4×2×5
= 49 – 40
= 9
x = -7 ± √9/2×2
= -7 ± √(3)²/4
= -7 ± ¾
x = -7+3/4 or x = -7-3/4
x = -4/4 or x = -10/4
x = -1/ or x = -5/2
(vii) 2/3 x = -1/6 x² -1/3
= Solution:
The given equation is 2/3 x = -1/3 x2 – 1/3
Transform the general quadratic equation.
-1/6 x² -2/3 x -1/3 = 0
1/6 x² + 2/3 x + 1/3 = 0 (multiply by (-1))
Multiply by 6 on both side,
6 × 1/6 x² + 6 × 2/3 x + 6 × 1/3 = 0
x² + 4x + 2 = 0
Comparing with ax² + bx + c = 0
Here, a = 1, b = 4, and c = 2
By using formula,
x = – b ± √b² – 4ac/2a
First we have to solve b² – 4ac
b² – 4ac = (4)² – 4×1×2
= 16 – 8
= 8
x = – 4 ± √8/2×1
x = -4± √4×2/2
x = -4± 2√2/2
x = -4± 2√2/2
x = 2 (2± √2)/2
x = 2 ± √2
x = 2 + √2 or x = 2 – √2
(viii) 1/15 x² + 5/8 = 2/3 x
= Solution:
The given equation is 1/15 x² + 5/3 = 2/3 x
Transform general equation,
1/15 x² – 2/3 x + 5/3 = 0
Multiply by 15 on both sides.
15 × 1/15 x² – 15 × 2/3 x + 15 × 5/3 = 0
x² – 5×2x + 5×5 = 0
x² – 10x + 25 = 0
Comparing with ax² + bx + c = 0
Here, a = 1, b = -10 and c = 25
By using formula,
x = -b ± √b² – 4ac/2a
First we have to solve b² – 4ac
b² – 4ac = (-10)² – 4 × 1 × (25)
= 100 – 100
= 0
x = – (-10) ± √0/2×1
x = 10 ±0/2
x = 10/2
x = 5
(ix) x² – 6 = 2√2 x
= Solution:
The given equation is x² – 6 = 2√2x
Transform the general quadratic equation.
x² – 2√2 x – 6 = 0
Comparing with ax² + bx + c = 0
Here, a = 1, b = – 2√2 and c = -6
By using formula,
x = – b ± √b² =- 4ac/2a
First we have to solve b² – 4ac
b² – 4ac = (-2√2)² – 4×1×(-6)
b² – 4ac = (4×2) + 24
= 8 + 24
= 32
x = – (-2√2) ± √32/2×1
= 2√2 ± √8×4/2
= 2√2 ± 2√8/2
= 2√2 ± 2√4×2/2
= 2√2 ± 4√2/2
= 2√2 (1 ± 2)/2
= √2 (1 ± 2)
x = √2 (1+2) or x = √2 (1-2)
x = 3√2 or x = -√2
(x) 4/x – 3 = 5/2x + 3
= Solution:
The given equation is 4/x – 3 = 5/2x + 3
4-3x/x = 5/2x+3
(4 – 3x) (2x + 3) = 5x
4 (2x + 3) – 3x (2x + 3) = 5x
8x + 12 – 6x² – 9x = 5x
-6x² – x – 5x + 12 = 0
-6x² + 6x – 12 = 0 (multiply by (-1)) (divide by 6)
6x²/6 + 6x/6 – 12/6 = 0
x² + x – 2 = 0
Comparing with ax² + bx + c = 0
Here, a = 1, b = 1 an c = -2
By using formula,
x = – b ± √b² – 4ac/2a
First we have to solve b² – 4ac
b² – 4ac = 12 – 4×1×(-2)
= 1 + 8
= 9
x = – 1 ± √9/2×1
= – 1 ± 3/3
x = -1+3/3 or x = -1-3/3
x = 2/3 or x = -4/3
(xi) 2x+3/x+3 = x+4/x+2
= Solution:
The given equation is 2x+3/x+3 = x+4/x+2
First we have to do cross multiplication.
