We have to find the value of a²+ b² + c², given that a+ b+ c=5 and ab + bc + ca 10,

Using the identity (a + b+ c)²

= a²+b² + c²+ 2(ab + bc + ca)

Substituting the given values:

(5)² = a² + b² + c²+2(10)

25 = a² + b² + c²+ 20

a²+b² + c²= 25- 20

a² + b²+ c²=5

∴ a²+b² +c²=5

\(27 b^3-\left(\frac{1}{64 b^3}\right)=(3 b)^3-\left(\frac{1}{(4 b)}\right)^3\)

\(=\left(\frac{3 b-1}{(4 b)}\right)\left[(3 b)^2+3 b\left(\frac{1}{(4 b)}\right)+\left(\frac{1}{(4 b)}\right)^2\right]\)

\(=\left(\frac{3 b-1}{(4 b)}\right)\left(9 b^2+\frac{3}{4}+\frac{1}{\left(16 b^2\right)}\right)\)

Factorise: 2x² +x-6

a x c=2 x (-6) =-12

Numbers: 4 and-3

2x² +x-6 = 2x²+ 4x- 3x-6

= 2x(x + 2) - 3(x+ 2)

= (2x-3)(x + 2)

Need p + q = -5, pq = +6.

Both p, g must be negative.

Negative factor-pairs of 6: (-1, -6), (-2, -3).

Only (-2, -3) sums to -5.

So,x²- 5x + 6 = (x- 2)(x - 3).

Need p + q=7, pq= 12.

Positive factor-pairs of 12: (1, 12), (2, 6), (3. 4).

Only (3, 4) sums to 7.

So, x² +7x+ 12 = (x+3)(x + 4).

Area of a square= (side)²

Given side: (2x + 3) m

So, Area = (2x + 3)²

Using the identity:

(a+ b)² = a²+ 2ab +b²

(2x + 3)² = (2x)²+2(2x)(3) + 3²

= 4x² + 12x +9

Use the identity (a - b)² = a²-2ab+ b²

So, 395 =400-5.

Then,

395² = (400 - 5)²

= 400² - 2(400)(5) + 5²

=160000 – 4000 + 25 = 156025

Hence, 395² = 156025.

Use the identity (a + b)² = a²+ 2ab + b².

Here, a = 4q and b=\(\frac{2}{s} .\)

\(\left(4 q+\frac{2}{s}\right)^2=(4 q)^2+2(4 q)\left(\frac{2}{s}\right)+\left(\frac{2}{s}\right)^2\)

\(=16 q^2+\frac{16 q}{s}+\frac{4}{s^2}\)

Hence, the expanded form is \(16 q^2+\frac{16 q}{s}+\frac{4}{s^2}\)

Use the identity (a + b)²= a²+ 2ab + b²

Here, a=1.4w and b =2

(1.4w + 2)² = (1.4w)² + 2(1.4w)(2) + (2)²

= 1.96w²+ 5.6w + 4

Hence, the expanded form is 1.96w² + 5.6w + 4.

119 = 100 + 10 +9.

Set a = 100, b = 10, c = 9:

119² = 100² + 10² + 9² + 2(100)(10) + 2(10)(9)+2(100)(9)

= 10000 + 100 + 81 + 2000 + 180+ 1800

= 14 161.

25z² - 4y² = (5z)² -(2y)²

Using the identity a²- b²= (a - b)(a + b), with a = 5z and b = 2y:

= (5z- 2y)(5z + 2y)

∴ The correct option is (b) (5z - 2y)(5z + 2y).

The standard identity for the difference of two cubes is:

a³- b³ = (a- b)(a² + ab + b²)

But option (a) gives a³-b³ = (a + b)(a² + ab + b³), where the first factor (a + b) is wrong; it should be (a - b). The other three options match the correct standard identities:

(b) a³+b³= (a + b)(a² - ab + b² )

(c) (a+ b)³ = a³+b³+ 3ab(a + b)

(d) (a- b)³ = a³-b³- 3ab(a -b)

⇒ The correct option is (a) a³-b³=(a + b)(a² + ab + b²), which is the incorrect identity.

Standard identity:

a³- b³ = (a - b) (a² + ab + b²).

