QUADRATIC EQUATIONS

Exercise 4.1

Question 1:

Check whether the following are quadratic equations :

(i) (x + 1)² = 2(x – 3)

Solution:

The given equation is:
(x + 1)² = 2(x – 3)

Step 1: Expand both sides

Left-hand side:
(x + 1)² = x² + 2x + 1

Right-hand side:
2(x – 3) = 2x – 6

Now substitute back into the equation:
x² + 2x + 1 = 2x – 6

Step 2: Move all terms to one side

Subtract 2x – 6 from both sides:
x² + 2x + 1 – 2x + 6 = 0

Simplify:
x² + 7 = 0

Step 3: Final simplified form

x² + 7 = 0

This is a quadratic equation because it is in the form:

ax² + bx + c = 0, where
a = 1, b = 0, c = 7

Final Answer: Yes, it is a quadratic equation.

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(ii) x² – 2x = (–2)(3 – x)

Solution:

The given equation is:
x² – 2x = (–2)(3 – x)

Step 1: Expand both sides

Right-hand side:
(–2)(3 – x) = –6 + 2x

Now substitute back into the equation:
x² – 2x = –6 + 2x

Step 2: Move all terms to one side

Subtract (–6 + 2x) from both sides:
x² – 2x – (–6 + 2x) = 0
x² – 2x + 6 – 2x = 0

Simplify:
x² – 4x + 6 = 0

Step 3: Final simplified form

x² – 4x + 6 = 0

This is a quadratic equation because it's in the standard form:

ax² + bx + c = 0, where
a = 1, b = –4, c = 6

Final Answer: Yes, it is a quadratic equation.

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(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

Solution:

The given equation is:
(x – 2)(x + 1) = (x – 1)(x + 3)

Step 1: Expand both sides

Left-hand side:
(x – 2)(x + 1) = x² + x – 2x – 2 = x² – x – 2

Right-hand side:
(x – 1)(x + 3) = x² + 3x – x – 3 = x² + 2x – 3

Now substitute back:
x² – x – 2 = x² + 2x – 3

Step 2: Move all terms to one side

Subtract the right-hand side from both sides:
x² – x – 2 – (x² + 2x – 3) = 0

Simplify:
x² – x – 2 – x² – 2x + 3 = 0
(x² – x²) + (–x – 2x) + (–2 + 3) = 0
–3x + 1 = 0

Step 3: Final simplified form

–3x + 1 = 0
This is a linear equation because the highest power of x is 1.

Final Answer: No, it is not a quadratic equation.

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(iv) (x – 3)(2x + 1) = x(x + 5)

Solution:

The given equation is:
(x – 3)(2x + 1) = x(x + 5)

Step 1: Expand both sides

Left-hand side:
(x – 3)(2x + 1)
= x(2x + 1) – 3(2x + 1)
= 2x² + x – 6x – 3
= 2x² – 5x – 3

Right-hand side:
x(x + 5) = x² + 5x

Now substitute back:
2x² – 5x – 3 = x² + 5x

Step 2: Move all terms to one side

Subtract x² + 5x from both sides:
2x² – 5x – 3 – x² – 5x = 0

Simplify:
(2x² – x²) + (–5x – 5x) – 3 = 0
x² – 10x – 3 = 0

Step 3: Final simplified form

x² – 10x – 3 = 0

This is a quadratic equation, since it's in the form:

ax² + bx + c = 0, where
a = 1, b = –10, c = –3

Final Answer: Yes, it is a quadratic equation.

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(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

Solution:

Step 1: Expand both sides.

Left side:
(2x – 1)(x – 3) = 2x² – 7x + 3

Right side:
(x + 5)(x – 1) = x² + 4x – 5

Step 2: Bring all terms to one side:

2x² – 7x + 3 – (x² + 4x – 5) = 0
= 2x² – 7x + 3 – x² – 4x + 5
= x² – 11x + 8 = 0

Step 3: The final equation is:

x² – 11x + 8 = 0

This is a quadratic equation because the highest power of x is 2.

Answer: Yes, it is a quadratic equation.

