JKBOSE 10th Class Maths Solutions chapter – 4 Quadratic Equations

In chapter 2, we study about various forms of the Quadratic Polynomials. Let the general form of the quadratic polynomial be:

p (x) = ax² + bx + c; a ≠ 0

Now, equate this polynomial to the zero, we get a Quadratic Equation i.e. p (x) = 0

⇒      ax² + bx + c = 0; a ≠ 0

Quadratic Equations come up when we deal with many real life situations.

Many people believe that Babylonians were the first to solve quadratic equations. For instance, they knew how to find two positive numbers with a given positive sum and a given positive product, and this problem is equivalent to solving a quadratic equation of the form x² – px + q = 0. Greek mathematician Euclid developed geometrical approach for finding out lengths which, in our present day terminology, are solutions of quadratic equations. Solving of quadratic equations, in general form, is often credited to ancient Indian mathematicians. In fact, Brahmagupta (A.D. 598-665) gave an explicit formula to solve a quadratic equation of the form ax² + bx = c. Later, Sridharacharya (A.D. 1025) derived a formula, now known as the quadratic formula, (as quoted by Bhaskara II) for solving quadratic equation by the method of completing the square. An Arab mathematician Al-Khwarizmi (about A.D. 800) also studied quadratic equations of different types. Abraham bar Hiyya Ha-Nasi, in his book! ‘Liber embadorum’ published in Europe in A.D. 1145 gave complete solutions of different quadratic equations.

In this chapter, you will study quadratic equations, and various ways of finding their roots. You will also see some applications of quadratic equations in daily life situations.

Remark: When we write the terms of a Quadratic Equation in descending order of their degrees, is known as standard form of a Quadratic Equation i.e. ax²+bx+c = 0; a ≠ 0

Solution. (i) Given that

(x + 1)² = 2 (x − 3)

or      x² + 1 + 2x = 2x – 6

or      x² + 1 + 2x – 2x + 6 = 0

or      x² + 7 = 0

or      x² + 0x + 7 = 0

which is in the form of ax² + bx + c = 0 ; a ≠ 0

∴ It is a quadratic equation Ans.

(ii) Given that

x² – 2x = (-2) (3 – x)

or      x² – 2x = – 6 + 2x

or      x² – 2x + 6 – 2x = 0

or      x² – 4x + 6 = 0

which is the form of ax² + bx + c = 0; a ≠ 0

∴  It is the quadratic equation Ans.

(iii) Given that

(x – 2) (x + 1) = (x − 1) (x + 3)

or       x² + x – 2x – 2 = x² + 3x – x – 3

or      x² – x – 2 – x² – 2x + 3 = 0

or      – 3x + 1 = 0,

which have no term of x².

So it is not a quadratic equation Ans.

(iv) Given that

(x – 3) (2x + 1) = x (x + 5)

or       2x² + x – 6x – 3 = x² + 5x

or       2x² – 5x – 3 – x² – 5x = 0

or       x² – 10x – 3 = 0

which is a form of ax² + bx + c = 0; a ≠ 0

∴ It is a quadratic equation Ans.

(v) Given that

(2x – 1) (x – 3) = (x + 5) (x – 1)

or       2x² – 6x – x + 3 = x² – x + 5x – 5

or       2x² – 7x + 3 = x² + 4x – 5

or       2x² – 7x + 3 – x² – 4x + 5 = 0

or       x² – 11x + 8 = 0

which is a form of ax² + bx + c = 0; a ≠ 0

∴ It is a quadratic equation Ans.

(vi) Given that

x² + 3x + 1 = (x – 2)²

or      x² + 3x + 1 = x² + 4 – 4x

or      x² + 3x + 1 – x² – 4 + 4x = 0

or      7x – 3 = 0

which have no term of x².

So it is not a quadratic equation Ans.

(vii) Given that

(x + 2)³ = 2x (x² – 1)

or      x³ + (2)³ + 3 (x)² 2 + 3 (x) (2)² = 2x³ – 2x

or      x³ + 8 + 6x² + 12x = 2x³ – 2x

or      x³ + 8 + 6x² + 12x – 2x³ + 2x = 0

or      x³ + 6x² + 14x – 8 = 0

Here the highest degree of x is 3.

which is a cubic equation.

∴ It is not a quadratic equation Ans.