99³ = (100– 1)³

Using identity:

(a- b)³ = a³ -3a²b+ 3ab² - b³

= 1000000-3 x 10000 x 1+ 3 x 100 x 1 -1

= 1000000 - 30000 + 300 -1

=970299

To factorise the quadratic expression x²+ 7x- 18, we compare it with the identity:

(x+ p)(x +q )-x²+ (p + q)x + pq

Thus, we require:

p+ q= 7 and pq= 18

Now, checking the given rows:

Row P: (7, -18) satisfies both conditions.

Row Q: (-7, 18) does not satisfy the required values.

Row R: (9, -2) does not satisfy the required values.

Row S: (2, -9) does not satisfy the required values.

Hence, the correct row is P.

To match each quadratic expression with its correct factors, we factorise each expression:

(P) x²+ 8x + 15

Two numbers whose sum is 8 and product is 15 are 3 and 5.

= (x+3)(x+ 5) →(ii)

(Q) x²+ 7x+ 10

Two numbers whose sum is 7 and product is 10 are 2 and 5.

= (x+ 2)(x+ 5)→(i)

(R) x²-9x + 18

Two numbers whose sum is -9and product is 18 are -3 and -6.

= (x-3)(x-6) →(iii)

(S) x²-9x + 20

Two numbers whose sum is -9 and product is 20 are -4 and -5.

= (x- 4)(x-5)→(iv)

Hence, the correct matching is:

P → (ii), Q→(i),R→(iii),S→(iv)

To determine the dimensions of the rectangle formed by the given tiles, we first express the total area as an algebraic expression.

The tiles consist of:

2 tiles of x² ⇒total area = 2x²

7 tiles of x ⇒ total area =7x

3 unit tiles ⇒ total area =3

Thus, the total area of the rectangle is:

2x² + 7x+3

To find the dimensions, we factorise this expression. We look for two numbers whose product is 2 x 3 = 6 and whose sum is 7. These numbers are 6 and 1.

Splitting the middle term:

2x²+ 7x+3= 2x² + 6x +x+3

= 2x (x + 3) + 1 (x+ 3)

= (2x + 1) (x+3)

Hence, the dimensions of the rectangle are (2x + 1) and (x+ 3).

A student arranges one x²-tile, five x-tiles and six-unit tiles to form a rectangle.

Total area of the rectangle:

x²+ 5x + 6

To find the sides of the rectangle, we factorise the quadratic expression.

We need two numbers whose product is 6 and sum is 5.

These numbers are 2 and 3.

Thus,

x²+ 5x +6 = x²+2x+3x+6

= x(x + 2) + 3(x + 2)

= (x+ 2) (x+ 3)

Hence, the sides of the rectangle are:

(x+ 2) and (x + 3)

We begin with the given inequality:

(a+ b)² > a²+b²

Expanding the left-hand side

(a+ b)² =a²+ 2ab + b²

Substituting this in the inequality.

a²+ 2ab +b² >a² + b²

Cancelling a² and b² from both sides, we get

2ab > 0

Dividing both sides by 2,

ab > 0

This happens when a and b have the same sign (both positive or both negative).

So, the correct answer is (a).

Given, side length of squared cricket ground = (30 + x)m

Area of square ground

= (30 + x)²

=30² + 2(30)(x) + x²

= (900 + 60x + x²) m²

Hence, required expression is (c) 900 + 60x + x²

Let us check all options one-by-one:

(P) (x + 5)(x- 5)

Applying the identity (a + b)(a - b) = a²- b², we obtain:

(x+ 5)(x - 5) = x²-25

This corresponds to (ii) in Column Il.

(Q) (x+ 4)²

Employing the identity (a + b)² = a² + 2ab + b² ,we get:

(x+ 4)² = x²+2(x)(4) + 4²

=x² +8x + 16

This corresponds to (i) in Column Il.

(R) (2x- 1)(2x + 1)

Again applying (a - b)(a + b) = a² - b², we obtain:

(2x- 1)(2x + 1) = 4x²-1

This corresponds to (iii) in Column II

So, the correct answer is (P)-(ii). (Q)-(i). (R)-(iii).

(a + b)²> a² + b²

⇒ 2ab > 0

⇒ a and b have the same sign.

Let us check all options one-by-one:

Row Q gives (-2, -5), both negative, so 2ab = 20 > 0.