(2x + 3) (x + 2) = (x + 4) (x + 3)
2x (x + 2) +3 (x + 2) = x (x + 3) + 4 (x + 3)
2x² + 4x + 3x + 6 = x² + 3x + 4x + 12
2x² – x² + 7x – 3x – 4x + 6 – 12 = 0
x² + 7x – 7x – 6 = 0
x² = 6 =0
x² = 6
x = ±√6
x = + √6 and x = -√6
(xii) √6x² – 4x – 2√6 = 0
= Solution:
The given equation is √6x² – 4x – 2√6 = 0
Comparing given equation with ax² + bx + c = 0
Here, a = √6, b = -4 and c = -2√6
By using formula,
x = -b ± √b² – 4ac/2a
First we have to solve b² – 4ac
b² – 4ac = (4)² – 4×(√6) (-2√6)
= 16 + 8 (√6×√6)
= 16 + 8 (6)
= 16 + 48
= 64
x = -(-4) ± √64/2 × √64
= 4 ± √64/2√6
= 4 ± 8/2√6
x = 4+8/2√6 or x = 4-8/2√6
x = 12/2√6 or x = -4/2√6
= 6/√6 or x = -2/√6
= √6×√6/√6 or x = -2/√6
x = √6 or x = -2/√6
(xiii) 2x/x – 4 + (2x – 5)/(x-3) = 8 1/3
= Solution:
The given equation is 2x/x – 4 + (2x-5)/(x-3) = 8 1/3
= 2x/(x – 4) + (2x – 5)/(x – 3) = 24+1/3
= (x-3) (2x) + (x-4) (2x-5)/(x-4) (x-3) = 25/3
= 2x² – 6x + 2x² – 5x – 8x/x2 – 3x – 4x + 12 = 25/3
= 4x² – 11x – 8x + 20/x² – 7x + 12 = 25/3
= 4x² – 19x + 20/x² – 7x + 12 = 25/3
= 3 (4x² – 19x + 20) = 25 (x² – 7x + 12)
= 12x² – 57x + 60 = 25x² – 175x + 300
= 12x² – 25x² – 57x + 175x + 60 – 300 = 0
= -13x² + 118x – 240 = 0
= 13x² – 118x + 240 = 0 (Multiply by (-1))
Comparing equation with ax² + bx + c = 0
Here, a = 13, b = -118, c = 240
By using formula
x = b ± √b²-4ac/2a
First we have to solve b² – 4ac
b² – 4ac = (-118)² – 4 × 13 × 240
= 13924 – 12480
= 1444
x = – (-118) ± √1444/2×13
= 118 ± 38/26
x = 118 + 38/26 or x = 118 – 38/26
x = 158/26 or x = 80/26
x = 6 or c = 40/13
(xiv) x-1/x-2 + x-3/x-4 = 31/3
=> Solution:
The given equation is x-1/x-2 + x-3/x-4 = 31/3
= x-1/x-2 + x-3/x-4 = 9+1/3
= (x-4) (x-1) + (x-3) (x-2)/(x-2) (x-4) = 10/3
= x² – x – 4x + 4 + x² – 2x – 3x + 6/x² – 4x – 2x + 8 = 10/3
= 2x² – 5x – 5x + 10/x² – 6x + 8 = 10/3
2x² – 10 + 10/x² – 6x + 8 = 10/3
= 3 (2x² – 10x + 10) = 10 (x² – 6x + 8)
= 6x² – 30x + 30 = 10x² – 60x + 80
= 6x² – 10x² – 30x + 60x + 30
= 0 – 4x² + 30x – 50 = 0
= 4x² – 30x + 50 = 0
Comparing with ax² + bx + c = 0
Here, a = 4, b = -30 and c = 50
By using formula,
x = -b ± √b² – 4ac/2a
First we have to solve b² – 4ac
b² – 4ac = (-30)² – 4 × 4 × (50)
= 900 – 800
= 100
x = – (-30) ± √100/2×4
x = 30±10/8
x = 30+10/8 or x = 30-10/8
x = 40/8 or x = 20/8
x = 5 or x = 5/2
(Q2) solve each of the following equation for x and give, in each case, your answer correct to one decimal
(i) x² – 8x + 5 = 0
5x² + 10x – 3 = 0
= (i) x² – 8x + 5 = 0
= Solution:
The given equation is x² – 8x + 5 = 0
Comparing given equation with ax² + bx + c = 0
Here, a = 1, b = -8 and c = 5
By using quadratic formula
x = b ± √b² – 4ac/2a
First we have to solve b² – 4ac
b² – 4ac = (-8)² – 4×1×5
