Class 10 Maths Chapter 4 Previous Year Questions - Quadratic Equations

Ans:
Let the speed of aircraft be x km/hr.
Time taken to cover 2800 km by speed of x km/hr = 2800/x  hrs.
New speed is (x – 100) km/hr
so time taken to cover 2800 km at the speed of (x - 100) km/hr = 2800x - 100 hrs

ATQ,
2800x - 100 - 2800x = 12

⇒ 2800 (x - x + 100)x (x - 100) = 12

⇒ 100x² - 100x = 12 × 2800

⇒ 560000 = x² – 100x
⇒ x² – 100x – 560000 = 0
⇒ x² – 800x + 700x – 560000 = 0
⇒ x(x – 800) + 700(x – 800) = 0
⇒ (x – 800) (x + 700) = 0
⇒ x = 800, – 700 (Neglect)
⇒ x = 800
Speed = 800 km/hr
Time = 2800/800
= 3 hr 30 min.

Ans: 
Let the numerator of the required fraction be x.
Then, its denominator = (2x + 1).

The fraction = x / (2x + 1) and its reciprocal = (2x + 1) / x.
We are given that:
(x / (2x + 1)) + ((2x + 1) / x) = 58 / 21

Multiplying both sides by 21x(2x + 1):
21x [x² + (2x + 1)²] = 58x(2x + 1)
21x [x² + (2x + 1)²] = 58x(2x + 1)
Simplifying this:
21x [5x² + 4x + 1] = 116x² + 58x
Expanding both sides:
11x² - 26x - 21 = 0
This is a quadratic equation.

Solving the quadratic equation:
11x² - 33x + 7x - 21 = 0
11x(x - 3) + 7(x - 3) = 0
(x - 3) (11x + 7) = 0

So, x = 3 or x = -7/11.
Since the numerator cannot be negative, x = 3.

Thus, the required fraction is x / (2x + 1) = 3 / 7.

Ans: (b) 25/16
Given that,
Roots of quadratic equation 4x² − 5x + k = 0 are real and equal
On comparing with ax² + bx + c = 0,
We get, a = 4, b = −5 and c = k
For real and equal roots,
⇒ b² − 4ac = 0
⇒ (−5)² − 4(4)(k) = 0
⇒ 16k = 25
⇒ k = 25/16

Ans: (i) Let the original side length of each tile be x units.
The area of the rectangular floor using 200 tiles = 200 x² unit² 
The area with increased side length (each side increased by 1 unit) using 128 tiles
= 128 (x + 1)² unit²
So, the required quadratic equation is: 200x² = 128 (x + 1 )²
(ii) We have,
200x² = 128 (x + 1)²
⇒ 200x² = 128 (x² + 2x + 1)
⇒ 200x² = 128x² + 256x + 128
⇒ 72x² − 256x − 128 = 0
which is the quadratic equation is standard form.
(iii) (a) We have,
72x² − 256x − 128 = 0
or, 9x² − 32x − 16 = 0
or, 9x² − 36x + 4x − 16 = 0
or, 9x (x − 4) + 4 (x − 4) = 0
or, (x − 4) (9x + 4) = 0

Since the side cannot be negative, thus x = 4 units.
OR
(b) We have 9x² − 32x − 16 = 0
On comparing with ax² + bx + c = 0, we get
a = 9, b = −32 and c = −16
Using quadratic formula,

Ans: (c) ac = b²/4
If the discriminant is equal to zero, i.e., b² − 4ac = 0 where a, b, c are real numbers and a ≠ 0, then roots of the quadratic equation ax² + bx + c = 0, are real and equal.
Thus, b² - 4ac = 0 or ac = b²/4

Ans: Let α and β be the roots of given quadratic equation  2x² - 9x + 4 = 0.
Sum of roots = α + β = -b/a = (-9)/2 = 9/2
and Product of roots, αβ = c/a = 4/2 = 2  

Ans: Let the first root be α, then the second root will be 6a
Sum of roots  = -b/a
⇒ a + 6a = 14/p
⇒ 7a = 14/p
⇒ a = 2/p
Product of roots = c/a
⇒ a x 6a  = 8/p
⇒ 6a² = 8/p

⇒ 6(2p)² = 8p

⇒ 6 x 4p² = 8p

⇒ p = 6 x 4/8
⇒ p = 3
Hence, the value of p is 3.