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(vi) x² + 3x + 1 = (x – 2)²

Solution:

Step 1: Expand the right side.

(x – 2)² = x² – 4x + 4

Step 2: Bring all terms to one side:

x² + 3x + 1 – (x² – 4x + 4) = 0
= x² + 3x + 1 – x² + 4x – 4
= 7x – 3 = 0

Step 3: The final equation is:

7x – 3 = 0

This is not a quadratic equation because the highest power of x is 1, not 2.

Answer: No, it is not a quadratic equation.

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(vii) (x + 2)³ = 2x(x² – 1)

Solution:

Step 1: Expand both sides

Left side:
(x + 2)³ = (x + 2)(x + 2)(x + 2)
First, (x + 2)(x + 2) = x² + 4x + 4
Now multiply by (x + 2):
(x² + 4x + 4)(x + 2)
= x³ + 2x² + 4x² + 8x + 4x + 8
= x³ + 6x² + 12x + 8

Right side:
2x(x² – 1) = 2x³ – 2x

Step 2: Bring all terms to one side

x³ + 6x² + 12x + 8 – (2x³ – 2x) = 0
= x³ + 6x² + 12x + 8 – 2x³ + 2x
= –x³ + 6x² + 14x + 8 = 0

Step 3: Final equation is:

–x³ + 6x² + 14x + 8 = 0

This is not a quadratic equation because the highest power of x is 3.

Answer: No, it is not a quadratic equation.

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(viii) x³ – 4x² – x + 1 = (x – 2)³

Solution:

Step 1: Expand the right side

(x – 2)³ = (x – 2)(x – 2)(x – 2)
First, (x – 2)(x – 2) = x² – 4x + 4
Now multiply by (x – 2):
(x² – 4x + 4)(x – 2)
= x³ – 2x² – 4x² + 8x + 4x – 8
= x³ – 6x² + 12x – 8

Step 2: Bring all terms to one side

x³ – 4x² – x + 1 – (x³ – 6x² + 12x – 8) = 0
= x³ – 4x² – x + 1 – x³ + 6x² – 12x + 8
= 0x³ + 2x² – 13x + 9 = 0

Step 3: Final equation is:

2x² – 13x + 9 = 0

This is a quadratic equation because the highest power of x is 2.

Answer: Yes, it is a quadratic equation.

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Question 2:

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.

Solution:

Let the breadth of the plot be x metres.
Then the length will be 2x + 1 metres (since it is one more than twice the breadth).

Area of rectangle = length × breadth

So,
x(2x + 1) = 528

Multiply:
2x² + x = 528

Bring all terms to one side:
2x² + x – 528 = 0

Answer:
The required quadratic equation is:
2x² + x – 528 = 0, where x is the breadth (in metres) of the plot.

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Q2

(ii) The product of two consecutive positive integers is 306.
We need to find the integers.

Solution:

Let the smaller integer be x.
Then the next consecutive integer is x + 1.

Their product is:
x(x + 1) = 306

Multiply:
x² + x = 306

Bring all terms to one side:
x² + x – 306 = 0

Answer:
The required quadratic equation is:
x² + x – 306 = 0, where x is the smaller integer.

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Q2:

(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

Solution:

Let Rohan’s present age be x years.
Then his mother’s present age is x + 26 years.

After 3 years:

Rohan’s age = x + 3
Mother’s age = x + 26 + 3 = x + 29

Their product = 360
(x + 3)(x + 29) = 360

Multiply:
x² + 29x + 3x + 87 = 360
x² + 32x + 87 = 360

Bring all terms to one side:
x² + 32x + 87 – 360 = 0
x² + 32x – 273 = 0

Answer:
The required quadratic equation is:
x² + 32x – 273 = 0, where x (in years) is the present age of Rohan.

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Q2:

(iv) 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.
We need to find the speed of the train.

Solution:

Let the speed of the train be u km/h.
Time taken at speed u = distance/speed = 480/u hours.

If speed is reduced by 8 km/h, speed = (u – 8) km/h.
Time taken then = 480/(u – 8) hours.

