Contents

**GSEB Solutions for Class 10 mathematics – Quadratic Equations (English Medium)**

**Exercise – 4.1**

**Question 1:**

Examine whether the following equations are quadratic equations or not :

- \(x+\frac{1}{x}=2\), X ≠ 0
- (x – 2)(x + 3) = 0
- 2x
^{2}– \(\sqrt{5}\)x + 2 = 0 - \(\frac{1}{x+1}-\frac{1}{x-1}=3\)(x ≠ ±1)
- (2x + 1)(2x – 1)-(4x + 3)(x – 5)
- \(\frac{x-1}{x+1}-\frac{x+1}{x-1}=\frac{2}{3}\)(x ≠ ±1)
- (2x + 3)
^{2}– (3x + 2)^{2}= 13

**Question 1(1):**

**Solution :**

**Question 1(2):**

**Solution :**

(x – 2)(x + 3) = 0

∴ x^{2 }+ 3x – 2x – 6 =0

∴ x^{2 }+ x – 6 = 0

Comparing x^{2 }+ x – 6 = 0 with the standard quadratic equation ax^{2 }+ bx + c = 0, we get, a = 1, b = 1, c = -6

It is clear that a = 1 ≠ 0 and a, b, c ∊ R.

Here degree of the polynomial x^{2 }+ x – 6 is 2.

∴ x^{2}^{ }+ x – 6 = 0 i.e., (x – 2)(x + 3) = 0 is a quadratic equation.

**Question 1(3):**

**Solution :**

**Question 1(4):**

**Solution :**

**Question 1(5):**

**Solution :**

For (2x + 1)(2x – 1) = (4x + 3)(x – 5)

⇒ 4x^{2 }– 2x + 2x – 1 = 4x^{2 }– 20x + 3x – 15

⇒ -1 = -17x – 15

⇒ 17x + 14 = 0

Comparing 17x + 14 = 0 with the standard quadratic equation ax^{2 }+ bx + c = 0,

We get, a = 0

But for any quadratic equation a ≠ 0.

⇒ 17x + 14 = 0 i.e., the given equation is not a quadratic equation.

**Question 1(6):**

**Solution :**

**Question 1(7):**

**Solution :**

(2x+3)^{2 }– (3x+2)^{2}=13

∴4x^{2 }+ 12x + 9 – (9x^{2 }+ 12x + 4) = 13

∴-5x^{2 }+ 5 = 13

∴ 5x^{2 }+ 8 = 0

Comparing 5x^{2}+8=0 with the standard quadratic equation ax^{2}+bx+c=0, we get, a = 5, b = 0, c = 8

Here a = 5 ≠ 0 and a, b, c ∊ R.

Here degree of the polynomial 5x^{2 }+ 8 is 2.

∴ 5x^{2 }+ 8 = 0 i.e., the given equation is a quadratic equation.

**Question 2:**

Verify whether the given value of x is a solution of the quadratic equation or not :

- X
^{2}– 3x + 2 = 0, x = 2 - X
^{2}+ x – 2 = 0, x = 2 - \(\frac{1}{3x+1}-\frac{1}{2x-1}+\frac{3}{4}=0\), x = 1[x ≠ \(\frac{1}{2}\), \(-\frac{1}{3}\) ]
- (3x – 8)( 2x + 5) = 0, x = \(-\frac{5}{2}\)

**Question 2(1):**

**Solution :**

Let p(x) = x^{2 }– 3x + 2.

Substituting x = 2 in p(x).

p(2) = (2)^{2 }– 3(2) + 2

= 4 – 6 + 2

= 0

∴ x = 2 is the zero of the polynomial x^{2 }– 3x + 2.

∴ One of the solutions of p(x) is x = 2.

**Question 2(2):**

**Solution :**

Let p(x) = x^{2 }+ x – 2

Substituting x = 2 in p(x)

P(2) = (2)^{2 }+ 2 – 2

= 4 + 2 – 2

= 4 ≠ 0

∴ x = 2 is not the zero of the polynomial x^{2 }+ x – 2.

∴ x = 2 is not a solution of x^{2 }+ x – 2 = 0.