Ans: (c)
Sol: Put k = 2,

⇒ 2x² + 2x - 4 = 0

⇒ 2x² + 4x - 2x - 4 = 0

⇒ 2x (x + 2) - 2(x + 2) = 0

⇒ x = 1, -2

Put k = -2,

⇒ 2x² - 2x - 4 = 0

⇒ 2x² - 4x + 2x - 4 = 0

⇒ 2x (x - 2) + 2 (x - 2) = 0

⇒ x = -1, 2
Hence, to get the rational values of x, that is, to get rational roots, k must be ±2.

Ans: Given quadratic equation is 4x² - 5 = 0
Discriminant, D = b² - 4ac = 0² - 4(4)(-5) = 80 > 0
Hence, the roots of the given quadratic equation are real and distinct.

Ans: The given quadratic equation is px(x - 2) + 6 = 0
⇒ px² - 2xp +6 = 0
On comparing with ax²+ bx + c = 0, we get
a = p, b = -2p and c = 6
Since, the quadratic equations has two equal real roots.

∴ Discriminant D = 0
⇒ b² - 4ac = 0
⇒ (-2p)² - 4 x p x 6 = 0
⇒ 4p² - 24p = 0
⇒ p² - 6p = 0
⇒  p(p - 6) = 0
⇒  p = 0 or p = 6
But p ≠ 0 as it does not satisfy equation
Hence, the value of p is 6.

Ans: Area = 18 x 12 cm = 216 cm²
Length (l) is increased by x cm
So,  new length =(18 + x ) cm
New width = (12 + x) cm

(i) Area of photo after increasing the length and width
= (18 + x)(12 + x) = 2 x 18 x 12
i.e., (18 + x) (12 + x) = 432 is the required algebraic equation.
(ii) From part (i) we get, (18 + x) (12 + x) = 432
⇒ 216 + 18x + 12x + x² = 432
⇒  x² + 30x - 216 = 0
(iii)  x² + 30x - 216 = 0
⇒ x²+ 36x - 6x - 216 = 0
⇒  x(x+ 36) - 6 (x+ 36) = 0 ⇒  x = 6, -36
-36 is not possible.
So, new length = (18 + 6) cm = 24 cm
New width = (12 + 6) cm = 18cm
So. new dimension = 24cm x 18 cm

According to question (18 + x) (12 + x) = 220
⇒ 216 + 30x + x² = 220
⇒  x² + 30x + 216 - 220 = 0
⇒  x² + 30x - 4 = 0
For rational value of x. discriminant (D) must be perfect square.
So, D = b² - 4ac
= (30)² - 4(1)(-4) = 900 + 16 = 916
∴  916 is not a perfect square.
So, no rational value of x is possible.

Ans: (a)
To find the roots of the quadratic equation x² + 3x − 10 = 0, we can use the quadratic formula:

For the equation x² + 3x − 10 = 0:
a = 1
b = 3
c = −10
Substitute these values into the formula:

Now, calculate the two roots:

(i) For x = (-3 + 7) / 2 = 4/2 = 2
(ii) For x  = (-3 - 7)/2 = -10/2 = -5

The roots of the equation are $\mathrm{<br\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">}$2 and -5.
So, the correct answer is: $(a)\; \backslash text\{(a)\; \}\; 2,\; -5$(a) 2,−5

Ans: Given, quadratic equation is ky² – 11y + (k – 23) = 0
Let the roots of the above quadratic equation be α and β.
Now, Sum of roots, α + β = -(-11)/k = 11/k ...(i)
and Product of roots, αβ = (k-23)/k  ...(ii)
According to the question,
α + β = αβ + 13/21

∴ 11k = k - 23k + 1321 ... [From equations (i) and (ii)]

⇒ 11k = k - 23k + 1321

⇒ 11 - k + 23k = 1321

⇒ 21(34 – k) = 13k
⇒ 714 – 21k = 13k
⇒ 714 = 13k + 21k
⇒ 34k = 714
⇒ k = 714/34
⇒ k = 21

Ans: x² – 2ax – (4b² – a²) = 0 
⇒ x² + (2b – a)x – (2b + a)x – (4b² – a²) = 0
⇒ x(x + 2b – a) – (2b + a)(x + 2b – a) = 0
⇒ (x + 2b – a)(x – 2b – a) = 0
⇒ (x + 2b – a) = 0, (x – 2b – a) = 0
∴ x = a − 2b, a + 2b 

