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1.↪ (i) 2x²–3x+5 = 0
Here, a = 2, b = –3 and c = 5
D = b² – 4ac
= (–3)² – 4(2)(5)
= 9 – 40
= –3 < 0
So, the given equation has no equal roots.
(ii) 3x²–4√3x+4 = 0
Here, a = 3, b = –4√3 and c = 4
D = b² – 4ac
= (–4√3)² – 4(3)(4)
= 48 – 48
= 0
So, the given equation has real roots.
x = –b ± √D / 2a
= –(–4√3) ± 0 / 2(2)
= √3
(iii) 2x² – 6x + 3 = 0
Here, a = 2, b = –6 and c = 3
D = b² – 4ac
= (–6)² – 4(2)(3)
= 36 – 24
= 12 > 0
So, the given equation has real roots.
x = –b ± √D / 2a
= –(–6) ± √12 / 2(2)
= 6 ± 2√3 / 4
= 3 ± √3 / 2
2. ↪ (i) 2x² + kx + 3 = 0
Here, a = 2, b = k and c = 3
D = b² – 4ac
= k² – 4(2)(3)
= k² – 24
The given equations will have real and equal roots, if
D = 0
=> k² – 24 = 0
=> k = ± √24
=> k = ± 2√6
(ii) kx (x–2) + 6 = 0
=> kx² – 2kx + 6 = 0
Here, a = k, b = –2k and c = 6.
D = b² – 4ac
= (–2k)² – 4 (k)(6)
= 4k² – 24k
= 4k( k– 6)
The equation will have real and equal roots, if
D = 0
=> 4k ( k –6 ) = 0
=> k = 0 or k = 6
3.↪ Let 2x be the length and x be the breadth of a rectangular mango grove.
Area = 2x × x
=> 800 = 2x²
=> x² = 800/2 = 400
=> x = ± √400
=> x = ± 20
The value of x is real so design of grove is possible.
Its length = 40m and breadth = 20m.
4.↪ Let 'x' years be the age of one friend.
Age of other friend = (20–x) years.
Four years ago,
Age of one friend = (x–4) years
Age of other friend = (20–x–4) = (16–x) years
A.T.Q.
(x–4) (x–16) = 48
=> 16x – x² – 64 + 4x = 48
=> x² – 20x + 112 = 0
Here, a = 1, b = –20 and c = 112
D = b² – 4ac
= (–20)² – 4(1)(112)
= 400 – 448
= –48 < 0
So, the given equation has no real roots.
Thus, the given situation is not possible.
5.↪ Let length be x metres and breadth be y metres.
Perimeter = 80 m
2 ( x + y ) = 80
x + y = 40 _(1)
Area = 400 m²
x(40–x) = 400
40x – x² = 400
x² – 40x + 400 = 0
Here, a = 1, b = –40 and c = 400
D = b² – 4ac
= (–40)² – 4(1)(400)
= 1600 – 1600
= 0
So, the given equation has equal roots
Its length and breadth is given by
x² – 40x + 400 = 0
(x–20)² = 0
x = 20, 20
Hence, Length = 20 m and breadth = 20 m
Thus, design is possible.
==================================
Hope my ans.'s satisfactory.☺
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1.↪ (i) 2x²–3x+5 = 0
Here, a = 2, b = –3 and c = 5
D = b² – 4ac
= (–3)² – 4(2)(5)
= 9 – 40
= –3 < 0
So, the given equation has no equal roots.
(ii) 3x²–4√3x+4 = 0
Here, a = 3, b = –4√3 and c = 4
D = b² – 4ac
= (–4√3)² – 4(3)(4)
= 48 – 48
= 0
So, the given equation has real roots.
x = –b ± √D / 2a
= –(–4√3) ± 0 / 2(2)
= √3
(iii) 2x² – 6x + 3 = 0
Here, a = 2, b = –6 and c = 3
D = b² – 4ac
= (–6)² – 4(2)(3)
= 36 – 24
= 12 > 0
So, the given equation has real roots.
x = –b ± √D / 2a
= –(–6) ± √12 / 2(2)
= 6 ± 2√3 / 4
= 3 ± √3 / 2
2. ↪ (i) 2x² + kx + 3 = 0
Here, a = 2, b = k and c = 3
D = b² – 4ac
= k² – 4(2)(3)
= k² – 24
The given equations will have real and equal roots, if
D = 0
=> k² – 24 = 0
=> k = ± √24
=> k = ± 2√6
(ii) kx (x–2) + 6 = 0
=> kx² – 2kx + 6 = 0
Here, a = k, b = –2k and c = 6.
D = b² – 4ac
= (–2k)² – 4 (k)(6)
= 4k² – 24k
= 4k( k– 6)
The equation will have real and equal roots, if
D = 0
=> 4k ( k –6 ) = 0
=> k = 0 or k = 6
3.↪ Let 2x be the length and x be the breadth of a rectangular mango grove.
Area = 2x × x
=> 800 = 2x²
=> x² = 800/2 = 400
=> x = ± √400
=> x = ± 20
The value of x is real so design of grove is possible.
Its length = 40m and breadth = 20m.
4.↪ Let 'x' years be the age of one friend.
Age of other friend = (20–x) years.
Four years ago,
Age of one friend = (x–4) years
Age of other friend = (20–x–4) = (16–x) years
A.T.Q.
(x–4) (x–16) = 48
=> 16x – x² – 64 + 4x = 48
=> x² – 20x + 112 = 0
Here, a = 1, b = –20 and c = 112
D = b² – 4ac
= (–20)² – 4(1)(112)
= 400 – 448
= –48 < 0
So, the given equation has no real roots.
Thus, the given situation is not possible.
5.↪ Let length be x metres and breadth be y metres.
Perimeter = 80 m
2 ( x + y ) = 80
x + y = 40 _(1)
Area = 400 m²
x(40–x) = 400
40x – x² = 400
x² – 40x + 400 = 0
Here, a = 1, b = –40 and c = 400
D = b² – 4ac
= (–40)² – 4(1)(400)
= 1600 – 1600
= 0
So, the given equation has equal roots
Its length and breadth is given by
x² – 40x + 400 = 0
(x–20)² = 0
x = 20, 20
Hence, Length = 20 m and breadth = 20 m
Thus, design is possible.
==================================
Hope my ans.'s satisfactory.☺
Anonymous:
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