Quadratic Equations Class 10 Mathematics — Competency Based Questions

20 Questions | 70 Marks

A rectangular park is to be designed such that its length is 8 metres more than its breadth. If the area of the park is 240 m², form a quadratic equation to find the dimensions of the park and solve it. 5marks

Two water taps connected to a tank, can together fill the tank in 9 hours and 20 minutes. If the tap of the larger diameter takes 3 hours less than the tap of the smaller diameter to fill the tank separately, find the time in which each tap can fill the tank separately. 5marks

The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages was 130. Find their present ages. 5marks

A train travels a distance of 400 km at a uniform speed. If the speed had been 10 km/h less, it would have taken 2 hours more to cover the same distance. Find the original speed of the train. 5marks

Assertion: The roots of the quadratic equation

x^{2} - 5 x + 6 = 0

are

x = 2

and

x = 3

.Reason: A quadratic equation

a x^{2} + b x + c = 0

can be solved by factorizing the polynomial into two linear factors

(p x + q) (r x + s) = 0

and setting each factor to zero.(A) Both assertion and reason are true and reason is the correct explanation of assertion.(B) Both assertion and reason are true but reason is not the correct explanation of assertion.(C) Assertion is true but reason is false.(D) Assertion is false but reason is true.1mark

Consider the quadratic equation

2 x^{2} - 7 x + 3 = 0

.(a) Find the discriminant of the equation.(b) Comment on the nature of the roots.(c) If real roots exist, find them. 3marks

A certain number of students planned a picnic. The total bill for the food arrangement was 2400. Two students failed to attend, and the cost for each student who attended increased by 40. How many students attended the picnic? 5marks

A rectangular garden is 10 m by 16 m. It is surrounded by a uniform path. If the area of the path is 120 m², find the width of the path. 5marks

Consider the quadratic equation

k x^{2} + 4 x + 1 = 0

.(a) For what value of

k

will the equation have equal roots?(b) For what values of

k

will the equation have distinct real roots?(c) For what values of

k

will the equation have no real roots? 3marks

Assertion: The quadratic equation

x^{2} + 1 = 0

has no real roots.Reason: The discriminant of a quadratic equation

a x^{2} + b x + c = 0

D = b^{2} - 4 a c

. If

D < 0

, the equation has no real roots.(A) Both assertion and reason are true and reason is the correct explanation of assertion.(B) Both assertion and reason are true but reason is not the correct explanation of assertion.(C) Assertion is true but reason is false.(D) Assertion is false but reason is true.1mark

The diagonal of a rectangular field is 26 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, find the sides of the field. 5marks

A craftsman manufactured a batch of articles in a day. The cost of production of each article (in rupees) was 3 more than twice the number of articles produced in that day. If the total cost of production was `270, find the number of articles produced and the cost of each article. 5marks

Find the value of k for which the quadratic equation

(k + 1) x^{2} - 2 (k - 1) x + 1 = 0

has equal roots. 3marks

Two numbers are such that the sum of the numbers is 27 and their product is 182. Find the numbers. 3marks

Assertion: The equation

(x - 2)^{2} + 1 = 2 x - 3

is a quadratic equation.Reason: An equation of the form

a x^{2} + b x + c = 0

where

a \neq 0

is a quadratic equation.(A) Both assertion and reason are true and reason is the correct explanation of assertion.(B) Both assertion and reason are true but reason is not the correct explanation of assertion.(C) Assertion is true but reason is false.(D) Assertion is false but reason is true.1mark

A boat travels upstream a distance of 32 km in 4 hours. If the speed of the boat in still water is 12 km/h, find the speed of the stream. 5marks

The sum of squares of two consecutive positive integers is 365. Find the integers. 3marks

A rectangular sheet has an area of 600 m². If the length is 10 m more than its breadth, find the length and breadth of the sheet. 3marks

Consider the quadratic equation

4 x^{2} - 12 x + 9 = 0

.(a) Calculate the discriminant.(b) Determine the nature of the roots.(c) Find the roots if they are real. 3marks

A page is to be designed such that it has a print area of 100 cm² and a margin of 2 cm on each side of the page. If the total width of the page is 10 cm more than the total length, find the dimensions of the printed area. 5marks

Answers

Let the breadth of the rectangular park be

x

metres. Then, the length is

(x + 8)

metres.Area = Length × Breadth

240 = (x + 8) \times x

240 = x^{2} + 8 x

x^{2} + 8 x - 240 = 0

Using the quadratic formula $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ :

x = \frac{- 8 \pm \sqrt{8^{2} - 4 (1) (- 240)}}{2 (1)}

x = \frac{- 8 \pm \sqrt{64 + 960}}{2}

x = \frac{- 8 \pm \sqrt{1024}}{2}

x = \frac{- 8 \pm 32}{2}

Since breadth must be positive, $x = \frac{- 8 + 32}{2} = \frac{24}{2} = 12$ metres.
Length = $x + 8 = 12 + 8 = 20$ metres.
The dimensions of the park are 12 m (breadth) and 20 m (length). [5 marks]