= 64 – 20
= 44
x = – (-8) ± √44/2×1
= 8 ± √(11×4)/2
= 8 ± 2√11/2
x = 8+2√11/2 or x = 8 – 2√11/2
x = 8+2× (3.31)/2 or x = 8 – 2×(3.31)/2
x = 8+6.62/2 or x = 8 – 6.62/2
x = 8+6.62/2 or x = 8 – 6.62/2
x = 14.62/2 or x = 1.38/2
x = 7.31 or x = 0.69
Or
x = 8/2 + 2√11/2 or x = 8/2 – 2√11/2
= 4 + √11 or x = 4 – √11
= 4 + 3.3 or x = 4 – 3.3
x = 7.7 or x = 0.7
(ii) 5x² + 10x – 3 = 0
= Solution:
The given equation is 5x² + 10x – 3 = 0
Comparing with ax² + bx + c = 0
Here, a = 5, b = 10 and c = – 3
By using formula,
x = – b ± √b² – 4ac/2a
First we have to solve b² – 4ac
b² – 4ac = (10)² – 4×5×(-3)
= 100 + 60
= 160
x = 10 ± √160/2×5
x = – 10 ± √16×10/10
x = – 10 ± 4√10/10
x = -10 + 4√10/10 or x = – 10 – 4√10/10
= – 10/10 + 4√10/10 or x = -10/10 – 4√10/10
= – 1 + 4√10/√10×√10 or x = – 1 – 4√10/√10×√10
= – 1 + 4/√10 or x = – 1 – 4/√10
= – 1 + 4/3.1 or x = -1-4/3.1
= – 1 + 1.29 or x = – 1 – 1.29
x = 0.3 or x = -2.3
(Q3) Solve each of the following equation for x and give, in each case, your answer correct to 2 decimal places
(i) 2x² – 10x + 5 = 0
Solution-
The given equation is 2x² – 10x + 5 = 0
Comparing with ax² + bx + c = 0
Here, a = 2, b = – 10 and c = 5
By using formula,
x = – b ± √b² – 4ac/2a
First we have solve b² – 4ac
b² – 4ac = (-10)² – 4×2×5
= 100 – 40
= 60
x = – (-10) ± √60/2×2
x = 10 ± √4×15/4
= 10 ± 2√15/4
x = 10/4 ± 2√15/4
x = 10/4 + 2√15/4 or = 10/4 – 2√15/4
x = 5/2 + √15/2 or x = 5/2 -√15/2
x = 2.5 + 3.87/2 or x = 2.5 – 3.87/2
x = 2.5 + 1.93 or x = 2.5 – 1.93
x = 4.44 or x = 0.57
(ii) 4x + 6/x + 13 = 0
= Solution:
The given equation is 4x + 6/x + 13 = 0
Multiply by ‘x’
x × 4x + x × 6/x + x × 13 = 0
4x² + 6 + 13x = 0
4x² + 13x + 6 = 0
Comparing with ax² + bx + c = 0
Here, a = 4, b = 13, c = 6
By using formula,
x = – b± √b² – 4ac/2a
First we have to solve b² – 4ac
b² – 4ac = (13)² – 4×4×6
= 169 – 96
= 73
x = – 13± √73/2×4
x = – 13 ± √73/8
x = -13/8 ± √73/8
x = -13/8 + √73/8 or x = -13/8 – √73/8
x = – 1.62 + 8.54/8 or x = – 1.62 – 8.54/8
x = – 1.62 + 1.06 or x = – 1.62 – 1.06
x = – 0.56 or x = – 2.68
(iii) 4x² – 5x – 3 = 0
= Solution:
The given equation is 4x² – 5x – 3 = 0
Comparing with ax² + bx + c = 0
Here, a = 4, b = – 5 and c = – 3
By using formula,
x = -b ± √b² – 4ac/2a
First we have to solve b² – 4ac
b² – 4ac = (-5)² – 4×4×(-2)
= 25 + 48
= 73
x = – (-5) ± √73/2×4
x = 5 ± √73/8
x = 5/8 ± √73/8
x = 5/8 + √73/8
x = 5/8 + √73/8 or x = 5/8 – √73/8
x = 0.62 + 8.54/8 or x = 0.62 – 8.54/8
x = 0.62 + 1.06 or x = 0.62 – 1.06
x = 1.68 or x = -0.44
(iv) x² – 3x- 9 = 0
= Solution:
The given equation is x² – 3x – 9 = 0
Comparing with ax² + bx + c = 0
Here, a = 1, b = -3, and c = -9
By using formula,
x = – b± √b² – 4ac/2a