Ans: Given, width of the sidewalk = x m.
Area of the pool = 36 sq.m
∴ Inner length of the pool
= (12 - 2x)m
Inner width of the pool = (7 - 2x)m
∴ Area of the pool = A = l x b
⇒ 36 = (12 - 2x) x  (7 - 2x)
⇒ 36 = 84 - 24x - 14x +4x²
⇒ 4x² - 38x + 48 = 0
⇒ 2x² - 19x + 24 = 0, is the required quadratic equation.

Area of the pool given by quadratic equation is 2x² - 19x + 24 = 0
⇒ 2x² - 16x - 3x + 24 = 0
⇒  2x(x - 8) - 3(x - 8) = 0
⇒ (x - 8)(2x - 3) = 0
⇒ x = 8 (not possible) 
Hence, x = 3/2 = 1.5
Width of the sidewalk =1.5m

Ans: Let one number be x and another number be y.
Since, x + y = 34 ⇒  y = 34 - x      (i)
Now. according to the question. (x - 3) (y + 2) = 260       (ii)
Putting the value or y from (i) in (ii), we get
⇒ (x - 3)(34 - x + 2) = 260
⇒ (x - 3)(36 - x) = 260
⇒ 36x - x² - 108 + 3x = 260
⇒ x² - 39x +368 = 0
⇒ x²- 23x - 16x + 368 = 0
⇒ x(x - 23) - 16(x - 23) = 0
⇒ (x - 23)(x - 16) =0
⇒ x = 23 or 16
Hence; when x  = 23 from (i),  y = 34 - 23 = 11
When x = 16. then y = 34 - 16 = 18
Hence the required numbers are 23 and 11 or 16 and 18.

Ans: 

Let ΔABC be the right angle triangle, right angled at B, as shown in the figure.
Also, let AB = c cm, BC = a cm and AC = b cm
Then, according to the given information, we have
b = 6 + 2a  .....(i) (Let a be the shortest side)
and c = 3a – 6  ...(ii)
We know that, b² = c² + a^{2   .....(Pythagoras Theorem)}
Putting the values of b and c from (i) and (ii) in above equation
⇒ (6 + 2a)² = (3a – 6)² + a² ...[Using (i) and (ii)]
⇒ 36 + 4a² + 24a = 9a² + 36 – 36a + a²
⇒ 60a = 6a²
⇒ 6a = 60  ...[∵ a cannot be zero]
⇒ a = 10 cm
Now, putting value of a in equation (i), we get
b = 6 + 2 × 10 = 26
and putting value of a in equation (ii), we get
c = 3 × 10 – 6 = 24
Thus, the dimensions of the triangle are 10 cm, 24 cm and 26 cm.

Ans: We have, x² – 2ax + (a² – b²) = 0
⇒ x² – ((a + b) + (a – b))x + (a² – b²) = 0
⇒ x² – (a + b)x – (a – b)x + (a + b)(a – b) = 0  ......[∵ a² – b² = (a + b)(a – b)]
⇒ x(x – (a + b)) – (a – b)(x – (a + b)) = 0
⇒ (x – (a + b))(x – (a – b) = 0
⇒ x = a + b, a – b 

Ans:  We have
(m - 1) x² + 2 (m - 1) x + 1 = 0 ----(i)
On comparing the given equation with ax² + bx + c = 0,
we have a = (m - 1), b = 2 (m - 1), c = 1
Discriminant, D = 0
⇒ b² - 4ac = 0 
⇒ (2 (m - 1))² - 4 (m - 1)(1)
⇒ 4m² + 4 - 8m - 4m + 4 = 0
⇒ 4m² -12m + 8 = 0
⇒ m² - 3m + 2 = 0 ⇒ m² - 2m - m + 2= 0
⇒ m(m - 2) - 1 (m - 2) = 0
⇒ (m - 1)(m - 2) = 0 ⇒ m = 1, 2