Let the time taken by the tap of smaller diameter to fill the tank be

x

hours.Then, the time taken by the tap of larger diameter is

(x - 3)

hours.In 1 hour, the smaller tap fills

\frac{1}{x}

part of the tank.In 1 hour, the larger tap fills

\frac{1}{x - 3}

part of the tank.Combined time = 9 hours 20 minutes =

9 + \frac{20}{60} = 9 + \frac{1}{3} = \frac{28}{3}

hours.In 1 hour, both taps together fill

\frac{1}{28 / 3} = \frac{3}{28}

part of the tank.So,

\frac{1}{x} + \frac{1}{x - 3} = \frac{3}{28}

\frac{(x - 3) + x}{x (x - 3)} = \frac{3}{28}

\frac{2 x - 3}{x^{2} - 3 x} = \frac{3}{28}

28 (2 x - 3) = 3 (x^{2} - 3 x)

56 x - 84 = 3 x^{2} - 9 x

3 x^{2} - 9 x - 56 x + 84 = 0

3 x^{2} - 65 x + 84 = 0

Using the quadratic formula:

x = \frac{- (- 65) \pm \sqrt{(- 65)^{2} - 4 (3) (84)}}{2 (3)}

x = \frac{65 \pm \sqrt{4225 - 1008}}{6}

x = \frac{65 \pm \sqrt{3217}}{6}

Since $\sqrt{3217}$ is not a perfect square and is not easily calculable without a calculator, let's recheck the problem. Assuming the problem intended for simpler numbers, or there's a mistake in the problem statement provided. However, if we proceed with the assumption that $\sqrt{3217}$ needs to be approximated, it's not standard for this level without a calculator.

Let's re-examine the problem for factorisation. We need two numbers that multiply to $3 \times 84 = 252$ and add up to $- 65$ . Factors of 252: (1, 252), (2, 126), (3, 84), (4, 63). Sum of 4 and 63 is 67. Let's try other factors of 252: (6, 42), (7, 36), (9, 28), (12, 21), (14, 18). None add up to 65.

Let's assume there's a typo and the problem meant something that results in integer or simple fractional roots.
If we assume the problem statement is correct as given, and requires using the formula:
The discriminant is $3217$ .
$x = \frac{65 \pm \sqrt{3217}}{6}$ .
Since $x$ is time, it must be positive.
Let's assume a different standard problem statement for a similar scenario where the numbers work out. For instance, if the combined time was 10 hours, then $\frac{1}{x} + \frac{1}{x - 3} = \frac{1}{10} ⟹ 10 (2 x - 3) = x^{2} - 3 x ⟹ 20 x - 30 = x^{2} - 3 x ⟹ x^{2} - 23 x + 30 = 0$ . This doesn't factor nicely either.

Let's try a very common type of problem: if the tank is filled in 6 hours, and the faster tap takes 3 hours less.

\frac{1}{x} + \frac{1}{x - 3} = \frac{1}{6}

6 (2 x - 3) = x^{2} - 3 x

12 x - 18 = x^{2} - 3 x

$x^{2} - 15 x + 18 = 0$ . Discriminant $15^{2} - 4 (18) = 225 - 72 = 153$ . Not perfect.

Let's go back to the original numbers and re-solve:
$3 x^{2} - 65 x + 84 = 0$ .
Factors of $3 \times 84 = 252$ . We are looking for two numbers that multiply to 252 and add to -65.
The pair is -4 and -61 (sum -65, product 244). Not 252.
The pair is -3 and -62 (sum -65, product 186). Not 252.
The pair is -2 and -63 (sum -65, product 126). Not 252.
The pair is -1 and -64 (sum -65, product 64). Not 252.

Let's consider the numbers whose sum is 65:
(1, 64), (2, 63), (3, 62), (4, 61), (5, 60), (6, 59), (7, 58), (8, 57), (9, 56), (10, 55), (11, 54), (12, 53), (13, 52), (14, 51), (15, 50), (16, 49), (17, 48), (18, 47), (19, 46), (20, 45), (21, 44), (22, 43), (23, 42), (24, 41), (25, 40), (26, 39), (27, 38), (28, 37), (29, 36), (30, 35), (31, 34), (32, 33).
Now check products for 252:
$4 \times 63 = 252$ . Sum is $4 + 63 = 67$ . Close.
$6 \times 42 = 252$ . Sum is $6 + 42 = 48$ .
$7 \times 36 = 252$ . Sum is $7 + 36 = 43$ .
$9 \times 28 = 252$ . Sum is $9 + 28 = 37$ .
$12 \times 21 = 252$ . Sum is $12 + 21 = 33$ .
$14 \times 18 = 252$ . Sum is $14 + 18 = 32$ .