First we have to solve b² – 4ac
b² – 4ac = – (-3)² – 4×1×(-9)
= 9+36
= 45
x = – (-3) ± √45/2×1
x = 3 ± √9×5/2
x = 3 ± 3√5/2
x = 3+3√5/2 or x = 3-3√5/2
x = 3/2 + 3√5/2 or x = 3/2 – 3√5/2
= 1.5 + 3×1.73/2 or x = 1.5 – 3×1.73/2
= 1.5 + 5.19/2 or x = 1.5 – 5.19/2
= 1.5 + 2.59 or x = 1.5 – 2.59
x = 4.09 or x = – 1.09
(v) x² – 5x – 10 = 0
= Solution:
The given equation is
x² – 5x – 10 = 0
Comparing with ax² + bx + c = 0
Here, a = 1, b = -5, c = -10
By using formula,
x = – b ± √b² – 4ac/2a
First we have to solve b² – 4ac
b² – 4ac = (-5)² – 4×1× (-10)
= 25 + 40
= 65
x = – (-5) + √65/2×1
x = 5 ± √65/2
x = 5/2 + √65/2 or x = 5/2 – √65/2
x = 2.5 + 8.06/2 or x = 2.5 – 8.06/2
x = 2.5 + 4.03 or = 2.5 – 4.03
x = 6.53 or x = – 1.53
(Q4) Solve each of the following equations for x and give, in each case, your answer correct to 3 decimal places
(i) 3x² – 12x – 1 = 0
= Solution:
The given equation is 3x² – 12x – 1 = 0
Comparing with ax² + bx + c = 0
Here, a = 3, b = -12 and c = – 1
By using formula,
x = – b ± √b² – 4ac/2a
First we have to solve b² – 4ac
b² – 4ac = (-12)² – 4×3×(-1)
= 144 + 12
= 156
x = – (-12) ± √156/2×3
= 12 ± √156/6
x = 12 ± √4×39/6
x = 12 ± 2√39/6
x = 12/6 + 2√39/6 or x = 12/6 – 2/6 √39
x = 2 + 1/3 √39 or x = 2 – 1/3 √39
x = 2 + 6.244/3 or x = 2 – 6.244/3
x = 2 + 2.081 or x = 2 – 2.081
x = 4.081 or x = – 0.081
(ii) x² – 16x + 6 = 0
= Solution:
The given equation is x² – 16x + 6 = 0
Comparing with ax² + bx + c = 0
Here, a = 1, b = – 16 and c = 6
By using formula,
x = – b ± √b² – 4ac/2a
First we have to solve b² – 4ac
b² – 4ac = (-16)² – 4×1×6
= 256 – 24
= 232
x = – (-16) ± √232/2×1
= 16 ± √4×58/2
x = 16 ± 2√58/2
x = 16/2 + 2√58/2 or x = 16/2 – 2√58/2
x = 8 + √58 or x = 8 – √58
x = 8 + 7.615 or x = 8 – 7.615
x = 15.615 or x = 0.385
(iii) 2x² + 11x + 4 = 0
= Solution:
The given equation is 2x² + 11x + 4 = 0
Comparing with ax² + bx + c = 0
Here, a = 2, b = 11 and c = 4
By using formula,
x = – b ± √b² – 4ac/2a
First we have to solve b² – 4ac
b² – 4ac = (11)² – 4×2×4
= 121 – 32
= 89
x = 11 ± √89/2×2
x = – 11±√89/4
x = – 11/4 + √89/4 or x = -11/4 -√89/4
x = -2.75 + 9.433/4 or = -2.75 – 9.433/4
x = -2.75 + 2.3558 or x = -2.75 – 2.358
x = – 0.392 or x = – 5.108
(Q5) solve:
(i) x⁴ – 2x² – 3 = 0
= Solution:
The given equation is
x⁴ – 2x² – 3 = 0
= x⁴ – 3x² + x² – 3 = 0 (∵ -2x² = -3x² + x²)
= x² (x²-3) + 1 (x² – 3) = 0
= (x² – 3) (x² + 1) = 0
= x² – 3 = 0 or x² + 1 = 0
∴ x² + 1 = 0 is not possible
∴ x² – 3 = 0
x² = 3
x = ± √3
(ii) x⁴ – 10x² + 9 = 0
= Solution:
The given equation is
x⁴ – 10x² + 9 = 0
x⁴ – x² – 9x² + 9 = 0
x² (x² – 1) – 9 (x² – 1) = 0
(x² – 1) (x² – 9) = 0
x² – 1 = 0 or x² – 9 = 0
x² – 1 or x² = +9
x = ± 1 or x = ± 3