According to the problem, the slower speed takes 3 hours more:
480/(u – 8) – 480/u = 3

Multiply both sides by u(u – 8) to clear denominators:
480u – 480(u – 8) = 3u(u – 8)

Simplify left side:
480u – 480u + 3840 = 3u² – 24u

Simplify:
3840 = 3u² – 24u

Bring all terms to one side:
3u² – 24u – 3840 = 0

Divide entire equation by 3:
u² – 8u – 1280 = 0

Answer:
The required quadratic equation is:
u² – 8u – 1280 = 0, where u (in km/h) is the speed of the train.

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Exercise 4.2

Question 1

(i) x² – 3x – 10 = 0

Solution:

Step 1: Find two numbers whose product is –10 and sum is –3.
These numbers are –5 and 2 because:
–5 × 2 = –10
–5 + 2 = –3

Step 2: Write the equation as:
x² – 5x + 2x – 10 = 0

Step 3: Factor by grouping:
x(x – 5) + 2(x – 5) = 0

Step 4: Take common factor (x – 5):
(x – 5)(x + 2) = 0

Step 5: Set each factor equal to zero:
x – 5 = 0 → x = 5
x + 2 = 0 → x = –2

Answer:
The roots are –2 and 5.

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Q1:

(ii) 2x² + x – 6 = 0

Solution:

Step 1: Multiply the coefficient of x² (which is 2) and the constant term (which is –6):
2 × (–6) = –12

Step 2: Find two numbers whose product is –12 and sum is the coefficient of x (which is 1).
These numbers are 4 and –3 because:
4 × (–3) = –12
4 + (–3) = 1

Step 3: Rewrite the middle term using these numbers:
2x² + 4x – 3x – 6 = 0

Step 4: Factor by grouping:
2x(x + 2) – 3(x + 2) = 0

Step 5: Take common factor (x + 2):
(2x – 3)(x + 2) = 0

Step 6: Set each factor equal to zero:
2x – 3 = 0 → x = 3/2
x + 2 = 0 → x = –2

Answer:
The roots are 3/2 and –2.

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(iii) √2 x² + 7x + 5√2 = 0

Solution:

We will solve this by the factorisation method.

Step 1: Multiply the coefficient of x² and the constant term:
√2 × 5√2 = 10

Step 2: Find two numbers whose product is 10 and whose sum is 7 (the middle term's coefficient).
These numbers are 5 and 2, because:
5 × 2 = 10 and 5 + 2 = 7

Step 3: Rewrite the equation:
√2 · x² + 5x + 2x + 5√2 = 0

Step 4: Group the terms:
(√2 · x² + 5x) + (2x + 5√2) = 0

Factor:
x(√2x + 5) + √2(√2x + 5) = 0

Take common factor:
(x + √2)(√2x + 5) = 0

Step 5: Solve each factor:

1. x + √2 = 0 → x = –√2

2. √2x + 5 = 0 → √2x = –5 → x = –5 / √2

Answer:
The roots are –√2 and –5 / √2

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(iv) 2x² – x + 1/8 = 0

Solution:

Step 1: Eliminate the fraction by multiplying the entire equation by 8 (LCM of denominators):
8 × (2x² – x + 1/8) = 0
→ 16x² – 8x + 1 = 0

Step 2: Factor the quadratic:
We need two numbers whose product is 16 × 1 = 16 and whose sum is –8.
These numbers are –4 and –4 (since –4 × –4 = 16 and –4 + –4 = –8)

So we rewrite the middle term:
16x² – 4x – 4x + 1 = 0
→ 4x(4x – 1) –1(4x – 1) = 0
→ (4x – 1)(4x – 1) = 0
→ (4x – 1)² = 0

Step 3: Solve the factor:
4x – 1 = 0 → x = 1/4

Answer:
The root is x = 1/4 (repeated root).

So, the roots are: 1/4, 1/4

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(v) 100x² – 20x + 1 = 0

Solution:

Step 1: Factor the quadratic.
We need two numbers whose product is:
100 × 1 = 100
and whose sum is –20.

