Important Questions: Quadratic Equations | Mathematics (Maths) Class 10 PDF Download

Formation and Solution of Quadratic Equations

Q1. In a flight of 2800 km, an aircraft was slowed down due to technical issues. Its average speed decreased by 100 km/h, which caused the flight time to increase by 30 minutes. Find the original duration of the flight.

Ans: 

Distance = 2800 km
Original speed = x km/h
Reduced speed = x−100 km/h
Increase in time = 30 minutes =1/2 hour

Let original time t₁=  hrs,

time at reduced speed t₂ =  hrs

Original Duration: 

$t\_1\; =\; \backslash dfrac\{2800\}\{x$

Q2. The denominator of a fraction is 1 more than twice the numerator. If the sum of the fraction and its reciprocal is , find the fraction.

Ans: Let the numerator of the required fraction be x.

Then, its denominator = (2x + 1).

∴ fraction = x/(2x + 1) and its reciprocal = (2x + 1)/x.

∴ x/(2x + 1) + (2x + 1)/x = 58/21

⇒ 21 × [x² + (2x + 1)²] = 58x(2x + 1)

⇒ 21 × [5x² + 4x + 1] = 116x² + 58x

⇒ 11x² - 26x - 21 = 0

⇒ 11x² - 33x + 7x - 21 = 0 ⇒ 11x(x - 3) + 7(x - 3) = 0

⇒ (x - 3)(11x + 7) = 0 ⇒ x - 3 = 0 or 11x + 7 = 0

⇒ x = 3 or x = -7/11

⇒ x = 3  [∵ numerator cannot be a negative fraction].

∴ required fraction = x/(2x + 1) = 3/7.

Q3. A rectangular floor area can be completely tiled using 200 square tiles. If the side length of each tile is increased by 1 unit, it would take only 128 tiles to cover the floor.
(i) Assuming the original length of each side of a tile is x units, form a quadratic equation from the above information.
(ii) Write the corresponding quadratic equation in standard form.
(iii)
(a) Find the value of x, the side of the tile, by factorisation.
OR
(b) Solve the quadratic equation using the quadratic formula.

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
or, x = 4,  −4/9
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, c = −16

Using quadratic formula,

x = [−(−32) ± √((−32)² − 4 × 9 × (−16))] / (2 × 9)
= [32 ± √(1024 + 576)] / 18
= [32 ± √1600] / 18
= [32 ± 40] / 18
= (32 + 40)/18 or (32 − 40)/18
= 72/18 or −8/18 = 4 or −4/9

Q4. The sum of two numbers is 20. If 2 is subtracted from one number and 1 is added to the other, the product of these two new numbers becomes 96. Find the numbers.

Ans: Let one number be x, then the other number will be:

y = 20 - x  → (1)

According to the question:

(x - 2)(y + 1) = 96  → (2)

Now substitute the value of y from (1) into (2):

(x - 2) × ((20 - x) + 1) = 96
⇒ (x - 2)(21 - x) = 96

Now expand the expression:
= 21x - x² - 42 + 2x
= -x² + 23x - 42

So,

-x² + 23x - 42 = 96

Bring all terms to one side:

-x² + 23x - 138 = 0
Multiply through by -1:

x² - 23x + 138 = 0

Now factor the quadratic:

x² - 23x + 138 = 0
⇒ (x - 6)(x - 23) = 0

So, x = 6 or x = 23

From equation (1), y = 20 - x:

If x = 6, then y = 14
If x = 23, then y = -3

We discard the negative value, since we want two positive numbers.

Speed, Distance, and Time Problems

Q5. A motorboat whose speed in still water is 20 km/h takes 1.5 hours more to go 30 km upstream than to return downstream to the same location. Find the speed of the stream.

Ans: Speed of boat in still water = 20 km/h
Distance = 30 km
Let the speed of the stream be x km/h
Then:

• Upstream speed = 20−x km/h 
• Downstream speed = 20+x km/h

Time taken upstream:

Time taken downstream:

According to the question:
t₁ = t₂ + 1.5
Substitute the values:

Now simplify:

Take LHS:

Now use identity:

So:

Multiply both sides by (400−x²):
60x = 1.5(400 − x²)
60x = 600 − 1.5x² 
Bring all terms to one side:
1.5x²+60x−600=0
Multiply through by 2 to remove decimal:
3x²+120x−1200=0
Divide by 3 to simplify:
x²+40x−400=0
Now factorise:
Split middle term: x² + 50x − 10x − 400 = 0
x(x+50)−10(x+50)=0
(x+50)(x−10)=0
So, x=−50 or x=10
The speed of the stream is 10 km/h

Q6. A train covers a distance of 540 km at a uniform speed. If the speed had been 6 km/h less, it would have taken 2 hours more to cover the same distance. Find the original speed of the train.

