Quadratic Equations

Let's Solve the Questions of

Class X.

4. Quadratic Equations.

Exercise 4.4 

Subject Mathematics 

Publication : NCERT.

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√3x + 4 = 0

(iii) 2x² – 6x + 3 = 0

Solution:

(i)

a = 2, b = -3, c = 5

D = b² -4ac

or, D = -3² -4×2×5

or, D = 9 -40

∴ D = -31

D<0

Hence, roots are not real.

(ii)

a = 3, b =-4√3, c = 4

D = b² -4ac

D = -4√3² -4×3×4

D = 48 - 48

D = 0

D = 0

Hence, roots are real and equal.

x = (-b ±√D)÷2a

or, x = (-(-4√3) ±√0)÷2×3

or, x = 4√3 ÷ 6

x = 2√3/3 Ans.

(iii)

a = 2, b = -6, c = 3

D = b² -4ac

or, D = -6² -4×2×3

or, D = 36 - 24

or, D = 12

D>0

Hence, roots are real and equal.

x = (-b ±√D)÷2a

x = (-(-6) ±√12)÷2×2

x = (6 ±2√3)÷4

x = (3 ±√3)÷2

∴ x = (3 +√3)/2 or (3 -√3)/2 Ans.

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

Solution:

2x² + kx + 3 = 0

a = 2, b = k , c = 3

For equal roots 

D = 0

b² - 4ac = 0

k² - 4.2.3 = 0

k² = 24

k = √24

k = ±2√6 Ans.

(ii)

kx(x - 2) + 6 = 0

kx² - 2kx + 6 = 0

a = k, b = -2k, c = 6

For equal roots

D = 0

b² - 4ac = 0

(-2k)² - 4.k.6 = 0

4k² - 24k = 0

4k(k - 6) = 0

Therefore,

k = 0 or 6 Ans.

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.

Solution:

Let the breadth be x m.

Length = 2x

Area = Length ×breadth

800 = 2x × x

or, 800 = 2x²

or, 2x² -800 = 0

or, x² -400 = 0

a = 1, b = 0, c = -400

D = b² -4ac

or, D = 0² -4×1×-400

or, D = 1600

Here, D > 0

So, roots are real and unequal.

Hence design is possible.

Therefore, 

x = (-b ±√D)÷2a

or, x = (-0 ±√1600)÷2

or, x = ±40÷2

or, x = ±20

Breadth not be in negative.

Therefore,

x = 20 m.

Length = 2x = 2×20 = 40 m.

Required Answers

40 m & 20 m Ans.

3. क्या एक एेसी आम की बगिया बनाना संभव है जिसकी लंबाई, चौड़ाई से दुगुनी हो और उसका क्षेत्रफल 800 m²  हो? यदि है, तो उसकी लंबाई और चौड़ाई ज्ञात कीजिए।

हल:

मान लिया कि चौडाई = x मीटर।

∴ लंबाई = 2x मीटर।

हम जानते है कि

क्षेत्रफल = लंबाई×चौड़ाई

800 = 2x × x

or, 800 = 2x²

or, 2x² -800 = 0

or, x² -400 = 0

a = 1, b = 0, c = -400

D = b² -4ac

or, D = 0² -4×1×-400

or, D = 1600

D > 0

अतः बगिया बनना संभव है।

x = (-b ±√D)÷2a

or, x = (-0 ±√1600)÷2×1

or, x = ±40÷2

चौडाई ऋणात्मक नही हो सकता है।

अतः x = 20 मीटर।

लंबाई = 2x = 2×20 = 40 मीटर।

अभिष्ट उत्तर

40 मीटर एवं 20 मीटर Ans.

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.

Solution:

Let the present age of one friend be x yrs.

Therefore,

Present age of second friend = 20 -x yrs.

4 yrs ago

(x -4)×(20 -x -4) = 48

or, (x -4)×(16 -x) = 48

or, 20x -x² -64 = 48

or, x² -20x + 112 = 0

a = 1, b = -20, c = 112

D = b² -4ac

or, D = (-20)² -4×1×112

or, D = 400 - 448

or, D = -48

Here D is less than 0

Hence, under this situation  age cannot be determined. 

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

Solution:

Let the length of the park is = x metre. 

Breadth = 80/2 -x 

or, Breadth = (40 -x)

Area = Length× breadth

400 = x(40 -x)

or, 400 = 40x - x²

ot, x² -40x + 400 = 0

a = 1, b = -40, c = 400

or, D = b² -4ac

or, D = -40² -4×1×400

or, D = 1600 -1600

or, D = 0

It is not possible to design rectangular park but possible to design square park.

x = -b ÷ 2a

or, x = -(-40) ÷ 2×1

or, x = 40 ÷ 2

or, x = 20 meter.

Breadth = 40 -x = 40 -20 = 20 meter

Hence, both Length and breadth are equal. 

So, design of square park is possible instead of rectangular park.