## Extra Questions for Class 10 Maths Quadratic Equations with Answers

Extra Questions for Class 10 Maths Chapter 4 Quadratic Equations. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

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### Quadratic Equations Class 10 Extra Questions Very Short Answer Type

Question 1.

For what value of V the quadratic equation 9x^{2} – 3ax + 1 = 0 has equal roots?

Or

If one root of the quadratic equation 2x^{2} + 2x + k = 0 is – \(\frac{1}{3}\)\frac{1}{3}, then find the value of k. [CBSE 2019]

Answer:

For equal roots D = 0

where D = b^{2} – 4ac

Here a = 9, b = – 3a, c = 1

∴ (- 3a)^{2} – 4(9) (1) = 0

⇒ 9a^{2} – 36 = 0

⇒ 9(a^{2} – 4) = 0

⇒ a^{2} = 4

⇒ a = ± 2

Or

Let, p(x) = 2x^{2} + 2x + k

Question 2.

For what values of k, the roots of the equation x^{2} + 4x + k = 0 are real?

Or

Find the value of k for which the roots of the equation 3x^{2} – 10x + k = k = 0 are reciprocal of each other. [CBSE 2019]

Answer:

For real roots D ≥ 0 i.e., b^{2} – 4ac ≥ 0

Here, a = 1, b = 4, c = k

∴ (4)^{2} – 4(1)k ≥ 0

⇒ 16 – 4k ≥ 0 ⇒ k ≤ A

Or

Roots of the equation 3x^{2} – 10x + k = 0 are reciprocal of each other

⇒ Product of roots = 1

⇒ \(\frac{c}{a}=\frac{k}{3}\) = 1 ⇒ k = 3

Question 3.

If the roots of quadratic equation px^{2} + 6x + 1 = 0 are real, then find p.

Answer:

Roots are real, D ≥ 0

b^{2} – 4ac ≥ 0

(6^{2}) – 4 × p × 1 ≥ o

9 ≥ p

Question 4.

If 7x^{2} – (2p^{2} – 8)x + 16 = 0 has two roots which are equal in magnitude but opposite in sign then find p.

Answer:

Let the roots be α and – α.

0 = 2p^{2} – 8

p^{2} = 4

p = ±2

### Quadratic Equations Class 10 Extra Questions Short Answer Type-1

Question 1.

Find the roots of the quadratic equation

√2x^{2} + 7x + 5√2 = 0 [CBSE Delhi 2017]

Answer:

√2x^{2} + 7x + 5√2 = 0

√2x^{2} + 2x + 5x + 5√2 = 0

⇒ (√2x + 5) (x + √2) = 0

⇒ x = \(\frac{-5}{\sqrt{2}}\), – √2

or \(\frac{-5 \sqrt{2}}{2}\), – √2

Question 2.

Find the value of k for which the equation x^{2} + k(2x + k – 1) + 2 = 0 has real and equal roots.

Answer:

Rewriting given equation as

x^{2} + 2kx + (k^{2} – k + 2) = 0

Here a = 1, b = 2k, c = k^{2} – k + 2

For equal roots, b^{2} – 4ac = 0

⇒ (2k)^{2} – 4(1) (k^{2} – k + 2) = 0

⇒ k = 2

Question 3.

If – 5 is a root of the quadratic equation 2x^{2} + px – 15 = 0 and the quadratic equation p(x^{2} + x) + k = 0 has equal roots, find the value of k. [CBSE Outside Delhi 2016]

Answer:

– 5 is root of the equation 2x^{2} + px – 15 = 0, then

2(-5)^{2} + p(-5) – 15 = 0

⇒ 50 + (- 5)p – 15 = 0

⇒ p = \(\frac{-35}{-5}\) = 7 ……….. (1)

p(x^{2} + x) + k = 0 has equal root

⇒ b^{2} – 4ac = 0

Here b = p, a = p, c = k

⇒ p^{2} = 4 pk ………….. (2)

(7)^{2} – 4 × 7k = 0

(1) and (2) gives,

∴ k = \(\frac{7}{4}\)

Question 4.

For what value of k, x^{2} + 4x + k is a perfect square?