Ans: (1 + a²)x² + 2abx + (b² - c²)= 0
Comparing on Ax² + Bx + C = 0
A = 1 + a², B = 2ab  &  C = (b² - c²)
Now, B² - 4AC = 0
⇒ (2ab)² - 4 × (1 + a²) × (b² - c²) = 0
⇒ 4a²b² - 4(b² - c² + a²b² - a²c²) = 0
⇒ 4a²b² - 4b² + 4c² - 4a²b² + 4 a²c² = 0
⇒ - b²+ c² + a²c² = 0
⇒ c² + a²c² = b²
∴ c² (1 + a²) = b²

Ans: Roots of quadratic equation are given as 2 and - 5.
Sum of roots = 2 + (-5) = -3
Product of roots = 2 (-5) = -10
Quadratic equation can he written as
x² - (sum of roots)x + Product of roots = 0
⇒ x² + 3x - 10 = 0

Ans: Let the sides of the two squares be x m and y m, where ; x > y.
Then, their areas are x² and y² and their perimeters are 4x and 4y respectively.
By the given condition, x² + y² = 544 --------(i)
and 4x - 4y = 32
⇒ x - y = 8
⇒ x = y + 8 ------------ (ii)
Substituting the value of x from (ii) in (i) we get
⇒ (y + 8)² + y² = 544
⇒ y² + 64 + 16y + y² = 544
⇒ 2y² + 16y - 480 = 0
⇒ y² + 8y - 240 = 0
⇒ y² + 20y - 12y - 240 = 0
⇒ y(y + 20) - 12(y + 20) = 0
⇒ (y - 12) (y + 20) = 0
⇒ y = 12    (∵ y ≠ - 20 as length cannot be negative)
From (ii), x = 12 + 8 = 20 Thus, the sides of the two squares are 20 m and 12 m.

Ans: 

Speed of boat = 18 km/hr

Distance = 24 km

Let x be the speed of the stream.

Let t₁ and t₂ be the time for upstream and downstream.

As we know that, SpeedDistance = TimeDistance ⇒ Time = DistanceSpeed

For upstream:

Speed = (18 - x) km/hr

Distance = 24 km

Time = t₁

t₁ = 2418 - x

For downstream:

Speed = (18 + x) km/hr

Distance = 24 km

Time = t₂

t₂ = 2418 + x

Now according to the question:

t₁ = t₂ + 1

Substitute the values:

2418 - x = 2418 + x + 1

Simplify:

118 - x - 118 + x = 124

Combine the fractions:

(18 + x) - (18 - x)(18 - x)(18 + x) = 124

⇒ 2x(18 - x)(18 + x) = 124

Cross-multiply:

48x(18 - x)(18 + x) = 1
Expand:

48x = 324 + 18x - 18x - x²

x² + 48x - 324 = 0

Rearrange:

x² + 54x - 6x - 324 = 0

(x + 54) - 6(x + 54) = 0

(x + 54)(x - 6) = 0

Solve for x:

x = -54 or x = 6

Since speed cannot be negative:

x = 6

Ans: (b)
Given Quadratic equation is 2x² + kx + 2 = 0
Since, the equation has equal roots.
∴ Discriminant = 0
= b² - 4ac
⇒ k² - 4 x 2 x 2 = 0
⇒ k²- 16 = 0
⇒ k² = 16
⇒ k = ±4

Ans: Given, 

⇒ 1x + 4 - 1x - 7 = 1130

⇒ (x -7) - (x + 4)(x + 4)(x - 7) = 1130

⇒ -11(x + 4)(x - 7) = 1130

⇒ (x + 4)(x – 7) = –30 
⇒ (x + 4) (x – 7) + 30 = 0 
⇒ x² + 4x – 7x – 28 + 30 = 0 
⇒ x² – 3x – 28 + 30 = 0 
⇒ x² – 3x + 2 = 0 
⇒ x² – 2x – x + 2 = 0 
⇒ x(x – 2) – 1(x – 2) = 0 
⇒ (x – 2) (x – 1) = 0 
i.e., x – 1 = 0 or x – 2 = 0 
⇒ x = 1 or 2

Ans: Let the original speed of the train be x km/h. 
Then, time taken to cover the journey of 480 km , t = 480 / x hours 
Time taken to cover the journey of 480 km with speed of (x – 8) km/h = 480 / x − 8 hours 
Now, according to question,

480x - 8 - 480x = 3

⇒ 480 [x - x + 8x(x - 8)] = 3
⇒ 480 (8) = 3x(x - 8)

⇒ 3x(x – 8) = 3840 
⇒ x(x – 8) = 1280 
⇒ x² – 8x – 1280 = 0 
⇒ x² – 40x + 32x – 1280 = 0 
⇒ x(x – 40) + 32(x – 40) = 0 
⇒ (x + 32) (x – 40) = 0 
⇒ x + 32 = 0 or x – 40 = 0 
∴ x = – 32 (not possible) 
∴ x = 40 Thus, the original speed of the train is 40 km/h.