There seems to be an issue with the numbers in the question as standard CBSE questions of this type usually yield factorable quadratics or result in cleaner square roots for the discriminant.

Let's assume for the sake of providing an answer that the discriminant was a perfect square. If the discriminant was 3249 instead of 3217, then $\sqrt{3249} = 57$ .
Then $x = \frac{65 \pm 57}{6}$ .
$x = \frac{65 + 57}{6} = \frac{122}{6} = \frac{61}{3}$ hours.
Or $x = \frac{65 - 57}{6} = \frac{8}{6} = \frac{4}{3}$ hours.
If $x = \frac{4}{3}$ hours, then $x - 3 = \frac{4}{3} - 3 = \frac{4 - 9}{3} = - \frac{5}{3}$ , which is not possible as time cannot be negative.
If $x = \frac{61}{3}$ hours, then $x - 3 = \frac{61}{3} - 3 = \frac{61 - 9}{3} = \frac{52}{3}$ hours.
This would mean the smaller tap takes $\frac{61}{3}$ hours and the larger takes $\frac{52}{3}$ hours.
Check: $\frac{3}{61} + \frac{3}{52} = \frac{3 \times 52 + 3 \times 61}{61 \times 52} = \frac{156 + 183}{3172} = \frac{339}{3172}$ . This should equal $\frac{3}{28}$ .
$\frac{339}{3172} \approx \frac{340}{3170} \approx 0.107$ . And $\frac{3}{28} \approx 0.107$ . This is close but not exact.

Given the constraints, and assuming the question implies a solvable scenario within typical CBSE patterns: Let's try to reverse-engineer a problem where roots are integer or simple fractions.
Suppose the times were 12 hours and 9 hours. Combined time = $\frac{1}{12} + \frac{1}{9} = \frac{3 + 4}{36} = \frac{7}{36}$ . So tank fills in $\frac{36}{7}$ hours, i.e., 5 hours 8.57 mins.

Let's stick to the provided problem and formula.
$x = \frac{65 \pm \sqrt{3217}}{6}$ .
The exact answer, as derived from the equation, is $x = \frac{65 + \sqrt{3217}}{6}$ hours for the smaller tap and $x - 3 = \frac{65 + \sqrt{3217}}{6} - 3 = \frac{65 + \sqrt{3217} - 18}{6} = \frac{47 + \sqrt{3217}}{6}$ hours for the larger tap. (Note: The smaller root for x, $\frac{65 - \sqrt{3217}}{6}$ , yields a negative value for x-3, hence it's rejected).
As this is highly unusual for CBSE, it is possible there's a typo in the question. However, based strictly on the math:
Time for smaller tap = $\frac{65 + \sqrt{3217}}{6}$ hours.
Time for larger tap = $\frac{47 + \sqrt{3217}}{6}$ hours.
[5 marks - This answer is based on strict interpretation but would likely be considered flawed in an exam setting due to the non-standard numbers].

Let the present age of the father be

F

years and the present age of the son be

S

years.Given:

F + S = 45

(Equation 1)Five years ago, father's age was

F - 5

and son's age was

S - 5

.Given:

(F - 5) (S - 5) = 130

F S - 5 F - 5 S + 25 = 130

F S - 5 (F + S) + 25 = 130

Substitute $F + S = 45$ from Equation 1:

F S - 5 (45) + 25 = 130

F S - 225 + 25 = 130

F S - 200 = 130

$F S = 330$ (Equation 2)
From Equation 1, $F = 45 - S$ . Substitute this into Equation 2:

(45 - S) S = 330

45 S - S^{2} = 330

S^{2} - 45 S + 330 = 0

Using the quadratic formula $S = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ :

S = \frac{- (- 45) \pm \sqrt{(- 45)^{2} - 4 (1) (330)}}{2 (1)}

S = \frac{45 \pm \sqrt{2025 - 1320}}{2}

S = \frac{45 \pm \sqrt{705}}{2}

Again, $\sqrt{705}$ is not a perfect square. This suggests a potential issue with the numbers in the question for a typical exam. If we assume a typo and the product was 504, then $S^{2} - 45 S + 504 = 0$ . Factors of 504 are $18 \times 28 = 504$ and $18 + 28 = 46$ . Factors $21 \times 24 = 504$ and $21 + 24 = 45$ .
So, if the product was 504:

(S - 21) (S - 24) = 0

$S = 21$ or $S = 24$ .
If $S = 21$ , then $F = 45 - 21 = 24$ . (Son younger than father).
If $S = 24$ , then $F = 45 - 24 = 21$ . (Son older than father - impossible for present ages).
So, assuming the product was 504: Son is 21 years old, Father is 24 years old. (This is unusual, father younger than son, implies we should reject this case and take S=24, F=21. Hmm, it implies Son is 24, Father is 21. This contradicts the context of father/son ages).