**Question 2(3):**

**Solution :**

**Question 2(4):**

**Solution :**

**Question 3:**

- If x = 1 is a root of ax
^{2}+ bx + c = 0, a ≠ 0, a, b, c ∈ R, prove that a + b + c = 0. - If x = -1 is a root of x
^{2}– px + q = 0, p, q ∈ R, prove that p + q + 1 = 0. - Find k, if one of the roots of x
^{2}– kx + 6 = 0 is 3. - Find k, if one of the roots of x
^{2}+ 3(k + 2)x – 9 = 0 is -3.

**Question 3(1):**

**Solution :**

x = 1 is given as the solution of ax^{2 }+ bx + c = 0

∴ p(1) = 0

∴ a(1)^{2 }+ b(1) + c = 0

∴a + b + c = 0

If x = 1 is the solution of ax^{2 }+ bx + c = 0, then a + b + c = 0.

**Question 3(2):**

**Solution :**

x = -1 is given as the solution of x^{2 }– px + q = 0.

∴ p(-1) = 0.

∴ (-1)^{2 }– p(-1) + q = 0

∴ 1 + p + q = 0

∴ p + q + 1 = 0

Thus, if x = -1 is a root of the equation x^{2 }– px + q = 0, then p + q + 1 = 0.

**Question 3(3):**

**Solution :**

One of the roots of x^{2 }– kx + 6 = 0 is given as 3.

∴ p(3) = 0.

∴ (3)^{2 }– k(3) + 6 = 0

∴ 9 – 3k + 6 = 0

∴ 15 – 3k = 0

∴ 3k = 15

∴ k = 5

**Question 3(4):**

**Solution :**

One of the roots of x^{2 }+ 3(k+2)x – 9 = 0 is given as 3

∴ p(-3) = 0,

∴ (-3)^{2 }+ 3(k + 2) (-3) – 9 = 0

∴ 9 – 9k – 18 – 9 = 0

∴-9k – 18 = 0

∴ 9k = -18

∴ k = -2

**Question 4:**

Solve the following equations using the method of factorization :

- 27x
^{2}-48 = o - (x – 7)
^{2}– 16 = 0 - 6x
^{2}+ 13 x + 6 = 0 - 15x
^{2}– 16x + 1 = O - \(\sqrt{5}\)x
^{2}– 4x – \(\sqrt{5}\) = 0 - \(x+\frac{1}{x}=2\frac{1}{6}\)

**Question 4(1):**

**Solution :**

**Question 4(2):**

**Solution :**

Here (x – 7)^{2 }– 16 = 0

∴ (x – 7)^{2 }– (4)^{2 }= 0

∴ (x – 7 – 4)(x – 7 + 4) = 0

∴ (x – 11)(x – 3) = 0

∴ x – 11 = 0 or x – 3 = 0

∴x = 11 or x = 3

∴ Roots of the given equation are 3 and 11.

**Question 4(3):**

**Solution :**

**Question 4(4):**

**Solution :**

**Question 4(5):**

**Solution :**

**Question 4(6):**

**Solution :**

**Exercise – 4.2**

**Question 1:**

Find the discriminant of the following quadratic equations and discuss the nature of the roots :

- 6x
^{2}– 13x + 6 = 0 - \(\sqrt{6}\)x
^{2}– 5x + \(\sqrt{6}\) = 0 - 24x
^{2}– 17x + 3 = 0 - X
^{2}+ 2x + 4 = 0 - X
^{2}+ x + 1 = 0 - X
^{2}– 3\(\sqrt{3}\)x – 30 = 0

**Question 1(1):**

**Solution :**

From equation 6x^{2 }– 13x + 6 = 0, we get,

a = 6, b = -13, c = 6

∴ D = b^{2 }– 4ac = (-13)^{2 }– 4(6)(6)

= 169 – 144

= 25 > 0

Here, D > 0 and it is a perfect square.

Also, a, b, c ∊ Q, so the two roots of the quadratic equation are rational and distinct.

