Practice Questions: Quadratic Equations

Q1. Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. What is the speed of the train?

Sol:
(i) Let us consider,
The breadth of the rectangular plot is x m.
Thus, the length of the plot = (2x + 1) m.
Area of a rectangle = length × breadth = 528 m².
Substituting length and breadth,
(2x + 1) × x = 528
⇒ 2x² + x - 528 = 0.
Hence, 2x² + x - 528 = 0 is the required quadratic equation representing the situation.

(ii) Let us consider,
Speed of train = x km/h (original speed).
Time taken to travel 480 km at speed x = 480/x hours.
If speed had been (x - 8) km/h, time would be 480/(x - 8) hours. As given, this time is 3 hours more than the original time.
Therefore,
480/(x - 8) = 480/x + 3
Multiply both sides by x(x - 8):
480x = 480(x - 8) + 3x(x - 8)
⇒ 480x = 480x - 3840 + 3x² - 24x
Cancel 480x from both sides and rearrange:
3x² - 24x - 3840 = 0
Divide by 3:
x² - 8x - 1280 = 0
Hence, x² - 8x - 1280 = 0 is the required quadratic equation for the train problem.

Q2. Find the roots of quadratic equations by factorisation:
(i) √2 x2 + 7x + 5√2=0
(ii) 100x² - 20x + 1 = 0

Sol: 

(i)  √2 x² + 7x + 5√2 = 0
We try to factor the left-hand side. Observe that
(√2 x + 5)(x + √2) 

= √2 x² + (5 + 2)x + 5√2 

= √2 x² + 7x + 5√2.
Thus the equation factors as (√2 x + 5)(x + √2) = 0.
So either √2 x + 5 = 0 or x + √2 = 0.
Therefore, x = -5/√2 = -(5√2)/2 (rationalised) or x = -√2.

(ii) Given, 100x² - 20x + 1 = 0
Notice that this is a perfect square:
(10x - 1)² = 100x² - 20x + 1.
Hence (10x - 1)² = 0, so 10x - 1 = 0.
Therefore x = 1/10 (a repeated root).

Q3. Find two consecutive positive integers, the sum of whose squares is 365.

Sol: 
Let the two consecutive positive integers be x and x + 1.
Then x² + (x + 1)² = 365.
⇒ x² + x² + 2x + 1 = 365
⇒ 2x² + 2x - 364 = 0
⇒ x² + x - 182 = 0
Factorising: x² + 14x - 13x - 182 = 0
⇒ (x + 14)(x - 13) = 0
Thus x = -14 or x = 13. Since the integers are positive, x = 13.
So the two consecutive positive integers are 13 and 14.

Q4. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x² - 7x +3 = 0
(ii) 2x² + x - 4 = 0
Sol:

(i) 2x² - 7x + 3 = 0
Divide both sides by 2 to make the coefficient of x² equal to 1:
x² - (7/2)x + 3/2 = 0
Move constant to the right: x² - (7/2)x = -3/2.
Add (7/4)² = 49/16 to both sides to complete the square:
(x - 7/4)²
= -3/2 + 49/16
= 25/16.
So x - 7/4 = ±5/4 
⇒ x = (7 ± 5)/4.
Therefore x = 12/4 = 3 or x = 2/4 = 1/2.

(ii) 2x² + x - 4 = 0
Divide by 2:
x² + (1/2)x - 2 = 0
Move constant to the right: x² + (1/2)x = 2.
Add (1/4)² = 1/16 to both sides to complete the square:
(x + 1/4)²
= 2 + 1/16 
= 33/16.
So x + 1/4 = ±(√33)/4 
⇒ x = -1/4 ± (√33)/4.
Thus x = (-1 ± √33)/4 (two real irrational roots).

Q5. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

Sol: 
Let the shorter side be x m. Then the longer side = x + 30 m and the diagonal = x + 60 m.
By Pythagoras: x² + (x + 30)² = (x + 60)²
Expand: x² + x² + 60x + 900 
= x² + 120x + 3600.
Simplify: x² - 60x - 2700 = 0.
Factor: (x - 90)(x + 30) = 0
⇒ x = 90 or x = -30. Discard negative value.
Thus the shorter side = 90 m, longer side = 90 + 30 = 120 m. (Diagonal = 150 m.)

Q6. Solve the quadratic equation 2x² - 7x + 3 = 0 by using quadratic formula.

Sol:  
Given 2x² - 7x + 3 = 0. Compare with ax² + bx + c = 0: a = 2, b = -7, c = 3.
Quadratic formula: x = [-b ± √(b² - 4ac)]/(2a).
So x = [7 ± √(49 - 24)]/4 
= [7 ± √25]/4 
= [7 ± 5]/4.
Therefore x = 12/4 = 3 or x = 2/4 = 1/2.

Q7. The sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

Sol:
Let the sides be x and y (in metres). Then x² + y² = 468.
Difference of perimeters: 4x - 4y = 24 ⇒ x - y = 6 ⇒ y = x - 6.
Substitute into areas: x² + (x - 6)² = 468
⇒ x² + x² - 12x + 36 = 468 
⇒ 2x² - 12x - 432 = 0
⇒ x² - 6x - 216 = 0 
⇒ (x + 12)(x - 18) = 0.
So x = 18 (reject x = -12). 
Then y = x - 6 = 12.
Hence the sides are 18 m and 12 m.

Q8. Find the values of k for each of the following quadratic equations, so that they have two equal roots.

(i) 2x² + kx + 3 = 0

(ii) kx (x - 2) + 6 = 0

Sol:

(i) 2x² + kx + 3 = 0. Here a = 2, b = k, c = 3. For equal roots the discriminant must be zero: b² - 4ac 
= k² - 4·2·3 
= k² - 24 = 0.
Thus k² = 24 
⇒ k = ±2√6.