(viii) Given that

x³ – 4x² – x + 1 = (x – 2)³

Or      x³ – 4x² – x + 1 = x³ – (2)³ + 3 (x)² (- 2) + 3 (x) (- 2)²

Or 2x² – 13x + 9 = 0

which is in the form of ax2 + bx+c = 0; a ≠ 0

∴ It is a quadratic equation Ans.

Solution. (i) Let breadth of rectangular plot = x m

Length of rectangular plot = (2x + 1) m

∴ Area of rectangular plot = [x (2x + 1)] m² = (2x² + x) m²

According to question,

Hence, breadth of rectangular plot = 16 m

Length of rectangular plot = (2 × 16 + 1) m = 33 m

and given problem in the form of Quadratic Equation are

2x² + x – 528 = 0 Ans.

(ii) Let two consecutive positive integers are x and x + 1.

Product of Integers = x (x + 1) = x² + x

According to question,

∵ We are to study about the positive integers, so we reject x = – 18.

∴                    x = 17

Hence, two consecutive positive integers are 17, 17 + 1 = 18

and given problem in the form of Quadratic Equation is

x² + x – 306 = 0 Ans.

(iii) Let present age of Rohan = x years

Rohan’s mother’s age = (x + 26) years

After 3 years, Rohan’s age = (x + 3) years

Rohan’s mother’s age = (x + 26 + 3) years

(x + 29) years

∴ Their product = (x + 3) (x + 29)

= x² + 29x + 3x + 87

= x² + 32x + 87

According to question,

∵ age of any person cannot be negative so, we reject x = – 39

∴               x = 7

Hence, Rohan’s present age = 7 years

and given problem in the form of Quadratic Equation is

x² + 32x – 273 = 0 Ans.

(iv) Let u km/hour be the speed of train.

Distance covered by train = 480 km

Hence speed of train is 40 km/hr Ans.

(v) Given quadratic equation

Solution. (i) Let the number of marbles John had be x.

Then the number of marbles Jivanti had = 45 – x

The number of marbles left with John, when he lost 5 marbles =x – 5

The number of marbles left with Jivanti, when she lost 5 marbles

= 45 – x – 5

= 40 – x

Hence, number of marbles they had to start with were 36 and 9 or 9 and 36 Ans.

(ii) Let the number of toys produced on that day be x.

Therefore, the cost of production (in rupees) of each toy that day

= 55 – x

So, the total cost of production (in rupees) that day = x (55 – x)

According to question,

Hence, two numbers are 13 and 14 or 14 and 13 Ans.

Solution. Let one positive integer = x

2nd positive integer = x + 1

According to question,

∴ One positive integer = 13

and 2nd positive integer = 13 + 1 = 14

Hence, required consecutive positive integers are 13 and 14 Ans.

Solution. Let base of right triangle = x cm

Altitude of right triangle = (x – 7) cm

and hypotenuse of right triangle = 13 cm (Given)

According to Pythagoras Theorem,

(Base)² + (Altitude)² = (Hypotenuse)²

∵ Length of any triangle cannot be negative. So, we reject x = – 5

∴                     x = 12

Hence, base of right triangle = 12 cm

Altitude of right triangle = (12 – 7) cm

= 5 cm Ans.

Solution. Let, number of pottery articles produced by industry in one day = x

Cost of production of each article = ₹ (2x + 3)

∴ Total cost of production in particular day

Hence, number of articles produced on certain day = 6

and cost of production of each article

= ₹ [2 x 6 + 3]

= ₹ 15 Ans. 

Solution. (i) Given quadratic equation is

∴ There is no real x which satisfied the given quadratic equation.

Hence, given quadratic equation has no real roots. Ans.

Solution. (i) Given quadratic equation is

(iii) Given quadratic equation is

Since the square of a real number cannot be negative, therefore x will not have any real value.

Hence, there are no real roots for the given quadratic equation. Ans.

(ii) Given equation is

Hence, 2 and 1 are the roots of given quadratic equation. Ans.

Solution. Let Rehman’s present age = x years

3 years ago Rehman’s age = (x – 3) years

5 years from now Rehman’s age (x + 5) years

According to question,

∵ age cannot be negative, so, we reject x = -3

∴      x = 7

Hence, Rehman’s present age = 7 years.