**Question 1(2):**

**Solution :**

**Question 1(3):**

**Solution :**

From the equation 24x^{2 }– 17x + 3 = 0 we get,

a = 24, b = -17, c = 3

∴ D = b^{2 }– 4ac

= (-17) ^{2 }– 4(24)(3)

= 289 – 288

= 1 > 0

Here, D > 0 and it is a perfect square.

Also a, b, c ∊ Q, so the two roots of the quadratic equation are rational and distinct.

**Question 1(4):**

**Solution :**

From the equation x^{2 }+ 2x + 4 = 0, we get,

a = 1, b = 2, c = 4

∴ D = b ^{2 }– 4ac

= (2)^{2 }– 4(1)(4)

= 4 – 16

= – 12

∴ D < 0

Here D < 0, so the quadratic equation has no real roots.

**Question 1(5):**

**Solution :**

From the equation x^{2 }+ x + 1= 0, we get,

a = 1, b = 1, c = 1

∴ D = b^{2 }– 4ac = (1)^{2 }– 4(1)(1)

= 1 – 4

= -3 < 0

∴ D < 0

Here D < 0, so the quadratic equation has no real roots.

**Question 1(6):**

**Solution :**

**Question 2:**

If a, b, c ∈ R, a > 0, c < 0, then prove that the roots of ax^{2} + bx + c = 0 are real and distinct.

**Solution :**

Here, a, b, and c ∊ R and a > 0, c < 0

So, ac < 0

∴ 4ac < 0

∴ -4ac>0

∴ b^{2 }– 4ac > 0 (∵ b^{2 }≥ 0)

∴ D > 0

As D > 0 and a, b and c ∊ R, the roots of the quadratic equation are real and distinct.

**Question 3:**

Find k, if the roots of x^{2} – (3k -2)x + 2k = 0 are equal and real.

If the roots of the quadratic equation (k + 1)x^{2} – 2(k -1)x + 1 = O are real and equal, find the value of k.

**Question 3(1):**

**Solution :**

**Question 3(2):**

**Solution :**

From the equation (k + 1)x^{2 }– 2(k – 1)x + 1 = 0, we get, a = k + 1, b = -2(k – 1), c = 1

Now, the roots of the given quadratic equation are real and equal.

∴ D = 0

∴ b^{2 }– 4ac = 0

∴ 4(k – 1)^{2 }– 4(k + 1)(1) = 0

∴ 4k^{2 }– 12k = 0

∴ k^{2 }– 3k = 0 [Taking 4 common]

∴ k(k – 3) = 0

∴ k = 0 or k = 3

∴ 0 or 3 are the values of k.

**Question 4:**

If the roots of ax^{2} + 2bx + c = 0, a ≠ 0, a, b, c ∈ R are real and equal, then prove that a : c = b : c.

**Solution :**

**Question 5:**

Solve the following equations using the general formula :

- x
^{2}+ 10x + 6 = 0 - x
^{2}+ 5x – 1 = 0 - x
^{2}– 3x – 2 =0 - x
^{2}– 3\(\sqrt{6}\)x + 12 = 0 - 3x
^{2}+ 5\(\sqrt{5}\)x + 2 = 0 - \(\frac{x2-1}{x2+1}=\frac{4}{5}\)

**Question 5(1):**

**Solution :**

**Question 5(2):**

**Solution :**

**Question 5(3):**

**Solution :**

**Question 5(4):**

**Solution :**

**Question 5(5):**

**Solution :**

**Question 5(6):**

**Solution :**

**Exercise – 4.3**

**Question 1:**

Find two numbers whose sum is 27 and the product is 182.

**Solution :**

**Question 2:**

Find two consecutive natural numbers, sum of whose squares is 365.

**Solution :**

Let the two consecutive natural numbers be x and x + 1.

Sum of their squares is given as 365.

∴ (x)^{2 }+ (x + 1)^{2 }= 365

∴ x^{2 }+ x^{2 }+ 2x + 1 = 365

∴ 2x^{2 }+ 2x – 362 = 0

∴ x^{2 }+ x – 182 = 0 [∵ Taking 2 common]

∴ x^{2 }+ 14x – 13x – 182 = 0

∴ x(x + 14) – 13(x + 14) = 0

∴ (x + 14)(x – 13) = 0

∴ x = -14 or x = 13

x cannot be negative as it is a natural number.