Ans: Let the original speed of the train be x km/h.
Then, time taken to cover 540 km at speed x =
If the speed is reduced by 6 km/h, then new speed = x−6 km/h New time taken =
According to the question:
Take LHS:

Cross-multiply:

Divide entire equation by 2: x²−6x−1620=0
Now factorise: 
Try splitting the middle term:

So,  x=−36 or x=45
Reject negative speed.
The original speed of the train is 45 km/h

Roots of Quadratic Equations

Q7. If the roots of the quadratic equation 9x² − 12x + m = 0 are real and equal, then the value of m is:
(a) 16/9
(b) 4/3
(c) 144/81
(d) 36/25
(e) 4

Ans: (e)
Given: Quadratic equation: 9x² − 12x + m = 0
Compare with standard form ax² + bx + c = 0, we get:

• a = 9
• b = −12
• c = m

For real and equal roots, the discriminant D = 0:
D = b² − 4ac = 0
(−12)²−4(9)(m)=0⇒144−36m=0⇒36m=144⇒ m = 144/36 = 4

Q8. If the roots of the quadratic equation px² + qx + r = 0,where p ≠= 0, are real and equal, then which of the following relations must be true?
(a)

(b)

(c) q² = 2pr
(d)

Ans: (b)
For real and equal roots, the discriminant must be zero:
D = q² − 4pr = 0 ⇒ q² = 4pr ⇒ r =

Q9. Find the discriminant of the quadratic equation 3x² + 7x − 4 = 0 and comment on the nature of the roots.

Ans: Given quadratic equation: 3x² + 7x − 4 = 0
For a quadratic equation ax² + bx + c = 0, the discriminant D is given by: D = b² − 4ac
Here, a = 3, b = 7, and c = −4.
Substitute these values into the discriminant formula:
D = 7² − 4(3)(−4) = 49 + 48 = 97
Since D>0, the roots of the quadratic equation are real and distinct.

Q10. Find the value of p for which the quadratic equation p⋅x(x − 3) + 4 = 0 has two equal real roots.

Ans: The given quadratic equation is: p⋅x(x−3)+4=0
Expanding this:
p⋅x²−3px+4=0
Comparing with the standard quadratic equation ax² + bx + c = 0, we have:

• a=p
• b=−3p
• c=4

For the equation to have two equal real roots, the discriminant must be zero:
D=b²−4ac=0
Substitute the values of a, b, and c:

Thus, p=0 or p= 16/9
Since p ≠= 0 (as it would make the equation linear), we conclude that the value of p is:

Q11. Find the sum and product of the roots of the quadratic equation 3x² − 7x + 2 = 0.

Ans: Let the roots of the quadratic equation 3x² − 7x + 2 = 0 be α and β.
Using the sum and product of roots formulas for a quadratic equation ax² + bx + c = 0:

• Sum of roots α + β = − b/a
• Product of roots αβ= c/a

For the given equation 3x² −7x + 2 = 0, we have:

• a = 3
• b = −7
• c = 2

So,
Sum of roots:

Product of roots:

Q12. Find the value of p, for which one root of the quadratic equation px² − 10x + 5 = 0 is 4 times the other.

Ans: Let the first root be α, and the second root will be 4α.
Sum of roots is given by:

For the equation px² −10x + 5 = 0, we have:

• a = p
• b = −10

Thus, the sum of the roots is:

Product of roots is given by:

For the equation px² − 10x + 5 = 0, we have:
c = 5
Thus, the product of the roots is:

Substitute α = 2/p:

Cross-multiply:
16p = 5p²
5p² − 16p = 0
Factor the quadratic equation:
p(5p−16)=0
Thus, p = 0 or p= 16/5.
Since p=0 is not valid for a quadratic equation, we conclude:
p=16​/5

Q13. The quadratic equation (1 + a²)x² + 2abx + (b² − c²) = 0 has only one root. What is the value of c²(1 + a²)?

Ans: Given quadratic equation:
(1 + a²)x² + 2abx + (b² − c²) = 0
Comparing it with the standard form Ax² + Bx + C = 0, we identify the following coefficients:

• A=1+a²
• B=2ab
• C = b² − c²​

For the quadratic equation to have only one root, the discriminant must be zero. The discriminant Δ is given by:
Δ = B² − 4AC
Substitute the values of A, B, and C:
Δ = (2ab)² − 4 × (1 + a²) × (b² − c²)
Simplifying:
Δ = 4a²b²−4(1+a²)(b²−c²)
Expanding:
Δ = 4a²b²−4(b²−c²+a²b²−a²c²)
Simplifying further:
Δ = 4a²b²−4b²+4c²−4a²b²+4a²c²
Now, cancel out the 4a² b² terms:
Δ = −4b² + 4c² + 4a²c² 
For the discriminant to be zero:
−4b²+4c²+4a²c²=0
Divide through by 4:
−b²+c²+a²c²=0
Rearranging:
c²+a²c²=b² 
Factor out c²:
c²(1+a²)=b² 
Thus, the value of c² (1+a²) is: b²​​​​