These numbers are –10 and –10, since:
–10 × –10 = 100
–10 + –10 = –20

Step 2: Rewrite the middle term:
100x² – 10x – 10x + 1 = 0

Group and factor:
10x(10x – 1) –1(10x – 1) = 0
(10x – 1)(10x – 1) = 0
→ (10x – 1)² = 0

Step 3: Solve:
10x – 1 = 0 → x = 1/10

Answer:
The roots are 1/10, 1/10

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Question 2

2. Solve the problems given in Example 1 (refer book).

(i) John and Jivanti together have 45 marbles. Both lost 5 marbles each, and the product of the number of marbles they now have is 124. We need to find how many marbles they had originally.

Solution:

Let John have x marbles.
Then Jivanti has (45 – x) marbles, because their total is 45.

After losing 5 marbles each:

• John has (x – 5) marbles

• Jivanti has (45 – x – 5) = (40 – x) marbles

According to the problem:
(x – 5)(40 – x) = 124

Step 1: Expand the expression:
x × (40 – x) – 5 × (40 – x) = 124
→ 40x – x² – 200 + 5x = 124
→ 45x – x² – 200 = 124

Step 2: Bring all terms to one side:
–x² + 45x – 324 = 0

Multiply through by –1 to simplify:
x² – 45x + 324 = 0

Step 3: Factor the quadratic:
Find two numbers whose product is 324 and sum is –45.
Those numbers are –9 and –36.

So:
x² – 9x – 36x + 324 = 0
x(x – 9) – 36(x – 9) = 0
(x – 9)(x – 36) = 0

Step 4: Solve:
x = 9 or x = 36

Answer:
John originally had 9 marbles, and Jivanti had 36 marbles.
(or vice versa)

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Q2.

(ii) A cottage industry produces a certain number of toys in a day.
The cost of production of each toy (in rupees) is 55 minus the number of toys produced. The total production cost for one day is ₹750. We need to find the number of toys produced that day.

Solution:

Let the number of toys produced be x.

Then, the cost of production per toy = (55 – x)

So, total cost = number of toys × cost per toy
= x × (55 – x)

According to the question:
x(55 – x) = 750

Step 1: Expand the equation:
55x – x² = 750

Step 2: Bring all terms to one side:
–x² + 55x – 750 = 0

Multiply through by –1 to simplify:
x² – 55x + 750 = 0

Step 3: Factor the quadratic:
We need two numbers whose product is 750 and sum is 55.
Those numbers are 25 and 30.

So:
x² – 25x – 30x + 750 = 0
x(x – 25) –30(x – 25) = 0
(x – 25)(x – 30) = 0

Step 4: Solve:
x = 25 or x = 30

Answer:
The number of toys produced was 25 or 30.

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Q3:
Find two numbers whose sum is 27 and product is 182.

Solution:

Let the two numbers be x and (27 – x)
(because their sum is 27)

Now, according to the question:
x(27 – x) = 182

Step 1: Expand the equation:
27x – x² = 182

Step 2: Rearrange all terms:
–x² + 27x – 182 = 0

Multiply through by –1 to simplify:
x² – 27x + 182 = 0

Step 3: Factor the quadratic:
We need two numbers whose product is 182 and sum is 27.
These numbers are 13 and 14, because:
13 × 14 = 182 and 13 + 14 = 27

So:
x² – 13x – 14x + 182 = 0
x(x – 13) –14(x – 13) = 0
(x – 13)(x – 14) = 0

Step 4: Solve:
x = 13 or x = 14

Answer:
The two numbers are 13 and 14.

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Q4:

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

Solution:

Let the smaller integer be x.
Then the next consecutive integer is x + 1.

