Second Degree Equations

Given area of reduced square = 49 square metres

We know that area of a square = a², where a is the side of square.

⇒ √49 = 7 (Side of reduced square)

Given, the original square’s side was reduced by 2 metres.

⇒ 7 + 2 = 9 metres

∴ The length of a side of the original square is 9 metres.

Given the square ground has 2 metre wide path.

We know that area of a square = a², where a is the side of square.

⇒ Area of path = 2² = 4 square metres

Given the total area of the ground and path = 1225 square metres

So, the Area of the ground = Total area of ground and path – Area of path

⇒ Area of ground = 1225 – 4 = 1221 square metres

∴ The area of the ground alone = 1221 square metres

The square of a term is 2500.

⇒ Term = √2500 = 50

Given, arithmetic sequence 2, 5, 8 …

Here first term = 2 = a₁

Common difference = 5 – 2 = 3 = d

We know that the expression on nth term in an arithmetic sequence is t_n = a₁ + (n – 1) d where n is the number of term.

⇒ 50 = 2 + (n – 1) (3)

⇒ 50 = 2 + 3n – 3

⇒ 50 = 3n – 1

⇒ 50 + 1 = 3n

⇒ 51 = 3n

⇒ n = 51/ 3

⇒ n = 17

∴ The position of the term 50 is 17.

We know that when interest is compounded annually,

Amount = P (1 + )ⁿ

Where P = principal, R = rate of interest and n = time in years

Given Amount = 2200 rupees, P = 2000, R = R and n = 2 years

⇒ 2200 = 2000 (1 + )²

⇒ = (1 + )²

⇒ ()² = (1 + )²

⇒ = 1 +

⇒ – 1 =

⇒ =

We know that √11 ≈ 3.316 and √10 ≈ 3.162.

⇒ =

⇒ =

⇒ 0.04881 =

⇒ 100 × 0.04881 = R

⇒ R = 4.881%

∴ The rate of interest is 4.881%.

Let the number be 2x. Therefore, the consecutive even number will be (2x + 2).

Now the product of these two numbers will be 2×(2x + 2)

Therefore, 1 added to the product will be x(x + 2) + 1 which is equal to 289

⇒ 2x (2x + 2) + 1 = 289

⇒ 4x² + 4x + 1 = 289

⇒ (2x + 1)² = 17²

⇒ √((2x + 1)²) = √(17²)

⇒ (2x + 1) = 17

⇒ 2x = 17 – 1

⇒ x = 8

One number is 16 and the other number 18

Let the numbers be 6x and (6x + 6)

Product of the two numbers is

Now adding 9 to the product gives 729

Therefore we can write it as

therefore the number is 24 and 30

If height is 2 m less than base

Let Base be x m

Then;

Height is (x–2)

In half of the triangle

⇒ By Pythagoras theorem:-

Base² + Height² = Side²

As in isosceles triangle

Altitude and median are the same

∴ in half of triangle

Base =

Height = (x–2)

Side is same as base because it is isosceles triangle

Side = x

Base² + Height² = Side²

= x²

+ x² + 4–4x = x²

+ 4–4x = x²–x² = 0

= 0

x² + 16–16x = 0

As comparing eq to ax² + bx + c = 0

x =

x =

The length cannot be negative

Hence Base is

Length of rod = 2.6 m

Floor Base = 1 m

Height of wall = ?

Now using Pythagoras Theorem

(Floor Base)² + (Height of wall)² = (Length of rod)²

(Height of wall)² = (Length of rod)² – (Floor Base)²

⇒ Height of the wall = √((Length of rod)² – (Floor Base)² )

⇒ Height of the wall = √((2.6)² + 1²)

using the formula

⇒ Height of the wall = √((2.6 + 1) (2.6 – 1)

= √(3.6 × 1.6)

⇒ √5.76

= 2.4

now we got the height of the wall but in the question it is given that it is further slided down, therefore we consider that as x.

Length of rod remains same.

Floor base becomes (1 + x)

Height of wall becomes (2.4 – x)

Now using Pythagoras theorem again

⇒ (floor base)² + (height of wall)² = ( lengthof rod)²

⇒ (2.4 – x)² + (1 + x)² = 2.6²

⇒ 5.76 + x² – 4.8 + 1 + x² + 2x = 6.76

⇒ 6.76 + 2x² – 2.8x = 6.76

⇒ 2x² – 2.8x = 6.76 – 6.76

⇒ 2x² – 2.8x = 0

⇒ 2x(x – 1.4) = 0

⇒ (x – 1.4) = 0

∴ x = 1.4

Using the given information we conclude

16 + S_n = 256

Using the formula S_n

Using arithmetic sequence equation

⇒ n² + 4n – 140 = 0

⇒ n² + 14n – 10n – 140 = 0

⇒ n(n + 14) – 10(n + 14) = 0

⇒ (n + 14) (n – 10) = 0

x = –14 as we cannot have negative number of terms.

Therefore x = 10, the number of terms is 10

Given.

Distance = 100km

If average speed been increased by 10 kilometres per hour, I could have reached here one hour earlier

Formula used.

Average speed =

Let the sum of time taken be x

Average speed =

New average speed = average speed + 10

+ 10

10x = (x–1)(10 + x)

10x = 10x + x²–10–x

x²–x–10 = 10x–10x = 0

x²–x–10 = 0

As comparing eq to ax² + bx + c = 0

x =

x = = =

As √41 is greater than 1

∴ time cannot get negative

Hence; time is hours

Average speed =

Let the number of kids be x.

Let the number of sweets be y.