Answer:

A quadratic expression is a perfect square, if and only if corresponding equation has equal roots.

i.e., D = 0

⇒ 16 – 4k = 0

⇒ k = 4

Question 5.

Solve: \(\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}\),

[x ≠ 0, x ≠ – (a + b)].

Answer:

⇒ – ab = x^{2} + x (a + b)

⇒ x^{2} + x (a + b) + ab = 0

⇒ x^{2} + ax + bx + ab = 0

⇒ x (x + a) + b (x + a) = 0

⇒ (x + a) (x + b) = 0

∴ x = – a or x = – b

Hence, – a and – b are roots of given equation.

Question 6.

The product of two consecutive positive integers is 306. Find the integers.

Answer:

Let the consecutive positive integers be x and x + 1.

Then, x (x + 1) = 306

or, x^{2} + x – 306 = 0

or, x^{2} + 18x – 17x – 306 = 0

or, x(x + 18) – 17(x + 18) = 0

or, (x + 18) (x – 17) = 0

x = – 18, 17

Neglecting x = – 18

∴ x = 17 and x + 1 = 17 + 1 = 18

Hence, two consecutive positive integers are 17 and 18.

### Quadratic Equations Class 10 Extra Questions Short Answer Type-2

Question 1.

If the equation (1 + m^{2})x^{2} + 2 mcx + c^{2} – a^{2} = 0 has equal roots then show than c^{2} = a^{2} (1 + m^{2}). [CBSE 2017]

Answer:

(1 + m^{2}) x^{2} + 2mcx + c^{2} – a^{2} = 0

Here A = (1 + m^{2}), B = 2mc, C = c^{2} – a^{2}

For equal roots; B^{2} – 4AC = 0

⇒ (2mc)^{2} – 4(1 + m^{2}) (c^{2} – a^{2}) = 0

⇒ c^{2} – a^{2}(1 + m^{2}) = 0

⇒ c^{2} = a^{2}(1 + m^{2})

Question 2.

Solve for x:

\(\frac{2 x}{x-3}+\frac{1}{2 x+3}+\frac{3 x+9}{(x-3)(2 x+3)}\) = 0, x ≠ 3, – \(\frac{3}{2}\). [CBSE Delhi 2016]

Answer:

Given:

⇒ 4×2 + 6x + x- 3 + 3x + 9 = 0

⇒ 4×2 + lOx + 6 = 0

⇒ 4×2 + 4x+ 6x + 6 = 0

⇒ 4x(x + 1) + 6(x + 1) = 0

⇒ (x +1) (4x + 6) = 0

⇒ x + 1 = 0

or 4x + 6 = 0

⇒ x = – 1, – \(\frac{3}{2}\)

But

x ≠ – \(\frac{3}{2}\)

Question 3.

A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.

Answer:

Let the usual speed of plane be x km/hr

∴ [Time taken to travel 1500 km with speed × km/hr] – [Time taken to travel 1500 km with speed (x + 100) km/hr] = half an hour

⇒ x^{2} + 100x – 300000 = 0

x^{2} + 600x – 500x – 300000 = 0

(x + 600) (x – 500) = 0

x = 500 as x can’t be negative so – 600 is rejected.

∴ usual speed of the plane is 500 km/hr.

Question 4.

Find the value of p for which the quadratic equation (2p + 1)x^{2} – (7p + 2)x + (7p – 3) = 0 has equal roots. Also find these roots. [S.A-II, 2014]

Answer:

(2p + 1)x^{2} – (7p + 2)x + (7p – 3) = 0 …(1)

For equal roots D = 0

b^{2} – 4ac = 0

⇒ [- (7p + 2)]^{2} – 4(2p + 1) (7p – 3) = 0

⇒ 49p^{2} + 4 + 28p – 4 (14p^{2} – 6p + 7p – 3) = 0

⇒ 49p^{2} + 28p + 4 – 56p2 + 24p – 28p + 12 = 0

⇒ – 7p^{2} + 24p + 16 = 0

⇒ 7p^{2} – 24p – 16 = 0

⇒ 7p^{2} – 28p + 4p – 16 = 0

⇒ 7p (p – 4) + 4 (p – 4) = 0

⇒ (p – 4) (7p + 4) = 0

∴ p = 4, \(\frac{-4}{7}\)

