1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
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(i) 2x2 – 3x + 5 = 0
Comparing with ax2 + bx + c = 0:
a = 2, b = –3, c = 5
Discriminant, D = b2 – 4ac = (–3)2 – 4(2)(5) = 9 – 40 = –31
Since D < 0, no real roots exist.
Hence, 2x2 – 3x + 5 = 0 has no real roots. -
(ii) 3x2 – 4√3x + 4 = 0
Comparing with ax2 + bx + c = 0:
a = 3, b = –4√3, c = 4
D = b2 – 4ac = (–4√3)2 – 4(3)(4) = 48 – 48 = 0
As D = 0, equal real roots exist.
Roots = –b / (2a) = –(–4√3) / (2×3) = 4√3 / 6 = 2√3 / 3 = 2/√3
Roots: x = 2/√3 and x = 2/√3 -
(iii) 2x2 – 6x + 3 = 0
Comparing with ax2 + bx + c = 0:
a = 2, b = –6, c = 3
D = b2 – 4ac = (–6)2 – 4(2)(3) = 36 – 24 = 12
Since D > 0, two distinct real roots exist.
x = (–b ± √D) / 2a = (6 ± √12) / 4 = (6 ± 2√3) / 4 = (3 ± √3)/2
Roots: x = (3 + √3)/2 and x = (3 – √3)/2
2. Find the values of k for each of the following quadratic equations so that they have two equal roots:
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(i) 2x2 + kx + 3 = 0
Comparing with ax2 + bx + c = 0:
a = 2, b = k, c = 3
D = b2 – 4ac = k2 – 4(2)(3) = k2 – 24
For equal roots, D = 0 ⇒ k2 – 24 = 0 ⇒ k = ±√24 = ±2√6
Values of k: ±2√6 -
(ii) kx(x – 2) + 6 = 0
Expanding: kx2 – 2kx + 6 = 0
Comparing with ax2 + bx + c = 0:
a = k, b = –2k, c = 6
D = b2 – 4ac = (–2k)2 – 4(k)(6) = 4k2 – 24k
For equal roots, D = 0 ⇒ 4k(k – 6) = 0
⇒ k = 0 or k = 6
But if k = 0, the equation is not quadratic.
Hence, k = 6
3. Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m²?
Let breadth = l, then length = 2l.
Area = (2l)(l) = 2l2 = 800 ⇒ l2 = 400
l2 – 400 = 0
Comparing with ax2 + bx + c = 0: a = 1, b = 0, c = –400
D = b2 – 4ac = 0 – 4(1)(–400) = 1600 > 0
Hence, real roots exist.
l = ±20 ⇒ breadth = 20 m (positive value)
Length = 2 × 20 = 40 m
Breadth = 20 m, Length = 40 m
Area = (2l)(l) = 2l2 = 800 ⇒ l2 = 400
l2 – 400 = 0
Comparing with ax2 + bx + c = 0: a = 1, b = 0, c = –400
D = b2 – 4ac = 0 – 4(1)(–400) = 1600 > 0
Hence, real roots exist.
l = ±20 ⇒ breadth = 20 m (positive value)
Length = 2 × 20 = 40 m
Breadth = 20 m, Length = 40 m
4. Is the following situation possible? If so, determine their present ages.
Let the age of one friend be x years, other = (20 – x) years.
Four years ago: (x – 4) and (16 – x)
Equation: (x – 4)(16 – x) = 48
Expanding: 16x – x2 – 64 + 4x = 48
⇒ –x2 + 20x – 112 = 0
⇒ x2 – 20x + 112 = 0
a = 1, b = –20, c = 112
D = b2 – 4ac = (–20)2 – 4(1)(112) = 400 – 448 = –48
As D < 0, no real roots exist.
Hence, this situation is not possible.
Four years ago: (x – 4) and (16 – x)
Equation: (x – 4)(16 – x) = 48
Expanding: 16x – x2 – 64 + 4x = 48
⇒ –x2 + 20x – 112 = 0
⇒ x2 – 20x + 112 = 0
a = 1, b = –20, c = 112
D = b2 – 4ac = (–20)2 – 4(1)(112) = 400 – 448 = –48
As D < 0, no real roots exist.
Hence, this situation is not possible.
5. Is it possible to design a rectangular park with perimeter 80 m and area 400 m²?
Let length = l, breadth = b.
Perimeter = 2(l + b) = 80 ⇒ l + b = 40 ⇒ b = 40 – l
Area = l × b = l(40 – l) = 40l – l2 = 400
⇒ l2 – 40l + 400 = 0
a = 1, b = –40, c = 400
D = b2 – 4ac = (–40)2 – 4(1)(400) = 1600 – 1600 = 0
Thus, equal real roots exist.
l = –b / 2a = –(–40) / (2×1) = 40 / 2 = 20
Hence, l = 20 m, b = 40 – 20 = 20 m
Length = 20 m, Breadth = 20 m
Perimeter = 2(l + b) = 80 ⇒ l + b = 40 ⇒ b = 40 – l
Area = l × b = l(40 – l) = 40l – l2 = 400
⇒ l2 – 40l + 400 = 0
a = 1, b = –40, c = 400
D = b2 – 4ac = (–40)2 – 4(1)(400) = 1600 – 1600 = 0
Thus, equal real roots exist.
l = –b / 2a = –(–40) / (2×1) = 40 / 2 = 20
Hence, l = 20 m, b = 40 – 20 = 20 m
Length = 20 m, Breadth = 20 m