Exercise 4.3 Solutions
Quadratic Equations
1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
We use the discriminant, D = b² – 4ac.
• If D > 0, roots are real and distinct.
• If D = 0, roots are real and equal.
• If D < 0, no real roots exist.
(i) 2x² − 3x + 5 = 0
D = (-3)2 – 4(2)(5) = 9 – 40 = -31.
Since D < 0, no real roots exist.
(ii) 3x² − 4√3x + 4 = 0
D = (-4√3)2 – 4(3)(4) = (16 × 3) – 48 = 48 – 48 = 0.
Since D = 0, roots are real and equal.
Roots = -b/2a = -(-4√3) / (2 × 3) = 4√3 / 6 = 2√3 / 3.
The roots are 2√3/3 and 2√3/3.
(iii) 2x² − 6x + 3 = 0
D = (-6)2 – 4(2)(3) = 36 – 24 = 12.
Since D > 0, roots are real and distinct.
Roots = -b ± √D2a = 6 ± √124 = 6 ± 2√34 = 3 ± √32.
The roots are (3 + √3)/2 and (3 – √3)/2.
2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.
For equal roots, the discriminant D must be 0 (D = b² – 4ac = 0).
(i) 2x² + kx + 3 = 0
D = k2 – 4(2)(3) = 0
k2 – 24 = 0 => k2 = 24 => k = ±√24 = ±2√6.
(ii) kx(x − 2) + 6 = 0
First, expand the equation: kx2 – 2kx + 6 = 0.
D = (-2k)2 – 4(k)(6) = 0
4k2 – 24k = 0
4k(k – 6) = 0
This gives k=0 or k=6. If k=0, the equation is not quadratic. So, we must have k = 6.
3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.
Let breadth be x m. Then length = 2x m.
Area = length × breadth = (2x)(x) = 2x².
Given Area = 800 m². So, 2x² = 800 => x² = 400 => x = ±20.
Since breadth cannot be negative, x = 20 m.
Yes, a real value for the breadth exists.
Breadth = x = 20 m.
Length = 2x = 40 m.
Answer: Yes, it is possible. Length = 40 m, Breadth = 20 m.
4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Let the age of one friend be x years. The age of the other friend is 20 - x years.
Four years ago, their ages were (x – 4) and (20 – x – 4) = (16 – x).
Product of their ages four years ago: (x – 4)(16 – x) = 48.
16x – x2 – 64 + 4x = 48
–x2 + 20x – 64 = 48
x2 – 20x + 112 = 0
Check if real roots exist using the discriminant:
D = b² – 4ac = (-20)² – 4(1)(112) = 400 – 448 = -48.
Since D < 0, no real roots exist for this equation.
Answer: No, the situation is not possible.
5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.
Let the length be l and breadth be b.
Perimeter = 2(l + b) = 80 => l + b = 40 => l = 40 – b.
Area = l × b = 400.
Substitute l: (40 – b) × b = 400
40b – b2 = 400
b2 – 40b + 400 = 0
Check if real roots exist using the discriminant:
D = b² – 4ac = (-40)² – 4(1)(400) = 1600 – 1600 = 0.
Since D = 0, real and equal roots exist. The situation is possible.
The root for b is -b/2a = -(-40)/(2*1) = 40/2 = 20 m.
Breadth = b = 20 m.
Length = l = 40 – b = 40 – 20 = 20 m.
(This means the park is actually a square).
Answer: Yes, it is possible. The park would have a length of 20 m and a breadth of 20 m.