Let's try another possibility for typo where the sum is 45 and product is 350.
$S^{2} - 45 S + 350 = 0$ .
Factors of 350: $(10, 35)$ . $10 + 35 = 45$ . So, $(S - 10) (S - 35) = 0$ .
$S = 10$ or $S = 35$ .
If $S = 10$ , then $F = 45 - 10 = 35$ . (Son 10, Father 35. Ages are reasonable).
Check: 5 years ago, Son was 5, Father was 30. Product $5 \times 30 = 150$ . Not 130.

Let's try a product of 126.
$S^{2} - 45 S + 126 = 0$ .
Factors of 126: $(3, 42)$ . $3 + 42 = 45$ . So, $(S - 3) (S - 42) = 0$ .
$S = 3$ or $S = 42$ .
If $S = 3$ , then $F = 45 - 3 = 42$ . (Son 3, Father 42. Reasonable).
Check: 5 years ago, Son was -2 (not possible). So, this case is invalid.

Let's go back to the original problem statement and accept the non-integer solution.
$S = \frac{45 \pm \sqrt{705}}{2}$ .
Since $S$ must be positive.
Let's calculate $\sqrt{705} \approx 26.55$ .
$S \approx \frac{45 \pm 26.55}{2}$ .
$S \approx \frac{45 + 26.55}{2} = \frac{71.55}{2} \approx 35.78$ years.
$F = 45 - S \approx 45 - 35.78 = 9.22$ years. (Son older than father. Not possible).

Let's try $S \approx \frac{45 - 26.55}{2} = \frac{18.45}{2} \approx 9.22$ years.
$F = 45 - S \approx 45 - 9.22 = 35.78$ years. (Son 9.22, Father 35.78. This is reasonable).
Check: 5 years ago, Son was $9.22 - 5 = 4.22$ . Father was $35.78 - 5 = 30.78$ .
Product $= 4.22 \times 30.78 \approx 129.7$ . This is very close to 130.
So, the exact ages are $S = \frac{45 - \sqrt{705}}{2}$ years and $F = \frac{45 + \sqrt{705}}{2}$ years.

[5 marks - again, an unusual answer for a typical exam, but mathematically correct based on the prompt].

Let the original speed of the train be

x

km/h.Distance = 400 km.Time taken at original speed =

\frac{400}{x}

hours.If speed is 10 km/h less, new speed =

(x - 10)

km/h.Time taken at new speed =

\frac{400}{x - 10}

hours.According to the problem, the new time is 2 hours more than the original time.

\frac{400}{x - 10} = \frac{400}{x} + 2

Divide the entire equation by 2:

\frac{200}{x - 10} = \frac{200}{x} + 1

\frac{200}{x - 10} - \frac{200}{x} = 1

\frac{200 x - 200 (x - 10)}{x (x - 10)} = 1

\frac{200 x - 200 x + 2000}{x^{2} - 10 x} = 1

2000 = x^{2} - 10 x

x^{2} - 10 x - 2000 = 0

Using the quadratic formula $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ :

x = \frac{- (- 10) \pm \sqrt{(- 10)^{2} - 4 (1) (- 2000)}}{2 (1)}

x = \frac{10 \pm \sqrt{100 + 8000}}{2}

x = \frac{10 \pm \sqrt{8100}}{2}

x = \frac{10 \pm 90}{2}

Since speed must be positive, we take the positive root:
$x = \frac{10 + 90}{2} = \frac{100}{2} = 50$ km/h.
The original speed of the train was 50 km/h. [5 marks]

(A) Both assertion and reason are true and reason is the correct explanation of assertion. 1mark

The given quadratic equation is

2 x^{2} - 7 x + 3 = 0

. Here

a = 2, b = - 7, c = 3

.(a) Discriminant

D = b^{2} - 4 a c = (- 7)^{2} - 4 (2) (3) = 49 - 24 = 25

.(b) Since

D = 25 > 0

, the equation has two distinct real roots.(c) The roots are given by

x = \frac{- b \pm \sqrt{D}}{2 a} = \frac{- (- 7) \pm \sqrt{25}}{2 (2)} = \frac{7 \pm 5}{4}

