Class X Math
Quadratic Equations
Quadratic Equations
EXERCISE 4.4
Q1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x2 – 3x + 5 = 0 (ii) 
(iii) 2x2 – 6x + 3 – 0
Solution.
(i) 2x2 – 3x + 5 = 0
Comparing the given quadratic equation with ax2 + bx + c = 0, we have:
a = 2
b= – 3
c = 5
∴ The discriminant = b2 – 4ac
= ( – 3)2 – 4(2)(5)
= – 9 – 40
= – 31 < 0
Since b2 – 4ac is negative.
∴ The given quadratic equation has no real roots.
(ii)
Comparing the given quadratic equation with ax2 + bx + c= 0, we get
Thus, the given quadratic equation has two real roots which are equal. Here, the roots are:
(iii) 2x2 – 6x + 3 = 0
Comparing it with the general quadratic equation, we have:
a = 2
b = –6
c = 3
∴ The given quadratic equation has two real and distinct roots, which are given by
Q2. Find the values of k for each of the fallowing quadratic equations, so that they have two equal roots:
(i) 2x2 + kx + 3 = 0
(ii) kx(x – 2) + 6 = 0
Solution. (i) 2x2 + kx + 3 = 0
Comparing the given quadratic equation with ax2 + bx + c = 0, we get
a = 2
b = k
c = 3
b2 – 4ac = ( – k)2 – 4 (2) (3)
= k2 – 24
∵ For a quadratic equation to have equal roots,
b2 – 4ac = 0
(ii) kx(x – 2) + 6 = 0
Comparing kx (x – 2) + 6 = 0 i.e., kx2 – 2kx + 6 = 0 with ax2 + bx + c = 0, we get
a = k
b = – a
c = 6
∴ b2 – 4ac = ( – 2k)2 – 4 (k) (6)
= 4k2 – 24k
Since, the roots are real and equal,
∴ b2 – 4ac = 0
⇒ 4k2 – 24k = 0
⇒ 4k(K – 6) = 0
⇒ 4k = 0 or k – 6=0
⇒ k = 0 or k = 6
But k cannot be 0, otherwise, the given equation is no more quadratic. Thus, the required value of k = 6.
Q3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m? If so, find its length and breadth.
Solution. Let the breadth be x metres.
∴ Length = 2x metres
Now, Area = Length × Breadth
= 2x × x metre2
= 2x2 sq. metre.
According to the given condition,
2x2 = 800
Therefore, x = 20 and x = – 20
But x = – 20 is possible (∵ breadth cannot be negative).
∴ x = 20
⇒ 2x = 2 × 20 = 40
Thus, length = 40 m and breadth = 20 m
Q4. 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.
Solution. Let the age of one friend = x years
∴ The age of the other friend = (20 – x – 4) years [∵ Sum of their ages is 20 years]
Four years ago
Age of one friend = (x – 4) years
Age of the other friend = (20 – x – 4) years
= (16 – x) years
According to the condition,
(x – 4) × (16 – x) = 48
⇒ 16x – 64 – x2 – 4x = 48
⇒ – x2 – 20x – 64 – 48 = 0
⇒ – x2 – 20x – 112 = 0
⇒ x2 + 20x + 112 – 0 (1)
Here, a = 1, b = 20 and c = 112
∴ b2 – 4ac = (20)2 – 4 (1) (112)
= 400 – 448
= – 48 < 0
Since b2 – 4ac is Is than O.
∴ The quadratic equation (1) has no real roots.
Thus, the given equation is not possible.
Q.5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.
Since, the parimetee of the rectangle = 80 m
∴ 2 [Length + Breadth] = 80
2 [Length + x] 80
⇒ Length = (40 – x) metres
∴ Area of the rectangle = Length × breadth
= (40 – x) × x sq. m
= 40x – x2
Now, according to the given condition,
Area of the rectangle = 400 m2
∴ 40x – x2 = 400
⇒ – x2 + 40x – 400 = 0
⇒ x2 – 40x + 400 = 0
Comparing (1) with axe + bx + c = 0, we get
a = 1
b = –40
c = 400
∴ b2 – 4ac = ( – 40)2 – 4 (1) (400)
= 1600 – 1600 – 0
Thus, the equation (1) has two equal and real roots.
∴ Breadth, x = 20 m
Quadratic Equations EXERCISE 4.4
Quadratic Equations EXERCISE 4.4
Quadratic Equations EXERCISE 4.4