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

⇒
⇒ 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 be x.
Denominator = 2x + 1


Let,
Then, the equation will be.

⇒
⇒ 21y² + 21 = 58y
⇒ 21y² – 58y + 21 = 0
⇒ 21y² – 49y – 9y + 21 = 0
⇒ 7y(3y – 7) – 3(3y – 7) = 0
⇒ (3y – 7) (7y – 3) = 0

∴ Required fraction will be 7/3 and 3/7.

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

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

Ans: (b)
Sol: For rational roots, D is a perfect square
D = k²- 32
If k = 4√2  then,
D = (4√2)² - 32 = 0. 

Which is a perfect square

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
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 algebric 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 (13 + 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: 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

⇒ 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 = xm.
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 is2x² - 19x + 24 = 0
⇒ 2x² - 16x  - 3x + 24 = 0
⇒  2x(x - 8) - 3(x - 8) = 0
⇒ (x - 8)(2x - 3) = 0
⇒ x = 8 (not possible) or x = 3/2 = 1.5
Width of the sidewalk =1.5m

Ans: Let one number be x and another number be y.
Since, x + 4y = 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
⇒ 4x²- 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 = 3+ - 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 + a²
⇒ (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, from equation (i),
b = 6 + 2 × 10 = 26
and from equation (ii),
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 ⇒ 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: Let the speed of the stream be x km/hr.
Speed of the boat upstream = (18 - x) km/hr
Speed of the boat downstream = (18 + x) km/hr
According to question,

(∵ x ≠ -54 as speed can’t be negative)
Hence, the speed of the stream is 6 km/hr.

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

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.