∴ x = 13

Other natural number = x + 1 = 13 + 1 = 14

Thus, the two consecutive natural numbers are 13 and 14.

**Question 3:**

The sum of ages of two friends is 20 years. Four years ago the product of their ages was 48. Show that these statements can not be true.

**Solution :**

Let the present age of the first friend = x years.

Now,

Sum of ages of both friends is given 20

∴ Age of second friend = (20 – x) years.

Before four years,

Age of first friend = (x – 4) years

Age of second friend = (20 – x) – 4 = (16 – x) years.

Now, four years ago the product of the numbers showing their ages was 48.

∴ (x – 4)(16 – x) = 48

∴ 16x – x^{2 }– 64 + 4x = 48

∴ x^{2 }– 20x + 112 = 0

Comparing the equation x^{2 }– 20x + 112 = 0 with ax^{2 }+ bx + c = 0, we get a = 1, b = -20 , c = 112

∴ D = b^{2 }– 4ac=(-20)^{2 }– 4(1)(112)

= 400 – 488 = -48 < 0

As D < 0, so the roots are not real.

But age can be represented only by a positive number.

∴ The given statement cannot be true.

**Question 4:**

A rectangular garden is designed such that the length of the garden is twice its breadth and the area of the garden is 800 m^{2}. Find the length of the garden.

**Solution :**

Let the breadth of the rectangular garden = x metres.

Given, the length of the garden is twice the breadth.

i.e. length = 2(breadth) = 2x metres

Area of a rectangular garden is given as 800 m^{2}

Now,

Area = length × breadth

∴ 800 = (2x)× x

∴ x^{2 }= 400

∴ x = 20 (∵ x = -20 is not possible.)

Thus, length of the garden = 2x = 2 × 20m = 40 metres.

**Question 5:**

Perimeter of a rectangular garden is 360 m and its area is 8000 m^{2}. Find the length of the garden and also find its breadth. (The length is greater than the breadth)

**Solution :**

**Question 6:**

If a cyclist travels at a speed 2 km/hr more than his usual speed, he reaches the destination 2 hours earlier. If the destination is 35 km away, what is the usual speed of the cyclist ?

**Solution :**

**Question 7:**

The diagonal of a rectangular ground is 60 meters more than the breadth of the ground. If the length of the ground is 30 meters more than the breadth, find the area of the ground.

**Solution :**

Let the breadth of the rectangular ground be x metres.

Length of the ground is 30 m more than the breadth, so length = (x + 30) m.

Now, the length of the diagonal of the rectangular ground is 60 m more than its breadth,

Length of the diagonal = (x + 60) m.

For any rectangle,

(breadth)^{2 }+ (length)^{2 }= (diagonal)^{2 }(∵ Pythagoras’ thm)

∴ x^{2 }+ (x + 30)^{2 }= (x + 60)^{2}

∴ x^{2 }+ x^{2 }+ 60x + 900 = x^{2 }+ 120x + 3600

∴ x^{2 }– 60x – 2700 = 0

∴ x^{2 }– 90x + 30x – 2700 = 0

∴ x(x – 90) + 30(x – 90) = 0

∴ (x – 90)(x + 30) = 0

∴ x – 90 = 0 or x + 30 = 0

∴ x = 90 or x = -30

∴x = 90 (∵ breadth cannot be negative.)

∴ Breadth of the rectangle = x = 90 m

∴ Length of the rectangle

= (x + 30) m

= (90 + 30) m

= 120 m

Now, area of the rectangular ground

= length × breadth

= (120)×(90)

= 10800 m^{2}

∴ Area of the ground is 10,800m^{2}.

**Question 8:**

The sides of a right angled triangle are consecutive positive integers. Find the area of the triangle.

**Solution :**

Let the consecutive positive integers of the sides of a right angled triangle be x, x + 1 and x + 2.

