Quadratic Equations Class 10 Extra Questions Maths Chapter 4 with Solutions

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 9x² – 3ax + 1 = 0 has equal roots?
Or
If one root of the quadratic equation 2x² + 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 = b² – 4ac
Here a = 9, b = – 3a, c = 1
∴ (- 3a)² – 4(9) (1) = 0
⇒ 9a² – 36 = 0
⇒ 9(a² – 4) = 0
⇒ a² = 4
⇒ a = ± 2
Or
Let, p(x) = 2x² + 2x + k

Question 2.
For what values of k, the roots of the equation x² + 4x + k = 0 are real?
Or
Find the value of k for which the roots of the equation 3x² – 10x + k = k = 0 are reciprocal of each other. [CBSE 2019]
Answer:
For real roots D ≥ 0 i.e., b² – 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 3x² – 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 px² + 6x + 1 = 0 are real, then find p.
Answer:
Roots are real, D ≥ 0
b² – 4ac ≥ 0
(6²) – 4 × p × 1 ≥ o
9 ≥ p

Question 4.
If 7x² – (2p² – 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 = 2p² – 8
p² = 4
p = ±2

Quadratic Equations Class 10 Extra Questions Short Answer Type-1

Question 1.
Find the roots of the quadratic equation
√2x² + 7x + 5√2 = 0 [CBSE Delhi 2017]
Answer:
√2x² + 7x + 5√2 = 0
√2x² + 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 x² + k(2x + k – 1) + 2 = 0 has real and equal roots.
Answer:
Rewriting given equation as
x² + 2kx + (k² – k + 2) = 0
Here a = 1, b = 2k, c = k² – k + 2
For equal roots, b² – 4ac = 0
⇒ (2k)² – 4(1) (k² – k + 2) = 0

⇒ k = 2

Question 3.
If – 5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, find the value of k. [CBSE Outside Delhi 2016]
Answer:
– 5 is root of the equation 2x² + px – 15 = 0, then
2(-5)² + p(-5) – 15 = 0
⇒ 50 + (- 5)p – 15 = 0
⇒ p = \(\frac{-35}{-5}\) = 7 ……….. (1)
p(x² + x) + k = 0 has equal root
⇒ b² – 4ac = 0
Here b = p, a = p, c = k
⇒ p² = 4 pk ………….. (2)
(7)² – 4 × 7k = 0
(1) and (2) gives,
∴ k = \(\frac{7}{4}\)

Question 4.
For what value of k, x² + 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 = x² + x (a + b)
⇒ x² + x (a + b) + ab = 0
⇒ x² + 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, x² + x – 306 = 0
or, x² + 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 + m²)x² + 2 mcx + c² – a² = 0 has equal roots then show than c² = a² (1 + m²). [CBSE 2017]
Answer:
(1 + m²) x² + 2mcx + c² – a² = 0
Here A = (1 + m²), B = 2mc, C = c² – a²
For equal roots; B² – 4AC = 0
⇒ (2mc)² – 4(1 + m²) (c² – a²) = 0

⇒ c² – a²(1 + m²) = 0
⇒ c² = a²(1 + m²)