If p = 4, then (1)

⇒ 9x^{2} – 30x + 25 = 0

⇒ 9x^{2} – 15x – 15x + 25 = 0

⇒ 3x(3x – 5) – 5(3x – 5) = 0

⇒ (3x – 5) (3x – 5) = 0

⇒ (3x – 5)^{2} = 0

⇒ 3x – 5 = 0

∴ x = \(\frac{5}{3}\)

∴ Roots are \(\frac{5}{3}\) and \(\frac{5}{3}\).

If P = – \(\frac{4}{7}\), then (1)

⇒ \(\frac{-x^2}{7}\) + 2x – 7 = 0

– x^{2} + 14x – 49 = 0

x^{2} – 14x + 49 = 0

(x – 7)^{2} = 0

∴ x = 7, 7

Question 5.

A journey of 192 km from a town A to town B takes 2 hours more by a ordinary passenger train than a super fast train. If the speed of the faster train is 16 km/h more, find the speeds of the faster and the passenger train.

Answer:

Let the speed of passenger train be x km/h.

Then, speed of faster train = (x + 16) km/h

According to question:

Time taken to complete the journey by faster train (t_{1}) = \(\frac{192}{x+16}\) hours and time taken by passenger train (t_{2}) = \(\frac{192}{x}\)

According to question,

∵ Speed cannot be negative.

∴ Speed of the passenger train = 32 km/h

and speed of faster train = 32 + 16 = 48 km/h

### Quadratic Equations Class 10 Extra Questions Long Answer Type 1

Question 1.

Speed of a boat in still water is 15 kn\h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream. [CBSE Delhi 2017]

Answer:

Let speed of boat be xkm/hr

∴ Speed of boat upstream = (15 – x) km/hr.

200 = 225 – x^{2}

⇒ x^{2} = 25

x = 5 (Rejecting – 5)

∴ Speed of stream = 5 km/hr

Question 2.

Find x in terms of a, b and c:

\(\frac{a}{x-a}+\frac{b}{x-b}=\frac{2 c}{x-c}\), x ≠ a, b, c [CBSE Delhi 2016]

Answer:

Consider the given equation:

⇒ (x – c)[a(x – b) + b(x – a)] = 2c(x – a)(x – b)

⇒ (x – c)[ax – ab + bx – ab] = 2c(x^{2} – bx – ax + ab)

⇒ ax^{2} – 2abx + bx^{2} – acx + 2abc – bcx = 2cx^{2} – 2bcx – 2acx + 2abc

⇒ ax^{2} + bx^{2} – 2cx^{2} = 2abx – acx – bcx

⇒ (a + b – 2c)x^{2} = x(2ab – ac – bc)

⇒ (a + b – 2c)x^{2} – x(2ab – ac – bc) = 0

⇒ x[(a + b – 2c)x – (2 ab – ac – bc) = 0

⇒ either x = 0

or (a + b – 2c)x – (2ab – ac – bc) = 0

or (a + b – 2c)x = (2ab – ac – bc)

or x = \(\frac{(2 a b-a c-b c)}{(a+b-2 c)}\)

Thus, the two roots of the given equation are x = 0 and \(\)

Question 3.

A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete total journey, what is the original average speed ? [CBSE 2018]

Answer:

Let original speed of train be x km/ hr

Distance = 63 km; time (t_{1}) = 63/x hrs

∴ Faster speed = (x + 6) km/hr

[∵ Time = \(\frac{\text { Distance }}{\text { Speed }}\)]

Distance = 72 km; time (t_{2}) = 72/ (x + 6) hrs.

(t_{1}) + (t_{2}) = 3 hrs

\(\frac{63}{x}+\frac{72}{(x+6)}\) = 3

⇒ 63(x + 6) + 72x = 3x (x + 6)

⇒ 63x + 378 + 72x = 3x^{2} + 18x

⇒ 3x^{2} – 117x – 378 = 0

⇒ x^{2} – 39x – 126 = 0

⇒ 7x^{2} – 42x + 3x + 126 = 0

⇒ (x – 42) (x + 3) = 0

⇒ x = -3

[Rejected as speed can’t be negative]

∴ x = 42

∴ Original speed of train is 42 km/hr.