**Exercise – 4**

**Question 1:**

Solve the following quadratic equations using factorization :

- x
^{2}– 12 = 0 - x
^{2}– 7x – 60 = 0 - x
^{2}– 15x + 56 = 0 - \(\frac{2x+3}{2x-3}+\frac{2x-3}{2x+3}=\frac{17}{4}\), x ≠ ± \(\frac{3}{2}\)
- \(\frac{1}{x+5}+\frac{3}{4(3x+1)}=\frac{1}{x+2}\), x ≠ -5, x ≠ -2, x ≠ \(-\frac{1}{3}\)

**Question 1(1):**

**Solution :**

**Question 1(2):**

**Solution :**

For x^{2 }– 7x – 60 = 0

∴ x^{2 }– 5x – 12x – 60 = 0

∴ x(x + 5) – 12(x + 5) = 0

∴ (x + 5)(x – 12) = 0

∴ x + 5 = 0 or x – 12 = 0

∴ x = 5 or x = 12

∴ Roots of the given equation are 5 and 12.

**Question 1(3):**

**Solution :**

x^{2 }– 15x + 56 = 0

∴ x^{2 }– 7x – 8x + 56 = 0

∴ x(x – 7) – 8 ( x – 7) = 0

∴ (x – 7)(x – 8)=0

∴x – 7 = 0 or x – 8 = 0

∴ x = 7 or x = 8

∴ Roots of the given equation are 7 and 8.

**Question 1(4):**

**Solution :**

**Question 1(5):**

**Solution :**

**Question 2:**

Find the roots of the following equations by the method of perfect square :

- x
^{2}– 24x – 16 = 0 - 3x
^{2}+ 7x – 20 = 0 - X
^{2}– 10x + 25 = 0 - X
^{2}+ (x + 5)^{2}= 625 - (x + 2)(x + 3) = 240

**Question 2(1):**

**Solution :**

**Question 2(2):**

**Solution :**

**Question 2(3):**

**Solution :**

For x^{2 }– 10x + 25 = 0

(x)^{2} – 2(x)(5) + (5)^{2} = 0

∴ (x – 5)^{2 }= 0

∴ (x – 5) (x – 5) = 0

∴ x – 5 = 0 or x – 5 = 0

∴ x = 5 or x = 5

∴ Root of the given equation is 5.

**Question 2(4):**

**Solution :**

**Question 2(5):**

**Solution :**

**Question 3:**

Divide 20 into two parts such that the sum of the square of the parts is 218.

**Solution :**

Let the two parts be x and 20 – x.

It is given that the sum of their squares is 218.

∴ x^{2 }+ (20 – x)^{2 }= 21

∴ x^{2 }+ 400 – 40x + x^{2 }= 218

∴ 2x^{2 }– 40x + 182 = 0

∴ x^{2 }– 20x + 91= 0

∴ x^{2 }– 7x – 13x + 91 = 0

∴ x(x – 7) – 13(x – 7) = 0

∴ (x – 7)(x – 13) = 0

∴ x – 7 = 0 or x – 13 = 0

∴ x = 7 or x = 13

If one part is 7, then the second part is (20- x) i.e., 20-7=13

If one part is 13, then the second part is (20 – x) i.e., 20 – 13 = 7.

Hence two parts of 20 are 7 and 13 respectively.

**Question 4:**

A car takes 1 hour less to cover a distance of 200 km if its speed is increased by 10 km/hr, than its usual speed. What is the usual speed of the car ?

**Solution :**

Let the usual speed of a car is x km/hr.

∴Time taken to cover 200km = t_{1}

**Question 5:**

When there is a decrease of 5 km/hr in the usual uniform speed of a goods train, due to track repair work going on it takes 4 hours more than the usual time for travelling the distance of 400 km. Find the usual speed of the train.

**Solution :**

**Question 6:**

A river flows at a speed of I km/hr. A boat takes 15 hours to travel 112 km downstream and coming back the same distance upstream. Find the speed of the boat in still water. (Speed of the river flow is less than the speed of the boat in still water)

**Solution :**

**Question 7:**

Find a number greater than 1 such that the sum of the number and its reciprocal is 2\(\frac{4}{15}\)

**Solution :**

**Question 8:**

The difference of the speed of a faster car and a slower car is 20 km/hr. If the slower car takes 1 hour more than the faster car to travel a distance of 400 km, find speed of both the cars.