Row P: 2ab = -24;

Row R: 2ab = 0

Row S: 2ab = -2

Hence, option (b) gives correct answer.

The four pieces, a² + ab + ab + b²

= a² + 2ab + b²

= (a + b)², this matches option (a).

Using the identity (a - b)² = a²- 2ab + b²

99² = (100 – 1)²

= 10000 - 200 + 1

= 9801 cm²

Only (x + 1)² =x²+ 2x+ 1 is true for every x.

The other three are equations that hold for specific values of x only.

(c) Assertion (A) is true but reason (R) is false.

Explanation:

The Assertion states that to simplify a rational expression, we must first factorise both the numerator and the denominator. This is true, since cancellation is possible only after expressing the numerator and denominator as products of factors. The Reason states that only common terms can be cancelled, which is wrong; only common factors can be cancelled.

Hence, Assertion is true but Reason is false.

(a) Both assertion A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Explanation:

We are given:

x² + 8x + 16

Using the identity:

(a + b)² = a² +b²+ 2ab

Thus,

x²+ 8x + 16 = (x + 4)²

Hence, it is a perfect square.

Now,

The reason states:

A trinomial ax² + bx + c is a perfect square when b² = 4ac.

For the given expression:

⇒ a =1,b=8, c=16

Check:

b² =8²= 64. 4ac = 4 x 1 x 16= 64

Since b² = 4ac, the condition is satisfied.

Let the required number be x.

According to the given condition:

\(x+\frac{1}{x}=\frac{10}{3}\)

Multiplying both sides by x:

\(x^2+1=\left(\frac{10}{3}\right) x\)

Rewriting the equation:

\(x^2-\left(\frac{10}{3}\right) x+1=0\)

Multiplying throughout by 3 to remove the fraction:

3x²-10x+3=0

Now, factorising:

3x²-10x+3 = 0

=3x²-9x - x+3=0

=3x(x -3)- 1(x-3) = 0

= (3x - 1)(x – 3) = 0

Therefore,

3x-1 =0 or x-3=0

So,

\(x=\frac{1}{3} \text { or } x=3\)

Hence, the required numbers are:

\(3 \text { and } \frac{1}{3}\)

Expressing the first and third terms as perfect squares:

\(25 a^2=(5 a)^2 \text { and } \frac{1}{a^2}=\left(\frac{1}{a}\right)^2\)

The middle term can be written as:

\(-2(5 a) \frac{1}{a}=-2 \times 5=-10\)

Using the identity a²- 2ab + b² = (a- b)²:

\(25 a^2-10+\frac{1}{a^2}=(5 a)^2-2(5 a)\left(\frac{1}{a}\right)+\left(\frac{1}{a}\right)^2\)

\(=\left(5 a-\frac{1}{a}\right)^2\)

\(\therefore 25 a^2-10+\frac{1}{a^2}=\left(5 a-\frac{1}{a}\right)^2\)

Expressing the first and third terms as perfect squares:

\(9 x^2=(3 x)^2 \text { and } \frac{1}{x^2}=\left(\frac{1}{x}\right)^2\)

The middle term can be written as:

\(2(3 x)\left(\frac{1}{x}\right)=2 \times 3=6\)

∴ Using the identity a² + 2ab + b²= (a + b)²:

\(9 x^2+6+\frac{1}{x^2}=(3 x)^2+2(3 x)\left(\frac{1}{x}\right)+\left(\frac{1}{x}\right)^2\)

\(=\left(3 x+\frac{1}{x}\right)^2\)

∴ \(9 x^2+6+\frac{1}{x^2}=\left(3 x+\frac{1}{x}\right)^2\)

We have to find the value of(135)² using the identity:

(a+ b)(a - b) + b² = a² -b² + b = a²

For a suitable choice, let us take a = 135 and b = 5, so that (a - b) and (a + b) become easy round numbers.

Then: a-b= 135 - 5=130

a+ b= 135 + 5 = 140

Now, applying the identity:

(135)² = (a+ b)(a – b) + b²

= (140)(130) + (5)²

= 18200 + 25 = 18225

⇒ (135)² = 18225

Outer square area = (x + 7)²

Inner tiled square area = (x+ 3)²

Area of the un-tiled border strip:

(x+ 7)²- (x+ 3)²

Using expansion,

(x + 7)² =x² + 14x+ 49

(x+ 3)² =x²+6x+9

So, (x² + 14x + 49) - (x²+ 6x + 9)

= 8x + 40

Therefore, area of the untiled border strip is 8x + 40 m².