Question 4.

A motor boat whose speed is 18 km/hr in still water takes 1 hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. [CBSE 2018]

Answer:

Let the speed of stream be x km/ hr

∴ The speed of the boat upstream = (18 – x) km/hr

and speed of the boat downstream = (18 + x) km/hr

According to the question:

⇒ 48x = 324 – x^{2}

⇒ x^{2} + 48x – 324 = 0

⇒ x^{2} + 54x – 6x – 324 = 0

⇒ x(x + 54) – 6(x + 54) = 0

⇒ (x + 54) (x – 6) = 0

⇒ x + 54 = 0

x = – 54

[Rejecting – 54, as speed can’t be negative]

and x – 6 = 0

∴ x = 6 km/h

∴ Speed of the stream = 6 km/hr.

Question 5.

A thief runs with a uniform speed of 100 m/ minute. After one minute, a policeman runs after the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases his speed by 10 m/minute every succeeding minute. After how many minutes the policeman will catch the thief? [CBSE 2016]

Answer:

Suppose the policeman catches the thief after t minutes.

Uniform speed of the thief = 100 metres/minute

∴ Distance covered by thief in (t + 1) = 100 (t + 1) metres [∵ Distance = Speed × Time]

Distance covered by policeman in t minutes = Sum of t terms of an AP with first term 100 and common difference 10.

\(\frac{t}{2}\) [2 × 100 + (t -1) × 10] m

= t[100 + 5(t – 1)]

= t(5t + 95) = 5t^{2} + 95t

When the policeman catches the thief.

5t^{2} + 95t = 100 (t + 1)

⇒ 5t^{2} + 95t = 100t + 100

⇒ 5t^{2} – 5t – 100 = 0

⇒ t^{2} – t – 20 = 0

⇒ t = – 4 and t = 5

Neglecting -ve time. We get t = 5

Thus, the policeman catches the thief after 5 minutes.

Question 6.

The difference of two natural numbers is 5. If the difference of their reciprocals is \(\frac{1}{10}\), find the two numbers.

Answer:

Let the number be x and y (x > y).

According to question; x – y = 5 …………… (i)

Putting the value of y in (z)

x – \(\frac{50}{x}\) =5

⇒ x^{2} – 50 = 5x

⇒ x^{2} – 5x – 50 = 0

⇒ x2^{2} – 10x + 5x – 50 = 0

⇒ x (x – 10) + 5 (x – 10) = 0

⇒ (x – 10) (x + 5) = 0

⇒ x = – 5, 10

When x = – 5, y = \(\frac{50}{-5}\) = – 10

When x = 10, y = \(\frac{50}{10}\) = 5

∴ Two natural numbers as 5,10.

### Quadratic Equations Class 10 Extra Questions HOTS

Question 1.

Solve \(\sqrt{5+\sqrt{5+\sqrt{5+\ldots}}}\)

Answer:

Let x = \(\sqrt{5+\sqrt{5+\sqrt{5+\ldots}}}\)

⇒ x = \(\sqrt{5+x}\)

Squaring both sides, we get

x^{2} = 5 + x

⇒ x^{2} – x – 5 = 0

Question 2.

Solve by factorisation: (x – 3) (x – 4) = \(\frac{34}{33^2}\)

Answer:

Question 3.

If the equation (1 + m^{2}) x^{2} + 2mcx + c^{2} – a^{2} = 0 has equal roots, show that c^{2} = a^{2} (1 + m^{2}).

Answer:

Given equation is,

(1 + m^{2}) x^{2} + 2mcx + c^{2} – a^{2} = 0

∵ Roots are equal.

∴ D = 0

B^{2} – 4AC = 0

(2mc)^{2} – 4 (1 + m^{2}) (c^{2} – a^{2}) = 0

or 4m^{2}c^{2} – 4 [c^{2} – a^{2} + m^{2}c^{2} – m^{2}a^{2}] = 0

or 4[m^{2}c^{2} – c^{2} + a^{2} – m^{2}c^{2} + m^{2}a^{2}] = 0

or – c^{2} + a^{2} (1 + m^{2}) = 0

or c^{2} = a^{2} (1 + m^{2}).