.The roots are

x_{1} = \frac{7 + 5}{4} = \frac{12}{4} = 3

and

x_{2} = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2}

. 3marks

Let the original number of students who planned the picnic be

n

.The total bill was

2400.   The original cost per student = $\frac{2400}{n}$.   Two students failed to attend, so the number of students who attended = $n-2$.   The cost for each student who attended increased by

40, so the new cost per student =

\frac{2400}{n} + 40

.The total bill is still

2400 for the students who attended.   So, $(n-2) \times (\frac{2400}{n} + 40) = 2400$.   $(n-2) \times \frac{2400 + 40n}{n} = 2400$.   $(n-2)(2400 + 40n) = 2400n$.   $2400n + 40n^2 - 4800 - 80n = 2400n$.   $40n^2 - 80n - 4800 = 0$.   Divide by 40:   $n^2 - 2n - 120 = 0$.   Factorize the quadratic equation. We need two numbers that multiply to -120 and add to -2. These are -12 and 10.   $(n-12)(n+10) = 0$.   So, $n=12$ or $n=-10$.   Since the number of students cannot be negative, $n=12$.   The original number of students who planned was 12.   The number of students who attended = $n-2 = 12-2 = 10$.   The cost of each student who attended = $\frac{2400}{10} =

240

. O r i g i n a l c o s t p e r s t u d e n t =

\frac{2400}{12} = 200$. The increase is $240 - 200 = 40, which matches the condition.Therefore, 10 students attended the picnic. 5marks

Let the width of the uniform path be

w

metres.The dimensions of the rectangular garden are 16 m (length) and 10 m (breadth).Area of the garden =

16 \times 10 = 160

m².When the path is added, the new dimensions become:New length =

16 + 2 w

New breadth =

10 + 2 w

Area of the garden including the path =

(16 + 2 w) (10 + 2 w)

.Area of the path = (Area of garden including path) - (Area of garden).

120 = (16 + 2 w) (10 + 2 w) - 160

120 = 160 + 32 w + 20 w + 4 w^{2} - 160

120 = 52 w + 4 w^{2}

4 w^{2} + 52 w - 120 = 0

.Divide by 4:

w^{2} + 13 w - 30 = 0

.Factorize the quadratic equation. We need two numbers that multiply to -30 and add to 13. These are 15 and -2.

(w + 15) (w - 2) = 0

.So,

w = - 15

w = 2

.Since the width of the path cannot be negative,

w = 2

metres.The width of the path is 2 metres. 5marks

The given quadratic equation is

k x^{2} + 4 x + 1 = 0

. Here

a = k, b = 4, c = 1

.The discriminant is

D = b^{2} - 4 a c = 4^{2} - 4 (k) (1) = 16 - 4 k

.(a) For equal roots,

D = 0

16 - 4 k = 0 ⟹ 4 k = 16 ⟹ k = 4

.(b) For distinct real roots,

D > 0

16 - 4 k > 0 ⟹ 16 > 4 k ⟹ k < 4

.(c) For no real roots,

D < 0

16 - 4 k < 0 ⟹ 16 < 4 k ⟹ k > 4

.Note: For the equation to be quadratic,

k \neq 0

. So, for distinct real roots,

k < 4

and

k \neq 0

. For no real roots,

k > 4

Summary for this question:
(a) $k = 4$
(b) $k < 4$ (and $k \neq 0$ )
(c) $k > 4$
[3 marks]

(A) Both assertion and reason are true and reason is the correct explanation of assertion. 1mark

Let the shorter side of the rectangular field be

x

metres.The diagonal is

(x + 26)

metres.The longer side is

(x + 14)

metres.According to the Pythagorean theorem, (shorter side)² + (longer side)² = (diagonal)².

x^{2} + (x + 14)^{2} = (x + 26)^{2}

x^{2} + (x^{2} + 28 x + 196) = (x^{2} + 52 x + 676)

2 x^{2} + 28 x + 196 = x^{2} + 52 x + 676

2 x^{2} - x^{2} + 28 x - 52 x + 196 - 676 = 0

x^{2} - 24 x - 480 = 0

.Using the quadratic formula

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

x = \frac{- (- 24) \pm \sqrt{(- 24)^{2} - 4 (1) (- 480)}}{2 (1)}

x = \frac{24 \pm \sqrt{576 + 1920}}{2}

x = \frac{24 \pm \sqrt{2496}}{2}

\sqrt{2496}

is not a perfect square. Let's check for factors.

2496 = 16 \times 156 = 16 \times 4 \times 39 = 64 \times 39

.So,

\sqrt{2496} = \sqrt{64 \times 39} = 8 \sqrt{39}

x = \frac{24 \pm 8 \sqrt{39}}{2} = 12 \pm 4 \sqrt{39}

.Since the side length must be positive, we take the positive value:

x = 12 + 4 \sqrt{39}

metres.Shorter side =

12 + 4 \sqrt{39}

m.Longer side =

x + 14 = (12 + 4 \sqrt{39}) + 14 = 26 + 4 \sqrt{39}

m.This problem, like others above, yields a solution with a square root, which is unusual for standard CBSE questions unless it's specifically designed for application of the formula.

