Notes | EduRev

Ques 1: Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

(i) 2x² − 3x + 5 = 0
(ii) 3x² − 4√3 x+ 4 = 0
(iii) 2x² − 6x + 3 = 0

Sol: (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) 3x² - 4√3x + 4 = 0

Comparing the given quadratic equation with ax² + bx + c = 0, we get

a = 3
b = − 4√3
c = 4
∴

= (16 × 3) − 48
= 48 − 48 = 0

Thus, the given quadratic equation has two real roots which are equal. Here, the roots are:

       
⇒
Thus,

(iii) 2x²−6x + 3 = 0

Comparing it with the general quadratic equation, we have:

a = 2
b = − 6
c = 3

∴ b² − 4ac =(− 6)² − 4 (2) (3)
= 36 − 24
= 12 > 0

∴ The given quadratic equation has two real and distinct roots, which are given by

⇒

Thus, the roots are:


Ques 2: Find the values of k for each of the following quadratic equations, so that they have two equal roots: 
(i) 2x² + kx + 3 = 0  
(ii) kx (x − 2) + 6 = 0
Sol: (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

∴ k² − 24 = 0 ⇒ k =
⇒
Thus, the required values of k are

(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 = − 2k
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.

Ques 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.
Sol: 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
⇒ 2 x = 2 × 20 = 40

Thus, length = 40 m and breadth = 20 m

Ques 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.
Sol: Let the age of one friend = x years

∴ The age of the other friend = (20 − x) 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 less than 0.
∴ The quadratic equation (1) has no real roots.

Thus, the given equation is not possible.

Ques 5: Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.

Sol: Let the breadth of the rectangle be x metres.

Since, the perimeter 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     ...(1)

Comparing (1) with ax² + 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
∴ Length = (40 − x) = (40 − 20) m = 20 m.