For x = 5:

Area = 8(5) + 40

= 40 + 40

= 80 m².

Factorising:

n³-n=n(n² - 1)

= n(n- 1) (n + 1)

Thus, n³ -n= n(n - 1)(n + 1), which is the product of three consecutive natural numbers.

Among any three consecutive natural numbers:

One number is divisible by 2 (since every second number is even).

One number is divisible by 3 (since every third number is a multiple of 3).

Therefore, the product n(n - 1)(n + 1) is divisible by 2 x 3 =6.

Hence, n³ -n is always divisible by 6 for every natural number n.

Using the identity:

a³+b³ +c³ -3abc= (a + b +c )(a² + b² + c²- ab - bc - ca)

First, find a² + b² +c ²:

a²+b² +c²=(a+ b+ c)² -2(ab + bc + ca)

= 4² - 2(7)

= 16- 14

=2

Now,

a³+b³+c³-3abc=4(2-7)

= 4(-5)

=-20

Hence, the required value is -20.

The given expression is 6x²+ 17x + 12.

To factorise a quadratic of the form ax² + bx + c by splitting the middle term, we need to find two numbers whose:

Sum = b = 17 (the coefficient of x)

Product = a x c=6 x 12 =72

Let us list the factor pairs of 72 to find a suitable pair:

1x 72, 2 x 36,3 x 24, 4 x 18,6 x 12, 8 x 9

Among these, the pair (8,9) satisfies our condition, since:

8+9 =17

8 x 9 =72

Now,

6x² + 17x + 12 = 6x²+8x+9x + 12

= 2x(3x + 4) + 3(3x + 4)

= (3x + 4)(2x + 3)

The arrangement is shown below:

Algebra-Tile Diagram (Confirmation):

Algebra tiles provide a visual proof of the factorisation.

The given expression requires three types of tiles:

(i) Large square tiles, each of area x² ____ total 6 tiles representing 6x²

(ii) Rectangular tiles, each of area x ___ total 17 tiles representing 17x

(iii) Unit square tiles, each of area 1___total 12 tiles representing 12

These tiles are arranged into a single large rectangle whose dimensions correspond to the two factors:

Length of the rectangle = (3x + 4) units

Breadth of the rectangle = (2x + 3) units

Sample problem (student may create different ones): "A square pizza is cut into one big square slice of side 6 in, one small square slice of side 2 in, and two rectangular slices of sides 6 in x 2 in. What is the total area of the pizza, and is it itself a square? If so, what is its side?"

Total =6²+ 2² + 2(6 x 2)

=36+ 4 + 24

= 64 sq. in.

Since, \(\sqrt{64}=8 ;\)

The pizza is indeed an 8-inch square, confirming (6 + 2)²= 6² + 2(6)(2) + 2².

This is the identity (a + b)²= a² + 2ab + b²

with a = 6, b = 2.

Given expression:

\(\frac{m^2}{9}+\frac{m k}{3}+\frac{k^2}{4}+3 n k+2 m n+9 n^2\)

We use the identity:

(a+ b+ c)² = a² + b²+ c²+ 2ab + 2bc + 2ca

Rewriting the given expression, we observe that:

\(\frac{m^2}{9}=\left(\frac{m}{3}\right)^2, \frac{k^2}{4}=\left(\frac{k}{2}\right)^2, 9 n^2=(3 n)^2\)

Also,

\(\frac{m k}{3}=2\left(\frac{m k}{3}\right)\left(\frac{k}{2}\right)\)

\(3 n k=2\left(\frac{k}{2}\right)(3 n)\)

\(2 m n=2\left(\frac{m}{3}\right)(3 n)\)

Thus, the given expression is of the form (a + b+ c )²

where:

\(a=\frac{m}{3}, b=\frac{k}{2}, c=3 n\)

\(\left(\frac{m}{3}+\frac{k}{2}+3 n\right)^2\)

Hence, the factorised form of the given expression is:

\(\left(\frac{m}{3}+\frac{k}{2}+3 n\right)^2 .\)