Let's assume there's a typo and the problem resulted in a perfect square. For example, if the numbers led to $x^{2} - 24 x - 480 = 0$ and the discriminant was $24^{2} - 4 (1) (- 480) = 576 + 1920 = 2496$ . If the discriminant was a perfect square, say 1024 ( $32^{2}$ ), then $x = \frac{24 \pm 32}{2}$ , giving $x = 28$ or $x = - 4$ . If $x = 28$ , shorter side = 28, longer side = 42, diagonal = 54. Check: $28^{2} + 42^{2} = 784 + 1764 = 2548$ . $54^{2} = 2916$ . Not matching.

Let's assume the problem meant to have integer sides. A common Pythagorean triple is 5, 12, 13. If the shorter side is 5, longer is 12, diagonal is 13. Longer side is 7 more than shorter. Diagonal is 8 more than shorter. This doesn't fit the $x + 14$ and $x + 26$ pattern.
Another common triple: 8, 15, 17. Shorter = 8, Longer = 15, Diagonal = 17. Longer is 7 more, diagonal is 9 more.
Another triple: 7, 24, 25. Shorter = 7, Longer = 24, Diagonal = 25. Longer is 17 more, diagonal is 18 more.
Another triple: 20, 21, 29. Shorter = 20, Longer = 21, Diagonal = 29. Longer is 1 more, diagonal is 9 more.

Let's go back to the equation $x^{2} - 24 x - 480 = 0$ and calculate the exact roots.
Shorter side $x = 12 + 4 \sqrt{39}$ m.
Longer side = $26 + 4 \sqrt{39}$ m.
[5 marks]

Let the number of articles produced on that day be

n

.The cost of production of each article was

(2 n + 3)

rupees.The total cost of production was `270.So,

n \times (2 n + 3) = 270

2 n^{2} + 3 n = 270

2 n^{2} + 3 n - 270 = 0

.Using the quadratic formula

n = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

n = \frac{- 3 \pm \sqrt{3^{2} - 4 (2) (- 270)}}{2 (2)}

n = \frac{- 3 \pm \sqrt{9 + 2160}}{4}

n = \frac{- 3 \pm \sqrt{2169}}{4}

.Let's check if 2169 is a perfect square.

40^{2} = 1600, 50^{2} = 2500

. Ends in 9, so unit digit could be 3 or 7.

43^{2} = (40 + 3)^{2} = 1600 + 240 + 9 = 1849

47^{2} = (50 - 3)^{2} = 2500 - 300 + 9 = 2209

. So, 2169 is not a perfect square.Let's check for factors of

2 \times 270 = 540

that add to 3.

18 \times 30 = 540

30 - 18 = 12

20 \times 27 = 540

27 - 20 = 7

15 \times 36 = 540

36 - 15 = 21

10 \times 54 = 540

54 - 10 = 44

24 \times 22.5 = 540

.Let's check other pairs whose sum is 3: (x, x+3). We need

x (x + 3)

to be negative of (product of first and last term).We need two numbers whose product is

2 \times (- 270) = - 540

and sum is +3.The numbers are 27 and -20. (

27 \times - 20 = - 540

27 + (- 20) = 7

. This is not 3).The numbers are 24 and -21. (

24 \times - 21 = - 504

. Not -540).The numbers are 30 and -27. (

30 \times - 27 = - 810

. Not -540).

Let's retry with the correct factors of -540 that sum to +3.
Factors of 540: (1, 540), (2, 270), (3, 180), (4, 135), (5, 108), (6, 90), (9, 60), (10, 54), (12, 45), (15, 36), (18, 30), (20, 27).
Difference of pairs: 539, 268, 177, 131, 103, 84, 51, 44, 33, 21, 12, 7.
None of these differences is 3. This implies that the quadratic equation $2 n^{2} + 3 n - 270 = 0$ does not have integer or simple fractional roots.

Let's use the discriminant result: $\sqrt{2169}$ . Is it possible that 2169 is divisible by a perfect square? $2169 / 9 = 241$ . So $\sqrt{2169} = \sqrt{9 \times 241} = 3 \sqrt{241}$ .
$n = \frac{- 3 \pm 3 \sqrt{241}}{4}$ .
Since $n$ must be positive, $n = \frac{- 3 + 3 \sqrt{241}}{4}$ .
This again results in a non-standard answer for a typical CBSE question.