Ans: Since x = 2 is a solution of kx² + 2x - 3 = 0
k(2)² + 2(2) - 3 = 0
= 4k + 4 - 3 = 0
⇒ k = -1/4

Ans: Let the length of the side of one square be x m and the length of the side of another square be y m.
Given, x² + y²= 157    _(i)
and 4x + 4y=68    _(ii)
 x + y = 17
y = 17 - x    _(iii)
On putting the value of y in (i), we get
x² + (17 - x)² = 157
⇒ x² + 289 + x² - 34x = 157
=> 2x² - 34x +132 = 0
⇒ x² - 17x + 66 = 0
⇒ x² - 11x - 6x + 66 =0
⇒ x(x - 11) - 6(x - 11) = 0
⇒ (x - 11) (x - 6) = 0
⇒ x = 6 or x = 11
On putting the value of x in (iii), we get
y = 17 - 6 = 11 or y = 17 - 11 = 6
Hence, the sides of the squares be 11 m and 6 m.

Ans: Given quadratic equation is: 
6x² – x – k = 0. 
It's one root is 2/3
If x = 2 / 3 is root of the given equation, then it will satisfy the given equation,

∴ 6 (23)² - 23 - k = 0

⇒ 6 × 49 - 23 - k = 0

⇒ 83 - 23 - k = 0

⇒ k = 2
Hence, the value of k is 2.

Ans: Given, quadratic equation is, 
(1 + m² )x² + 2mc x + c² – a² = 0 
On comparing it with Ax² + Bx + C = 0, 
we get A = 1 + m², B = 2mc and C = c² – a²
The roots of the given equation are equal, then 
Discriminant, D = 0 
∴ B² – 4AC = 0 
(2mc)² – 4 × (1 + m²) × (c² – a²) = 0 
⇒ 4m²c² – 4(c² + c²m² – a² – a²m²) = 0 
⇒ 4m²c² – 4c² – 4c²m² + 4a² + 4m²a² = 0 
⇒ m²a² + a² – c² = 0 
⇒ c² = m²a2 + a²
⇒ c² = a² (1 + m²) 
Hence, proved

Ans: Let the length of AC be x 
Then, BC = (2 – x) m 
Now, AC² = AB × CB (given) 
Putting the value of AC and BC in above equation, we get 
∴ x² = 2 × (2 – x)     [∴ AB = 2 m (given)] 
⇒ x² = 4 – 2x 
⇒ x² + 2x – 4 = 0 
Now, if we compare the above equation with ax² + bx + c = 0. 
Then, a = 1, b = 2, c = –4 
Roots of the equation are
-b ± √b² - 4ac2a 
= -2 ± √(2)² - 4 × 1 × (-4)2 × 1

= -2 ± √202

= -2 ± 2√52

= -1 ± √5

When x = -1 + √5, or x = -1 - √5

Then, BC = 2 - (- 1 + √5) = 3 - √5

BC = 3 - 2.24 = 0.76 m

When x = -1 - √5

BC = 2 - (-1 - √5) = 3 + √5

BC = 3 + 2.24 = 5.24 m

(which is not possible)

Hence, the length of BC is 3 − √5 or 0.76 m.

Ans: (b)
To find the roots of the equation $x^2\; -\; 3x\; -\; m(m\; +\; 3)\; =\; 0$x² − 3x − m(m + 3) = 0, we can use the quadratic formula:
x = -b ± √b² - 4ac2a

For the equation $x^2\; -\; 3x\; -\; m(m\; +\; 3)\; =\; 0$x² − 3x − m(m + 3) = 0:
a = 1
b = −3
c = −m(m + 3)
Substitute these values into the formula:

Since  , we get two cases for the roots based on the sign:
Case 1:
Case 2:
Thus, the roots of the equation are -m and m + 3.
So, the correct answer is: (b) −m, m + 3