Total number of sweets = 30

…eqn 1

…eqn 2

using eqn 1

⇒ x² – x – 30 = 0

⇒ x² – 6x + 5x – 30 = 0

⇒ x(x – 6) + 5 (x – 6) = 0

⇒ (x – 6) (x + 5) = 0

∴ x = 6 as we cannot take –5

There were 6 kids.

Let the number be x.

Now, According to the question

(x)(x + 2) = 168

= x² + 2x = 168

= x² + 2x – 168 = 0

It’s a second degree equation.

and we can use splitting the middle term method.

= x² + 14x – 12x – 168 = 0

= x(x + 14) – 12(x + 14) = 0

= (x–12) (x + 14)

There could be two values of x which are 12 and –14.

and the two number in each case would be (12,14) and (–14,–12).

Let the numbers be x and y.

According to the question

x + y = 4

= y = 4 – x ---- (1)

xy = 2 ---- (2)

by (1) and (2)

x(4–x) = 2

4x – x² = 2

x² – 4x + 2 = 0

We can use the formula

x = 2 + √(2) , 2 – √(2)

sum of an AP =

In this question we need to find n and a = 99, d = –2

we’ll put the values in the above equation.

⇒ n[198 – 2n + 2] = 1800

⇒ 2n² – 200n + 1800 = 0

⇒ n² – 100n + 900 = 0

⇒ n² – 10n – 90n + 900 = 0

⇒ n(n–10) – 90(n–10) = 0

⇒ (n–90)(n–10) = 0

⇒ n = 90,10

If we take 10 then after ten terms its sum would become 900.

If we take 90, then after 10 terms its sum would be 900 and then it will increase until a certain point and then again the sum will start decreasing because of negative values which will continue till 90^th term making the sum 900 again.

let the number be x.

⇒

⇒

⇒ 6x² + 6 = 13x

⇒ 6x² + 6 – 13x = 0

⇒ (3x–2)(2x–3)

⇒ x =

Let the time taken by the larger tap alone to fill the tank be x minutes.

∴ Tank filled in 1 minute =

time taken by smaller tap to fill the tank alone = x + 10.

∴ Tank filled in 1 minute =

time taken by both of the taps to fill the tank = 12 minute.

Now in 1 minute they both will fill ^th of the tank.

⇒

⇒ (2x + 10)12 = x² + 10x

⇒ x² + 10x – 24x – 120 = 0

⇒ x² – 14x – 120 = 0

⇒ (x–20)(x + 6)

time cannot be negative so 20 minutes for larger tap

and 30 minutes for smaller tap.

Given.

Perimeter = 42 m

Diagonal = 15 m

Formula used/Theory

⇒ Pythagoras theorem:-

Base² + Height² = Hypotenuse²

⇒ Perimeter of rectangle = 2(L + B)

If Perimeter of rectangle = 2(L + B)

Then;

2(L + B) = 42m

(L + B) =

(L + B) = 21m

Let L be x

Then, B is (21–x)

Then by Pythagoras theorem:-

Base² + Height² = Hypotenuse²

x² + (21–x)² = 15²

x² + (21)² + x²–2×x× 21 = 225

2x² – 42x + (441 – 225) = 0

2x² – 42x + 216 = 0

2(x² – 21x + 108) = 0

x² – 21x + 108 = 0

As comparing eq to ax² + bx + c = 0

x = = 12

x = = 9

if length is 12m

then, breadth = (21–12) = 9m

if length is 9m

then, breadth = (21–9) = 12m

Conclusion/Result.

Length of sides can be either (12,9) or (9,12)

Formula used/Theory

Sum of ‘n’ natural numbers = n(n + 1)/2

Sum = 300

∴ 300 = n(n + 1)/2

300×2 = n(n + 1)

600 = n(n + 1)

n² + n–600 = 0

As comparing eq to ax² + bx + c = 0

x = = 24

x = = –25

Counting of numbers cannot be negative

∴ x = 24

Conclusion/Result. Sum of 24 natural numbers is 300

Let the number be x

Then reciprocal of number is

If reciprocal number, subtracted from the number itself is 1

Then;

(x–) = = 1

∴ comparing both

x = 2

x²–1 = 3

x² = 4

x = √4 = 2

∴ the number is 2

Conclusion/Result. Number comes out to be 2

Let the number be x

Then reciprocal of number is

If reciprocal number, added to the number itself

Then;

(x + ) = = 1

∴ comparing both

⇒ 2x² + 2 = 3x

2x²–3x + 2 = 0

As comparing eq to ax² + bx + c = 0

x =

x =

⇒ Both values of x comes with non-real number

Their can’t be negative sign in square root

∴ this number cannot be possible

Conclusion/Result. This type of number cannot be possible

Given.

Perimeter was wrongly written as 24 instead of 42.

The length of a side was then computed as 10 metres

Formula used/Theory

⇒ Perimeter of rectangle = 2(L + B)

⇒ Area of rectangle = (L×B)

If Perimeter of rectangle is taken as 24 m

And the length computed was 10 m

2(L + B) = 24m

(L + B) = = 12m

(10 + B) = 12m

B = 12m–10m = 2m

Then Area computed in problem was = (L×B)

= 10m×2m

= 20m²

Corrected perimeter = 42m

2(L + B) = 42m

(L + B) = = 21m

If area computed is 20m² and sum of length and breadth is 21m

Then;

The sides comes out to be 20m and 1m

Conclusion/Result. The sides comes out to be 20m and 1m

Given.

The number without x was written as 24 instead of –24

If c = 24

Means it will have factors AS 6 and 4

If c = –24

Means it will have factors either (–6,4) or (6, –4)

Conclusion/Result. Factors are either (–6,4) or (6, –4)