Assuming a typo, let's consider what could lead to integer roots. If the total cost was 250, and articles $n$ , cost $2 n + 5$ . $n (2 n + 5) = 250 ⟹ 2 n^{2} + 5 n - 250 = 0$ . $D = 25 - 4 (2) (- 250) = 25 + 2000 = 2025 = 45^{2}$ .
$n = \frac{- 5 \pm 45}{4}$ . $n = \frac{40}{4} = 10$ or $n = \frac{- 50}{4}$ (reject).
If $n = 10$ , cost of each article = $2 (10) + 5 = 25$ . Total cost = $10 \times 25 = 250$ .
This implies the original problem statement had numbers that do not result in easily calculable roots.

Based on the prompt's numbers:
Number of articles $n = \frac{- 3 + 3 \sqrt{241}}{4}$ .
Cost of each article = $2 n + 3 = 2 (\frac{- 3 + 3 \sqrt{241}}{4}) + 3 = \frac{- 3 + 3 \sqrt{241}}{2} + 3 = \frac{- 3 + 3 \sqrt{241} + 6}{2} = \frac{3 + 3 \sqrt{241}}{2}$ .
[5 marks - non-standard answer]

The given quadratic equation is

(k + 1) x^{2} - 2 (k - 1) x + 1 = 0

.For equal roots, the discriminant

D = b^{2} - 4 a c = 0

.Here

a = k + 1

b = - 2 (k - 1)

c = 1

D = [- 2 (k - 1)]^{2} - 4 (k + 1) (1) = 0

4 (k - 1)^{2} - 4 (k + 1) = 0

.Divide by 4:

(k - 1)^{2} - (k + 1) = 0

(k^{2} - 2 k + 1) - k - 1 = 0

k^{2} - 3 k = 0

k (k - 3) = 0

.So,

k = 0

k = 3

.However, for the equation to be a quadratic equation, the coefficient of

x^{2}

cannot be zero, i.e.,

k + 1 \neq 0

, so

k \neq - 1

.Also, if

k = 0

, the equation becomes

x^{2} + 2 x + 1 = 0

, which has equal roots

(x + 1)^{2} = 0

.If

k = 3

, the equation becomes

4 x^{2} - 2 (2) x + 1 = 0 ⟹ 4 x^{2} - 4 x + 1 = 0

, which has equal roots

(2 x - 1)^{2} = 0

.Both values are valid.The values of k are 0 and 3. 3marks

Let the two numbers be

x

and

y

.Given:

x + y = 27

(Equation 1)Given:

x y = 182

(Equation 2)We can form a quadratic equation whose roots are

x

and

y

t^{2} - (sum of roots) t + (product of roots) = 0

t^{2} - 27 t + 182 = 0

.We need to find two numbers that multiply to 182 and add to -27.Factors of 182: (1, 182), (2, 91), (7, 26), (13, 14).

13 + 14 = 27

. So, the numbers are 13 and 14.

(t - 13) (t - 14) = 0

.The numbers are 13 and 14. 3marks

(A) Both assertion and reason are true and reason is the correct explanation of assertion. 1mark

Let the speed of the boat in still water be

v_{b} = 12

km/h.Let the speed of the stream be

v_{s}

km/h.Speed upstream =

v_{b} - v_{s} = 12 - v_{s}

.Speed downstream =

v_{b} + v_{s} = 12 + v_{s}

.Distance upstream = 32 km.Time taken upstream = 4 hours.Speed upstream = Distance / Time =

32 / 4 = 8

km/h.So,

12 - v_{s} = 8

v_{s} = 12 - 8 = 4

km/h.The speed of the stream is 4 km/h.The question asks to find the speed of the stream. We have already found it. The information about the boat traveling upstream at a certain speed is enough. The problem does not require setting up a quadratic equation. This question might be intended to check understanding of upstream/downstream speeds.

To ensure it's a quadratic problem, let's assume the question implies that the boat travels upstream and downstream and the total time or some other condition leads to a quadratic equation. However, as stated, it is not a quadratic problem. If the question was, for instance, "A boat travels 32 km upstream and returns to the starting point in a total of 6 hours. Find the speed of the stream."
Time upstream = $\frac{32}{12 - v_{s}}$
Time downstream = $\frac{32}{12 + v_{s}}$
Total time = $\frac{32}{12 - v_{s}} + \frac{32}{12 + v_{s}} = 6$ .
$32 (12 + v_{s}) + 32 (12 - v_{s}) = 6 (12 - v_{s}) (12 + v_{s})$ .
$384 + 32 v_{s} + 384 - 32 v_{s} = 6 (144 - v_{s}^{2})$ .
$768 = 6 (144 - v_{s}^{2})$ .
$128 = 144 - v_{s}^{2}$ .
$v_{s}^{2} = 144 - 128 = 16$ .
$v_{s} = \pm 4$ . Since speed is positive, $v_{s} = 4$ km/h.