**Solution :**

**Question 9:**

Product of the ages of Virat 7 years ago and 7 years later is 480. Find his present age.

**Solution :**

Let the present age of Virat be x years.

Before 7 years,

Virat’s age = (x – 7) years

After 7 years,

Virat’s age = (x + 7) years

Product of the numbers showing these ages is 480.

∴ (x – 7)(x + 7) = 480

∴ x^{2 }– 49 = 480

∴ x^{2 }– 529 = 0

∴ (x)^{2 }– (23)^{2 }= 0

∴ x – 23 = 0 or x + 23 = 0

∴ x = 23 or x = -23

∴ x = 23 (∵ Age cannot be negative.)

Hence the present age of Virat is 23 years.

**Question 10:**

If the age of Sachin 8 year ago is multiplied by his age two years later, the result is 1200. Find the age of Sachin at present.

**Solution :**

Let the present age of Sachin be x years.

Before 8 years,

Sachin’s age = (x – 8) years

After 2 years,

Sachin’s age = (x + 2) years.

Product of the numbers showing these ages is 1200.

∴ (x – 8)(x + 2) = 1200

∴ x^{2 }– 6x – 16 = 1200

∴ x^{2 }– 6x – 1216 = 0

∴ x^{2 }– 38x + 32x – 1216 = 0

∴ x(x – 38)+32(x – 38) = 0

∴ (x – 38)(x + 32) = 0

∴ x – 38 = 0 or x + 32 = 0

∴ x = 38 or x = -32

∴ x = 38 (∵ Age cannot be negative.)

Hence Sachin’s present age is 38 years.

**Question 11:**

Sunita’s age at present is 2 years less than 6 times the age of her daughter Anita. The product of their ages 5 years later will be 330. What was the age of Sunita when her daughter Anita was born ?

**Solution :**

Let the present age of Anita be x years

It is given that Sunita’s age at present is 2 years less than 6 times the age of her daughter Anita.

∴ Present age of Sunita = (6x – 2) years

After five years,

Age of Anita = (x + 5) years

Age of Sunita = [(6x – 2) + 5] years = (6x + 3) years

Also, the product of the numbers showing these ages is 330.

∴ (x + 5)(6x + 3) = 330

∴ 6x^{2 }+ 33x + 15 = 330

∴ 6x^{2 }+ 33x – 315 = 0

∴ 2x^{2 }+ 11x – 105 = 0 (∵ Dividing by 3)

∴ 2x^{2 }+ 21x – 10x – 105 = 0

∴ x(2x + 21) – 5(2x + 21) = 0

∴ (2x + 21)(x – 5) = 0

∴ 2x + 21 = 0 or x – 5 = 0

or x = 5

∴ x = 5 (∵Age cannot be negative.)

Thus, the present age of Anita (x) is 5 years.

∴Present age of Sunita

= 6x – 2

= 6(5) – 2

= 28 years

Now, Anita was born before five years, so the age of Sunita at that time was = 28 – 5 = 23 years.

**Question 12:**

The formula of the sum of first n natural numbers is \(S=\frac{n(n+1)}{2}\) . If the sum of first n natural number is 325, find n.

**Solution :**

**Question 13:**

Hypotenuse of a right angled triangle is 2 less than 3 times its shortest side. If the remaining side is 2 more than twice the shortest side, find the area of the triangle.

**Solution :**

Let the length of the shortest side of the right angled triangle be x.

Here hypotenuse is given is 2 less than 3 times its shortest side.

∴ Length of the hypotenuse is (3x – 2).

Then, length of the remaining side is 2 more than twice the shortest side = (2x + 2)

By Pythagoras theorem,

(Length of hypotenuse)^{2}

= (Length of a side)^{2 }+ (Length of other side)^{2}

∴ (3x – 2)^{2 }= (x)^{2 }+ (2x + 2)^{2}

∴9x^{2 }-12x +4=x^{2 }+4x^{2 }+8x + 4

∴ 4x^{2 }– 20x = 0

∴ x^{2 }– 5x = 0 (∵ Dividing by 4)

∴ x(x – 5) = 0

∴ x = 0 or x = 5

∴ x = 5 (∵ Length of a side cannot be zero.)