(ii) kx(x - 2) + 6 = 0 ⇒ kx² - 2kx + 6 = 0. 
Here a = k, b = -2k, c = 6.
Discriminant: b² - 4ac 
= (-2k)² - 4·k·6 
= 4k² - 24k 
= 4k(k - 6).
Set equal to zero: 4k(k - 6) = 0 ⇒ k = 0 or k = 6.
If k = 0 the equation is no longer quadratic (it loses x² and x terms). Therefore the value that gives two equal roots for a quadratic is k = 6.

Q9. Is it possible to design a rectangular park of perimeter 80 and area 400 sq.m.? If so find its length and breadth.

Sol:
Let length = L and breadth = B. Perimeter 2(L + B) = 80 ⇒ L + B = 40 ⇒ B = 40 - L.
Area L × B = 400 ⇒ L(40 - L) = 400 
⇒ -L² + 40L - 400 = 0 
⇒ L² - 40L + 400 = 0.
Compare a = 1, b = -40, c = 400. 
Discriminant b² - 4ac = 1600 - 1600 = 0, so the equation has equal real roots. 
Hence the situation is possible.
For equal roots L = -b/(2a) = 40/(2) = 20. 
Then B = 40 - 20 = 20.
Therefore the park can be designed as a square of side 20 m (length and breadth both 20 m).

Q10. Find the discriminant of the equation 3x²- 2x +1/3= 0 and hence find the nature of its roots. Find them, if they are real.

Sol:
Given 3x² - 2x + 1/3 = 0. 
Here a = 3, b = -2, c = 1/3.
Discriminant Δ = b² - 4ac 
= (-2)² - 4·3·(1/3) 
= 4 - 4 = 0.
Since Δ = 0, the equation has two equal real roots. 
Each root = -b/(2a) = 2/(6) = 1/3.
Hence the roots are 1/3 and 1/3.

Q11. In a flight of 600 km, an aircraft was slowed due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. Find the original duration of the flight.

Sol:
Let the original duration be x hours. 
Then original average speed = 600/x km/h.
When slowed, the speed becomes 600/x - 200 and the time becomes x + 1/2 hours (30 minutes = 1/2 hour).
So distance = speed × time gives 600 = (600/x - 200)(x + 1/2). 
It is simpler to use the relation between times:
The slower time x + 1/2 equals 600/(600/x - 200). 
Equivalently, set up difference of speeds:
A standard and simpler equation is: (600/x) - 600/(x + 1/2) = 200.
Compute left-hand side: [600(2x + 1) - 1200x] / [x(2x + 1)] = 200 
⇒ (1200x + 600 - 1200x)/[x(2x + 1)] = 200 
⇒ 600 = 200x(2x + 1).
Thus x(2x + 1) = 3 
⇒ 2x² + x - 3 = 0.
Factor: (2x + 3)(x - 1) = 0 
⇒ x = -3/2 or x = 1. 
Time is positive, so x = 1 hour.
Hence the original duration of the flight was 1 hour.

Q12. If x = 3 is one root of the quadratic equation x² - 2kx - 6 = 0, then find the value of k.

Sol:

Substitute x = 3 into x² - 2kx - 6 = 0:
9 - 6k - 6 = 0 
⇒ 3 - 6k = 0 
⇒ 6k = 3 
⇒ k = 1/2.
Therefore, k = 1/2.

Q.13: Find the value of p, for which one root of the quadratic equation px2 - 14x + 8 = 0 is 6 times the other.

Sol:
Given px² - 14x + 8 = 0. Let the roots be α and 6α.
Sum of roots = α + 6α = 7α = -(-14)/p = 14/p ⇒ α = 2/p.
Product of roots = α·6α = 6α² = 8/p.
Substitute α = 2/p: 6(2/p)² = 8/p 
⇒ 6·4/p² = 8/p 
⇒ 24/p² = 8/p.
Multiply both sides by p²: 24 = 8p ⇒ p = 3.
Therefore, p = 3.

Q14. Solve for x: [1/(x + 1)] + [3/(5x + 1)] = 5/(x + 4); x ≠ -1, -⅕, -4

Sol:
Given (with exclusions) 1/(x + 1) + 3/(5x + 1) = 5/(x + 4).
Combine the left-hand terms over the common denominator (x + 1)(5x + 1):
[1(5x + 1) + 3(x + 1)] / [(x + 1)(5x + 1)] = 5/(x + 4).
Numerator simplifies to 5x + 1 + 3x + 3 = 8x + 4 and (x + 1)(5x + 1) = 5x² + 6x + 1.
Cross-multiply: (8x + 4)(x + 4) = 5(5x² + 6x + 1).
Expand: 8x² + 36x + 16 = 25x² + 30x + 5.
Bring all terms to one side: 17x² - 6x - 11 = 0.
Factor: (17x + 11)(x - 1) = 0 
⇒ x = 1 or x = -11/17.
Both values are allowed (not equal to -1, -1/5, or -4). 
Therefore the solutions are x = 1 and x = -11/17.

Q.15: 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.

Sol:
Since x = -5 is a root of 2x² + px - 15 = 0, substitute x = -5:
2(25) + p(-5) - 15 = 0 ⇒ 50 - 5p - 15 = 0 ⇒ 35 - 5p = 0 ⇒ p = 7.
Now substitute p = 7 into p(x² + x) + k = 0 to get 7x² + 7x + k = 0. 
For equal roots the discriminant must be zero:
b² - 4ac 
= 7² - 4·7·k 
= 49 - 28k = 0 
⇒ 28k = 49 
⇒ k = 49/28 = 7/4.
Therefore, k = 7/4.