Given the exact wording of Q16: "A boat travels upstream a distance of 32 km in 4 hours. If the speed of the boat in still water is 12 km/h, find the speed of the stream."
Speed upstream = Distance / Time = 32 km / 4 hours = 8 km/h.
Speed upstream = Speed of boat in still water - Speed of stream.
8 km/h = 12 km/h - Speed of stream.
Speed of stream = 12 km/h - 8 km/h = 4 km/h.
This is a direct calculation, not a quadratic equation problem. [5 marks - for calculation based on problem statement, even if not quadratic]

Let the two consecutive positive integers be

x

and

x + 1

.The sum of their squares is 365.

x^{2} + (x + 1)^{2} = 365

x^{2} + (x^{2} + 2 x + 1) = 365

2 x^{2} + 2 x + 1 = 365

2 x^{2} + 2 x - 364 = 0

.Divide by 2:

x^{2} + x - 182 = 0

.We need two numbers that multiply to -182 and add to +1. These are 14 and -13.

(x + 14) (x - 13) = 0

.So,

x = - 14

x = 13

.Since the integers are positive, we take

x = 13

.The first integer is 13.The second consecutive integer is

x + 1 = 13 + 1 = 14

.The two consecutive positive integers are 13 and 14. 3marks

Let the breadth of the rectangular sheet be

b

metres.The length is

b + 10

metres.The area is 600 m².Area = Length × Breadth.

600 = (b + 10) \times b

600 = b^{2} + 10 b

b^{2} + 10 b - 600 = 0

.We need two numbers that multiply to -600 and add to +10. These are 30 and -20.

(b + 30) (b - 20) = 0

.So,

b = - 30

b = 20

.Since breadth must be positive,

b = 20

metres.Length =

b + 10 = 20 + 10 = 30

metres.The length is 30 m and the breadth is 20 m. 3marks

The given quadratic equation is

4 x^{2} - 12 x + 9 = 0

.Here

a = 4, b = - 12, c = 9

.(a) Discriminant

D = b^{2} - 4 a c = (- 12)^{2} - 4 (4) (9) = 144 - 144 = 0

.(b) Since

D = 0

, the equation has two real and equal roots.(c) The roots are given by

x = \frac{- b \pm \sqrt{D}}{2 a} = \frac{- (- 12) \pm \sqrt{0}}{2 (4)} = \frac{12}{8} = \frac{3}{2}

.The roots are

\frac{3}{2}

and

\frac{3}{2}

. 3marks

Let the dimensions of the printed area be length

l

cm and breadth

b

cm.Area of printed area =

l \times b = 100

cm².The page has a margin of 2 cm on each side.Total length of the page =

l + 2 (margin on top/bottom) = l + 2 (2) = l + 4

cm.Total breadth of the page =

b + 2 (margin on sides) = b + 2 (2) = b + 4

cm.Given: The total width (breadth) of the page is 10 cm more than the total length of the page.

(b + 4) = (l + 4) + 10

b + 4 = l + 14

b = l + 10

.Now we have two equations:

$l b = 100$
$b = l + 10$
Substitute (2) into (1):
$l (l + 10) = 100$ .
$l^{2} + 10 l = 100$ .
$l^{2} + 10 l - 100 = 0$ .
Using the quadratic formula $l = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ :
$l = \frac{- 10 \pm \sqrt{10^{2} - 4 (1) (- 100)}}{2 (1)}$ .
$l = \frac{- 10 \pm \sqrt{100 + 400}}{2}$ .
$l = \frac{- 10 \pm \sqrt{500}}{2}$ .
$\sqrt{500} = \sqrt{100 \times 5} = 10 \sqrt{5}$ .
$l = \frac{- 10 \pm 10 \sqrt{5}}{2} = - 5 \pm 5 \sqrt{5}$ .
Since length must be positive, $l = - 5 + 5 \sqrt{5} = 5 (\sqrt{5} - 1)$ cm.
Now find $b$ :
$b = l + 10 = (- 5 + 5 \sqrt{5}) + 10 = 5 + 5 \sqrt{5} = 5 (\sqrt{5} + 1)$ cm.
The dimensions of the printed area are $5 (\sqrt{5} - 1)$ cm (length) and $5 (\sqrt{5} + 1)$ cm (breadth).
Check printed area: $l \times b = [5 (\sqrt{5} - 1)] \times [5 (\sqrt{5} + 1)] = 25 [(\sqrt{5})^{2} - 1^{2}] = 25 (5 - 1) = 25 \times 4 = 100$ cm². This matches.
[5 marks]

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Quadratic Equations — Competency Based Questions

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