Solution. Let Shefali get marks in Mathematics = x

∴ Shefali’s marks in English = 30 – x

According to 1st condition,

Shefali’s marks in Mathematics = x + 2

and Shefali’s marks in English

= 30 – x – 3

= 27 – x

∴ Their product = (x + 2) (27 – x)

= 27x – x² + 54 – 2x

= x² + 25x + 54

According to 2nd condition,

– x² + 25x + 54 = 210

Or     – x² + 25x + 54 – 210 = 0

Or     – x² + 25x – 156 = 0

Or       x² – 25x + 156 = 0

Compare it with ax² + bx + c = 0

∴         a = 1, b = 25, c = 156

Now, b² – 4ac = (- 25)² – 4 × 1 × 156

     = 625 – 624

     = 1 > 0

                     

then Shefali’s marks in Maths = 13

Shefali’s marks in English = 30 – 13 = 17

then Shefali’s marks in Maths = 12

Shefali’s marks in English = 30 – 12 = 18

Hence, Shefali’s marks in two subjects are 13 and 17 Or 12 and 18 Ans.

Solution. Let shorter side of rectangular field = AD = x m

Longer side of rectangular field = AB = (x + 30) m

and diagonal of rectangular field = DB = (x + 60) m

In rectangle, the angle between the length and breadth is right angle.

∴      ∠DAB = 90°

Now, in right angled triangle DAB, using Pythagoras Theorem,

(DB)² = (AD)² + (AB)²

(x + 60)² = (x)² + (x + 30)²

or       x² + 3600 + 120x = x² + x² +900 + 60x

or       x² + 3600 + 120x – 2x² – 900 – 60x = 0

or    – x² + 60x + 2700 = 0

or       x² – 60x – 2700 = 0

Compare it with ax² + bx + c = 0

∴      a = 1, b = – 60, c = – 2700

and  b² – 4ac = (- 60)² – 4 . 1 . (- 2700)

= 3600 + 10800

= 14400 > 0

∵ length of any side cannot be negative so, we reject x = − 30

∴                    x = 90

Hence, shorter side of rectangular field = 90 m

Longer side of rectangular field = (90 + 30) m = 120 m Ans.

Solution. Let larger number = x

Smaller number = y

According to 1st condition,

x² – y² = 180               ….(1)

According to 2nd condition,

y² = 8x                         …(2)

From (1) and (2), we get

x² – 8x = 180

or      x² – 8x – 180 = 0

Compare it with ax² + bx + c = 0

∴      a = 1, b = – 8, c = – 180

and  b² – 4ac = (- 8)² – 4 x 1 x (- 180)

= 64 + 720

= 784 > 0

When x = – 10 then from (2),

y² = 8 (- 10) = − 80, which is impossible. So, we reject x = – 10

When x = 18 then from (2),

y² = 8 (18) = 144

or      y = ± √144

or      y = ± 12

Hence, required numbers are 18 and 12 or 18 and – 12 Ans.

Solution. Let constant speed of the train = x km/hour

Distance covered by the train = 360 km

Increased speed of the train = (x + 5) km/hour

∴ Time taken by the train with increased speed = 360/x+5 hour

According to question,

∵ speed of any train cannot be negative. So, we reject x = – 45

∴                 x = 40

Hence, speed of train = 40 km/hour Ans.

Solution. Let time taken by larger tap to fill the tank = x hours.

Time taken by smaller tap to fill the tank = (x + 10) hours

From (1) and (2), we get

∵ time cannot be negative. So, we reject x = – 25/4

∴       x = 15

Hence, larger water tap fills the tank = 15 hours and smaller water tap fills the tank

= (15 + 10) hours

= 25 hours Ans.

Solution. Let average speed of passenger train = x km/hour

Average speed of express train

= (x + 11) km/hour

Distance between Mysore and Bangalore

= 132 km

∵ speed of any train cannot be negative

∴      x = 33

Hence, speed of passenger train

= 33 km/hour

and speed of express train

= (33 + 11) km/hour

= 44 km/hour Ans.