According to the question:
x² + (x + 1)² = 365

Step 1: Expand the terms:
x² + (x² + 2x + 1) = 365
→ x² + x² + 2x + 1 = 365
→ 2x² + 2x + 1 = 365

Step 2: Bring all terms to one side:
2x² + 2x + 1 – 365 = 0
→ 2x² + 2x – 364 = 0

Step 3: Simplify the equation by dividing all terms by 2:
x² + x – 182 = 0

Step 4: Factor the quadratic:
We need two numbers whose product is –182 and sum is 1.
These numbers are 14 and –13, because:
14 × (–13) = –182
14 + (–13) = 1

So:
x² + 14x – 13x – 182 = 0
x(x + 14) –13(x + 14) = 0
(x + 14)(x – 13) = 0

Step 5: Solve:
x = –14 or x = 13

Since we want positive integers, we take x = 13

Answer:
The two consecutive positive integers are 13 and 14.

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Q5:

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Solution:

Let the base be x cm
Then the altitude (height) is x – 7 cm

Using the Pythagoras Theorem:
(base)² + (altitude)² = (hypotenuse)²
x² + (x – 7)² = 13²

Step 1: Expand (x – 7)²
x² + (x² – 14x + 49) = 169
→ x² + x² – 14x + 49 = 169
→ 2x² – 14x + 49 = 169

Step 2: Bring all terms to one side:
2x² – 14x + 49 – 169 = 0
→ 2x² – 14x – 120 = 0

Step 3: Simplify the equation by dividing all terms by 2:
x² – 7x – 60 = 0

Step 4: Factor the quadratic:
We need two numbers whose product is –60 and sum is –7
Those numbers are –12 and +5

So:
x² – 12x + 5x – 60 = 0
x(x – 12) + 5(x – 12) = 0
(x – 12)(x + 5) = 0

Step 5: Solve:
x = 12 or x = –5

Since the base must be positive, we take x = 12

Then altitude = x – 7 = 12 – 7 = 5

Answer:
The base is 12 cm, and the altitude is 5 cm

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Q6:

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ` 90, find the number of articles produced and the cost of each article.

Solution:

Let the number of articles produced be x.

According to the question,
Cost of production of each article = 2x + 3 (in ₹)
Total cost of production = Number of articles × Cost per article
So, total cost = x × (2x + 3)

Given that the total cost is ₹90:
x × (2x + 3) = 90

Now, solve the equation:
2x² + 3x = 90

Bring all terms to one side:
2x² + 3x - 90 = 0

Now solve this quadratic equation:

Find factors of the equation:
2x² + 15x - 12x - 90 = 0
(2x² - 12x) + (15x - 90) = 0
2x(x - 6) + 15(x - 6) = 0
(2x + 15)(x - 6) = 0

So, x = -15/2 or x = 6
We reject the negative value because number of articles can't be negative.

So, number of articles = 6
Cost of each article = 2x + 3 = 2(6) + 3 = ₹15

Final Answer:
Number of articles = 6
Cost of each article = ₹15

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EXERCISE 4.3

1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

(i) 2x² – 3x + 5 = 0

Solution:

We are given the quadratic equation: 2x² – 3x + 5 = 0

To find the nature of the roots, we use the discriminant formula:

Discriminant (D) = b² – 4ac

Here,
a = 2,
b = -3,
c = 5

Now calculate D:
D = (-3)² – 4(2)(5)
D = 9 – 40
D = -31

Since the discriminant is negative, the equation has no real roots.

Final Answer:
Real roots do not exist

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(ii) 3x² – 4√3 x + 4 = 0

Solution:

Given quadratic equation:
3x² – 4√3 x + 4 = 0

We need to verify the roots:
2/√3 , 2/√3

Let’s solve the equation using the quadratic formula:
x = (-b ± √(b² - 4ac)) / 2a

Here:
a = 3,
b = –4√3,
c = 4

Step 1: Find the discriminant
= (–4√3)² – 4 × 3 × 4
= 48 – 48
= 0

Since discriminant = 0, the roots are equal.