Question 4.

A person on tour has ₹ 360 for his expenses. If he extends his tour for 4 days, he has to cut down his daily expenses by ₹ 3. Find the original duration of the tour.

Answer:

Let, original duration of tour = x days

Expenditure on tour = ₹ 360

So, expenditure per day = ₹ \(\frac{360}{x}\)

According to 1st condition,

Duration of extended tour = (x + 4) days

Expenditure per day = ₹ \(\frac{360}{x+4}\)

According to 2nd condition,

\(\frac{360}{x}-\frac{360}{x+4}\) = 3

360 (x + 4) – 360x = 3x (x + 4)

360x + 1440 – 360x = 3 (x^{2} + 4x)

x^{2} + 4x – 480 = 0, S = 4

x^{2} + 24x – 20x – 480 = 0, P = – 480

x(x + 24) – 20 (x + 24) = 0

(x + 24) (x – 20) = 0

Either x + 24 = 0 or x – 20 = 0

x = – 24 or x = 20

∵ x cannot be negative so, x = 20

Hence, original duration of tour has 20 days.

Question 5.

The hypotenuse of a right triangle is 1 m less than twice the shortest side. If the third side is 1 m more than the shortest side, find the sides of the triangle.

Answer:

Let, shortest side of triangle = x m

Hypotenuse of triangle = (2x – 1) m

Third side of triangle = (x + 1) m By

Pythagoras theorem,

(Hyp.)^{2} = (Base)^{2} + (Alt.)^{2}

(2x – 1)^{2} = (x + 1)^{2} + (x)^{2}

4x^{2} + 1 – 4x = x^{2} + 1 + 2x + x^{2}

2x^{2} – 6x = 0

2x (x – 3) = 0

Either 2x = 0 or x – 3 = 0

x = 0 or x = 3

But side cannot be zero, so . x = 3

∴ Shortest side = x = 3 m

Hypotenuse = (2x – 1) = [2 (3) – 1] = 5 m

Third side = x + 1 = (3 + 1) = 4 m.

Multiple Choice Questions

Choose the correct option for each of the following:

Question 1.

The condition on ‘a’ for ax^{2} + bx + c = 0 to represent a quadratic equation is:

(a) a = 0

(b) a > 0

(c) a < 0

(d) a ≠ 0

Answer:

(d) a ≠ 0

Question 2.

The value of k for which the expression x^{2} + 4x + k is a perfect square is :

(a) 4

(b) – 4

(c) 16

(d) 3

Answer:

(a) 4

Question 3.

Which of the following equation has 3 as a root?

(a) x^{2} – 4x + 2 = 0

(b) x^{2} + x – 3 = 0

(c) 2x^{2} – 8x + 6 = 0

(d) 3x^{2} – 6x – 2 = 0

Answer:

(c) 2x^{2} – 8x + 6 = 0

Question 4.

If \(\frac{1}{3}\) is a root of x^{2} + kx – 1 = 0, then value of k is:

(a) \(\frac{4}{3}\)

(b) \(\frac{8}{3}\)

(c) \(\frac{1}{3}\)

(d) – 3

Answer:

(b) \(\frac{8}{3}\)

Question 5.

Which of the following is true about the equation \(\frac{1}{x^2-3 x+5}\) = 1

(a) Its discriminant is not defined

(b) Its discriminant is – 7

(c) Its discriminant is -11

(d) Its discriminant is same as coefficient of x

Answer:

(b) Its discriminant is – 7

Question 6.

Which of the following is not a quadratic equation?

(a) \(\frac{1}{2 x^2-x+1}\) = – 2

(b) (x^{2} – 1)^{2} = x^{4&} + 31 + 4x^{2}

(c) 2(x – 1)^{2} = 4x^{2} – 3x + 1

(d) (√3x + √5)^{2} + x^{2} = 4x^{2} – 5x

Answer:

(d) (√3x + √5)^{2} + x^{2} = 4x^{2} – 5x

Question 7.