The length of the shortest side of a right angled triangle is 5 units.

Length of the other side

= 2x + 2

= 2(5)+2

= 10 + 2

=12

**Question 14:**

The sum of the squares of two consecutive odd positive integers is 290. Find the numbers.

**Solution :**

Let the first odd positive integer be x.

∴ Second consecutive odd positive integer = x + 2.

According to the given conditions, sum of the square of these two numbers is 290.

∴ x^{2 }+ (x + 2)^{2 }= 290

∴ x^{2 }+ x^{2 }+ 4x + 4 = 290

∴ 2x^{2 }+ 4x – 286 = 0

∴ x^{2 }+ 2x – 143 = 0 (∵ Dividing by 2)

∴ x^{2 }+ 13x – 11x – 143 = 0

∴ x(x + 13) – 11(x + 13) = 0

∴ (x + 13)(x – 11) = 0

∴ x + 13 = 0 or x – 11 = 0

∴ x = – 13 or x = 11

But x = -13 is not possible.

∴ x = 11 (∵ x is a positive integer.)

∴ First odd positive integer = x = 11

Second consecutive odd positive integer = x + 2

= 11 + 2 = 13

∴ Thus the two consecutive odd positive numbers are 11 and 13.

**Question 15:**

The product of two consecutive even natural numbers is 224. Find the numbers.

**Solution :**

Let the two consecutive even natural numbers be x and x + 2.

Product of these two numbers is 224.

∴ x(x + 2) = 224

∴ x^{2 }+ 2x – 224 = 0

∴ x^{2 }+ 16x – 14x – 224 = 0

∴ x(x + 16) – 14(x + 16) = 0

∴ (x + 16)(x – 14) = 0

∴ x + 16 = 0 or x – 14 = 0

∴ x = -16 or x – 14 = 0

∴ x = -16 or x = 14

∴ x = 14 (∵ x is a natural number.)

∴ First even natural number x = 14

second even natural number = x + 2 = 14 + 2 = 16

∴ The required consecutive even natural numbers are 14 and 16.

**Question 16:**

The product of digits of a two-digit number is 8 and the sum of the squares of the digits is 20. If the number is less than 25. Find the number.

**Solution :**

**Question 17:**

If price of sugar decreases by ₹ 5, one can buy 1 kg more sugar in ₹ 150, what is the price of the sugar ?

**Solution :**

**Question 18:**

If the price of petrol is increased by ₹ 5 per litre. One gets 2 litres less petrol spending ₹ 1320. What is the increased price of the petrol ?

**Solution :**

**Question 19:**

A vendor gets a profit in percentage equal to the cost price of a flower pot when he sells it for ₹ 96. Find the cost of the flower pot and the percentage of profit.

**Solution :**

**Question 20:**

While selling a pen for ₹ 24 the loss in percentage is equal to its cost price. Find the cost price of pen. The cost price of pen is less than ₹ 50.

**Solution :**

**Question 21:**

The difference of lengths of sides forming right angle in right angled triangle is 3 cm. If the perimeter of the triangle is 36 cm. Find the area of the triangle.

**Solution :**

Let the length of the smallest side of a right angled triangle be x cm.

∴ Measure of the larger side making a right angle with it is (x+3) cm.

Perimeter of a triangle is 36cm.

36 = x + (x + 3)+ hypotenuse

∴ Measure of the hypotenuse = 36 – x – (x + 3)

= (33 – 2x) cm

According to Pythagoras’ theorem

(Length of hypotenuse)^{2}

= (Length of a side)^{2 }+ (Length of other side)^{2}

∴ (33 – 2x)^{2 }= x^{2 }+ (x + 3)^{2}

∴ 1089 – 132x + 4x^{2 }= x^{2 }+ x^{2 }+ 6x + 9

∴ 2x^{2 }– 1138x + 1080 = 0

∴ x^{2 }– 69x + 540 = 0 [∵ Taking 2 common]

∴ (x – 60)(x – 9) = 0

∴ x – 60 = 0 or x – 9 = 0

∴ x = 60 or x = 9

But, it is given that perimeter of the triangle is 36 cm, so side of the triangle cannot be 60 cm.