Step 2: Use the formula
x = –(–4√3) / (2 × 3)
= 4√3 / 6
= 2√3 / 3
Now rationalize:

Multiply numerator and denominator by √3:
= (2√3 / 3) × (√3 / √3)
= 2 × 3 / (3√3)
= 6 / (3√3)
= 2 / √3

So, the roots are:
2 / √3 , 2 / √3

Answer:
2 / √3 , 2 / √3

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(iii) 2x² – 6x + 3 = 0

Solution

Given quadratic equation:
2x² – 6x + 3 = 0

We will use the quadratic formula:
x = (–b ± √(b² – 4ac)) / 2a

Here:
a = 2,
b = –6,
c = 3

Step 1: Find the discriminant
= (–6)² – 4 × 2 × 3
= 36 – 24
= 12

Since the discriminant is not zero, the roots are real and distinct.

Step 2: Apply the formula
x = –(–6) ± √12 / (2 × 2)
= (6 ± √12) / 4

Now simplify √12 = √(4×3) = 2√3
= (6 ± 2√3) / 4
= (2 × (3 ± √3)) / 4
= (3 ± √3) / 2

Answer:
(3 ± √3) / 2

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

(i) 2x² + kx + 3 = 0

Solution

Step 1: Use the discriminant condition

For equal roots, the discriminant D = 0
Discriminant D = b² – 4ac

Here:
a = 2,
b = k,
c = 3

So,
k² – 4 × 2 × 3 = 0
k² – 24 = 0
k² = 24
k = ±√24
k = ±2√6

Answer:
k = ±2√6

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Q2.

(ii) kx(x – 2) + 6 = 0

Solution:

Step 1: Expand the equation

kx² – 2kx + 6 = 0

This is a quadratic equation:
kx² – 2kx + 6 = 0

To have equal roots, discriminant must be zero.

Step 2: Use discriminant

Discriminant D = b² – 4ac

Here:
a = k,
b = –2k,
c = 6

D = (–2k)² – 4 × k × 6
= 4k² – 24k

Set D = 0 for equal roots:
4k² – 24k = 0
Divide by 4:
k² – 6k = 0
k(k – 6) = 0

So,
k = 0 or k = 6

But if k = 0, the equation becomes:
0 + 6 = 0 → Not a quadratic equation

So, only valid value:
k = 6

Answer:
k = 6

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3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2 ? If so, find its length and breadth.

Solution

Let the breadth be x meters.
Then the length is 2x meters (since length is twice the breadth).

Step 1: Use the area formula

Area of rectangle = length × breadth
So,
x × 2x = 800
⇒ 2x² = 800

Step 2: Solve for x

Divide both sides by 2:
x² = 400
x = √400
x = 20

Step 3: Find length and breadth

Breadth = 20 m
Length = 2 × 20 = 40 m

Answer: Yes.
Length = 40 m, Breadth = 20 m

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Question 4

Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Solution

Let the present ages of the two friends be x and 20 – x (since their sum is 20 years).

Step 1: Use the second condition

Four years ago, their ages were:
x – 4 and 20 – x – 4 = 16 – x

Their product was:
(x – 4)(16 – x) = 48

Step 2: Expand and simplify

(x – 4)(16 – x) = 48
= 16x – x² – 64 + 4x = 48
⇒ –x² + 20x – 64 = 48
⇒ –x² + 20x – 112 = 0
Multiply by –1:
x² – 20x + 112 = 0

Step 3: Check discriminant

D = b² – 4ac = (–20)² – 4 × 1 × 112
= 400 – 448
= –48

Since the discriminant is negative, the equation has no real solutions.

Answer:
No, this situation is not possible.

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5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2 ? If so, find its length and breadth.

Solution

Given:

• Perimeter = 80 m

• Area = 400 m²

Let length = l, breadth = b

Step 1: Use perimeter formula

2(l + b) = 80
⇒ l + b = 40 → (Equation 1)

Step 2: Use area formula

l × b = 400 → (Equation 2)

Step 3: From equation 1:

l = 40 – b

Substitute in equation 2:
(40 – b) × b = 400
⇒ 40b – b² = 400
⇒ b² – 40b + 400 = 0

Step 4: Solve the quadratic

b² – 40b + 400 = 0

Discriminant:
D = 1600 – 1600 = 0 → Equal roots

So,
b = 40 / 2 = 20
l = 40 – 20 = 20

Final Answer:

Yes. Length = 20 m, Breadth = 20 m