The roots of the equation ax^{2} + bx + c = 0 are non- real, if

(a) b^{2} – 4ac = 0

(b) b^{2} – 4ac < 0

(c) b^{2} – 4ac > 0

(d) coefficient of x is zero

Answer:

(b) b^{2} – 4ac < 0

Question 8.

The roots of equation ax^{2} + bx + c = 0, a ≠ 0 are real, if b^{2} – 4ac is

(a) = 0

(b) ≥ 0

(c) ≤ 0

(d) none of these

Answer:

(b) ≥ 0

Question 9.

If n denotes the number of roots of a quadratic equation, then

(a) n is always equal to 2

(b) n < 2 (c) n ≤ 2 (d) n > 2

Answer:

(c) n ≤ 2

Question 10.

The positive value of /c for which the equation x^{2} + kx + 9 = 0 and x^{2} – 12x + k = 0 will both have real roots

(a) 36

(b) 6

(c) 16

(d) 12

Answer:

(b) 6

Question 11.

The equation (a^{2} + b^{2})x^{2} – 2 (ac + bd) x + c^{2} + d^{2} = 0 has equal roots, then

(a) ab = cd

(b) ad = bc

(c) ad = √bc

(d) ad = √cd

Answer:

(d) ad = √cd

Question 12.

The sum of a number and its reciprocal is 10/3, then the number is:

(a) 1

(b) 2

(c) 3

(d) 5

Answer:

(c) 3

Question 13.

The value of k for which – 5 is a root of 2x^{2} + px – 15 = 0 and the quadratic equation p(x^{2} + x) + k = 0 has equal roots is

(a) \(\frac{3}{2}\)

(b) \(\frac{3}{7}\)

(c) \(\frac{7}{4}\)

(d) \(\frac{1}{4}\)

Answer:

(c) \(\frac{7}{4}\)

Question 14.

What constant should be added or subtracted to solve the quadratic equation x^{2} – \(\frac{\sqrt{5}}{2}\)x – 4 = 0 by the method of completing the square?

(a) \(\frac{5}{4}\)

(b) \(\frac{5}{16}\)

(c) \(\frac{\sqrt{5}}{4}\)

(d) \(\frac{25}{4}\)

Answer:

(b) \(\frac{5}{16}\)

Question 15.

The value of k for which roots of kx^{2} – 3x + 1 = 0 are real is .

(a) k ≥ \(\frac{9}{4}\)

(b) k ≤ \(\frac{- 9}{4}\)

(c) k ≤ \(\frac{9}{4}\)

(d) k ≤ \(\frac{3}{2}\)

Answer:

(c) k ≤ \(\frac{9}{4}\)

Fill in the blanks:

Question 1.

Quadratic equation ax^{2} + bx + c = 0 has two _____________ and _____________ roots if b^{2} – 4ac > 0.

Answer:

real, unequal

Question 2.

Quadratic equation ax^{2} + bx + c = 0 has two real and _____________ roots if b^{2} – 4ac = 0.

Answer:

equal

Question 3.

For a quadratic equation ax^{2} + bx + c = 0, the quantity b^{2} – 4ac is called _____________ .

Answer:

discriminant

Question 4.

Quadratic equation ax^{2} + bx + c = 0 has two _____________ roots if b^{2} – 4ac < 0. Answer: imaginary Question 5. If D > 0 and is a perfect square then roots are _____________ (rational/irrational/imaginary).

Answer:

rational

Question 6.

If D > 0 and it is not a perfect square then roots are _____________ (rational/irrational/imaginary).

Answer:

irrational

Question 7.

If the value of mα^{2} + nα + p is zero, then a is said to be the _____________ of equation mx^{2} + nx + p = 0.

Answer:

root

Question 8.

The quadratic formula for finding roots of equation ax^{2} + bx + c = 0 is given by _____________ .

Answer:

x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

Question 9.

A quadratic equation can have at most _____________ roots.

Answer:

two

Question 10.

The quadratic equation x^{2} – 3x + 5 = 0 has _____________ roots.

Answer:

imaginary