Let length of each side of square = x m

Area of square = x² m²

Perimeter of square = 4x m

Let length of each side of square = y m

Area of square = y² m²

Perimeter of square = 4y m

According to 1st condition,

x² + y² = 468            ….(1)

According to 2nd condition,

4x – 4y = 24

or               4 (x – y) = 24

or                     x – y = 6

or                          x = 6 + y          …(2)

From (1) and (2), we get

 (6 + y)² + y² = 468

or       36 + y² + 12y + y² = 468

or    2y² +12y + 36 – 468 = 0

or      2y² + 12y – 432 = 0

or      y² + 6y – 216 = 0

Compare it with ay² + by + c = 0

∴       a = 1, b = 6, c = – 216

and  b² – 4ac = (6)² – 4 × 1 × (- 216)

= 36 + 864

= 900 > 0

∵ length of square cannot be negative. So, we reject y = – 18

∴                y = 12

From (2), x = 6 + 12 = 18

Hence, sides of two squares are 12 m and 18 m Ans.

Solution. (i) Given quadratic equation is,

2x² 3x + 5 = 0

Compare it with ax² + bx + c = 0

∴     a = 2, b = 3, c = 5

      D = b² – 4ac

= (-3)² – 4 x 2 x 5

= 9 – 40

= – 31 < 0

Hence, given quadratic equation has no real roots. Ans.

(ii) Given quadratic equation is,

3x² – 4√3x + 4 = 0

Compare it with ax² + bx + c = 0

∴     a = 3, b = – 4√3, c = 4

      D = b² – 4ac

= (-4√3)² – 4 × 3 × 4

= 48 – 48 = 0

∴ given equation has real and equal roots.

Hence, roots of given quadratic equation are  Ans.

(iii) Given quadratic equation is :

2x² – 6x + 3 = 0

Compare it with ax² + bx + c = 0

∴      a = 2, b=-6, c = 3

       D = b² – 4ac =

= (- 6)² – 4 × 2 × 3

= 36 – 24

= 12 > 0

∴ given equation has real and distinct roots.

Hence, roots of given quadratic equation are Ans.

Solution. (1) Given quadratic equation is :

2x² + kx + 3 = 0

Compare it with ax² + bx + c = 0

∴       a = 2, b = k, c = 3

∵ roots of the given quadratic equation are equal.

∴      D = 0

b² – 4ac = 0

or (k)² – 4 × 2 × 3 = 0

or k² – 24 = 0

or k² = 24

or k = ±√24

or k = +±√6 Ans.

(ii) Given quadratic equation is :

kx (x – 2) + 6 = 0

or kx² – 2kx + 6 = 0

Compare it with ax² + bx + c = 0

∴ a = k, b = – 2k, c = 6

∵ roots of the given quadratic equation are equal

∴       D = 0

b² – 4ac = 0

or (- 2k)² – 4 x k x 6 = 0

or 4k² – 24k = 0

or 4k [k – 6] = 0

Either 4k = 0          or     k – 6 = 0

k = 0                         or     k = 6

k = 0, 6 Ans.

Solution. Let breadth of rectangular grove = x m

and length of rectangular grove = 2x m

Area of rectangular grove = length x breadth

= [x × 2x] m²

= 2x² m²

According to question

2x² = 800

x² = 800/2 = 400

x = ± √400

x = ±20

∵ length of rectangle cannot be negative. So, we reject x = – 20

∴ x = 20

∴ breadth of rectangular grove = 20 m

and length of rectangular grove = (2 × 20) m

= 40 m Ans.

Solution. Let age of one friend = x years

and age of 2nd friend = (20-x) years

Four years ago,

Age of 1st friend = (x – 4) years

Age of 2nd friend = (20 – x – 4) years

= (16 – x) years

∴ Their product = (x – 4) (16 – x)

= 16x – x² – 64 + 4x

= – x² + 20x – 64

According to Question

– x² + 20x – 64 = 48

Or     – x² + 20x – 64 – 48 = 0

Or     – x² + 20x – 112 = 0

Or       x² – 20x + 112 = 0                 …..(1)

Compare it with ax² + bx + c = 0

∴        a = 1, b = – 20, c = 112

D = b² – 4ac

= (- 20)² – 4 × 1 × 112

= 400 – 448

= – 48 < 0

∴ roots are not real

then no real value of x satisfies the quadratic equation (1).

Hence, given situation is not possible. Ans.

Let length of rectangular park = x m

Breadth of rectangular park = y m

∴ Perimeter of rectangular park = 2(x + y) m

and area of rectangular park = xy m²

According to 1st condition

2 (x + y) = 80

When          x = 20 then from (1)

y = 40 – 20 = 20

∴ Length and breadth of rectangular park are equal of measure 20 m.

Hence, given rectangular park exist and it is a square. Ans.