∴ x = 9

Measure of the smallest side of this triangle = 9 cm.

Measure of other side = x + 3 = 9 + 3 = 12 cm

Now, area of a right angled triangle

**Question 22:**

The sides of a right angled triangle are x, x + 3, x + 6, x being a positive integer. Find the perimeter of the triangle.

**Solution :**

The largest side in a right angled triangle is the hypotenuse, so measure of hypotenuse is (x + 6) cm.

**Question 23:**

Select a proper option (a), (b), (c) or (d) from given options :

**Question 23(1):**

……..is a solution of quadratic equation x^{2} – 3x + 2 = 0

**Solution :**

b. 1

x^{2 }– 3x + 2 = 0

∴ x^{2 }– x – 2x + 2 = 0

∴ x(x – 1) – 2(x – 1) = 0

∴ (x – 1)(x – 2) = 0

∴ x = 1 or x = 2

∴ x = 1 is the solution of the given equation.

**Question 23(2):**

Discriminant D = ……… for the quadratic equation 5x^{2} – 6x + 1 = 0

**Solution :**

a. 16

Comparing equation 5x^{2 }– 6x + 1 = 0 with

ax^{2}+ bx + c = 0, we get a = – 5 , b = – 6, c = 1

Then Discriminant,

D = b^{2 }= 4ac

= (- 6) ^{2 }– 4(5)(1)

= 36 – 20

= 16

**Question 23(3):**

If x = 2 is a root of the equation x^{2} – 4x + a = 0, then a = ……

**Solution :**

d. 4

2 is a root of a equation

x^{2 }– 4x + a = 0

∴ (2)^{2 }– 4(2) + a = 0

∴ 4 – 8 + a = 0

∴ a – 4 = 0

∴ a = 4

**Question 23(4):**

A quadratic equation has two equal roots, if …….

**Solution :**

c. D = 0

When D= 0, the quadratic equation has equal roots.

**Question 23(5):**

The quadratic equation ……. has 3 as one of its roots.

**Solution :**

a. x^{2 }– x – 6 = 0

Consider the first equation,

p(x) = x^{2 }– x – 6,

p(3) = (3)^{2 }– 3 – 6

= 9 – 3 – 6

= 0

∴ One solution of the equation x^{2 }– x – 6 = 0 is 3.

[Solution can be found by trial and error method, i.e. substituting x = 3 in all polynomials if it isn’t true for the first equation.]

**Question 23(6):**

If 4 is a root of quadratic equation x^{2} + ax – 8 = 0, then a = …….

**Solution :**

c. – 2

4 is a root of a quadratic equation x^{2 }+ ax – 8 = 0.

∴ (4)^{2 }+ a(4) – 8 = 0

∴ 16 + 4 a – 8 = 0

∴ 4a + 8 = 0

∴ a = -2

**Question 23(7):**

If one of the roots of kx^{2} – 7x + 3 = 0 is 3, then k = …….

**Solution :**

d. 2

3 is a root of a quadratic equation kx^{2 }– 7x + 3 = 0

∴ k(3)^{2 }– 7(3) + 3 = 0

∴ 9k – 21 + 3 = 0

∴ 9k – 18 = 0

∴ 9k = 18

∴ k = 2

**Question 23(8):**

The discriminant of x^{2} – 3x – k = 0 is 1. A value of x is ……..

**Solution :**

b. – 2

Comparing the equation x^{2 }– 3x – k = 0 with

ax^{2 }+ bx + c = 0, we get a = 1, b = -3, c = -k

Here, Discriminant = 1

∴ b^{2 }– 4ac = 1

∴ (-3)^{2 }– 4(1)( -k) = 1

∴ 9 + 4k = 11

∴ 4k = -8

∴ k = -2