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

⇒ x² + 100x – 300000 = 0
x² + 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)x² – (7p + 2)x + (7p – 3) = 0 has equal roots. Also find these roots. [S.A-II, 2014]
Answer:
(2p + 1)x² – (7p + 2)x + (7p – 3) = 0 …(1)
For equal roots D = 0
b² – 4ac = 0
⇒ [- (7p + 2)]² – 4(2p + 1) (7p – 3) = 0
⇒ 49p² + 4 + 28p – 4 (14p² – 6p + 7p – 3) = 0
⇒ 49p² + 28p + 4 – 56p2 + 24p – 28p + 12 = 0
⇒ – 7p² + 24p + 16 = 0
⇒ 7p² – 24p – 16 = 0
⇒ 7p² – 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)
⇒ 9x² – 30x + 25 = 0
⇒ 9x² – 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
– x² + 14x – 49 = 0
x² – 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 (t₁) = \(\frac{192}{x+16}\) hours and time taken by passenger train (t₂) = \(\frac{192}{x}\)
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 – x²
⇒ x² = 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(x² – bx – ax + ab)
⇒ ax² – 2abx + bx² – acx + 2abc – bcx = 2cx² – 2bcx – 2acx + 2abc
⇒ ax² + bx² – 2cx² = 2abx – acx – bcx
⇒ (a + b – 2c)x² = x(2ab – ac – bc)
⇒ (a + b – 2c)x² – 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 (t₁) = 63/x hrs
∴ Faster speed = (x + 6) km/hr
[∵ Time = \(\frac{\text { Distance }}{\text { Speed }}\)]
Distance = 72 km; time (t₂) = 72/ (x + 6) hrs.
(t₁) + (t₂) = 3 hrs
\(\frac{63}{x}+\frac{72}{(x+6)}\) = 3
⇒ 63(x + 6) + 72x = 3x (x + 6)
⇒ 63x + 378 + 72x = 3x² + 18x
⇒ 3x² – 117x – 378 = 0
⇒ x² – 39x – 126 = 0
⇒ 7x² – 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 – x²
⇒ x² + 48x – 324 = 0
⇒ x² + 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) = 5t² + 95t
When the policeman catches the thief.
5t² + 95t = 100 (t + 1)
⇒ 5t² + 95t = 100t + 100
⇒ 5t² – 5t – 100 = 0
⇒ t² – 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
⇒ x² – 50 = 5x
⇒ x² – 5x – 50 = 0
⇒ x2² – 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
x² = 5 + x
⇒ x² – x – 5 = 0

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

Question 3.
If the equation (1 + m²) x² + 2mcx + c² – a² = 0 has equal roots, show that c² = a² (1 + m²).
Answer:
Given equation is,
(1 + m²) x² + 2mcx + c² – a² = 0
∵ Roots are equal.
∴ D = 0
B² – 4AC = 0
(2mc)² – 4 (1 + m²) (c² – a²) = 0
or 4m²c² – 4 [c² – a² + m²c² – m²a²] = 0
or 4[m²c² – c² + a² – m²c² + m²a²] = 0
or – c² + a² (1 + m²) = 0
or c² = a² (1 + m²).

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 (x² + 4x)
x² + 4x – 480 = 0, S = 4
x² + 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
4x² + 1 – 4x = x² + 1 + 2x + x²
2x² – 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 ax² + 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 x² + 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) x² – 4x + 2 = 0
(b) x² + x – 3 = 0
(c) 2x² – 8x + 6 = 0
(d) 3x² – 6x – 2 = 0
Answer:
(c) 2x² – 8x + 6 = 0

Question 4.
If \(\frac{1}{3}\) is a root of x² + 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) (x² – 1)² = x^4& + 31 + 4x²
(c) 2(x – 1)² = 4x² – 3x + 1
(d) (√3x + √5)² + x² = 4x² – 5x
Answer:
(d) (√3x + √5)² + x² = 4x² – 5x

Question 7.
The roots of the equation ax² + bx + c = 0 are non- real, if
(a) b² – 4ac = 0
(b) b² – 4ac < 0
(c) b² – 4ac > 0
(d) coefficient of x is zero
Answer:
(b) b² – 4ac < 0

Question 8.
The roots of equation ax² + bx + c = 0, a ≠ 0 are real, if b² – 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 x² + kx + 9 = 0 and x² – 12x + k = 0 will both have real roots
(a) 36
(b) 6
(c) 16
(d) 12
Answer:
(b) 6

Question 11.
The equation (a² + b²)x² – 2 (ac + bd) x + c² + d² = 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 2x² + px – 15 = 0 and the quadratic equation p(x² + 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 x² – \(\frac{\sqrt{5}}{2}\)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 kx² – 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 ax² + bx + c = 0 has two _____________ and _____________ roots if b² – 4ac > 0.
Answer:
real, unequal

Question 2.
Quadratic equation ax² + bx + c = 0 has two real and _____________ roots if b² – 4ac = 0.
Answer:
equal

Question 3.
For a quadratic equation ax² + bx + c = 0, the quantity b² – 4ac is called _____________ .
Answer:
discriminant

Question 4.
Quadratic equation ax² + bx + c = 0 has two _____________ roots if b² – 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 mα² + nα + p is zero, then a is said to be the _____________ of equation mx² + nx + p = 0.
Answer:
root

Question 8.
The quadratic formula for finding roots of equation ax² + 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 x² – 3x + 5 = 0 has _____________ roots.
Answer:
imaginary

Extra Questions for Class 10 Maths

NCERT Solutions for Class 10 Maths