Let us first split the middle term - 5x as -2x -3x [because (-2x) x (-3x) = 6x² = (2x² ) x 3].
So, 2x² - 5x + 3 = 2x² - 2x - 3x + 3 = 2x (x - 1) -3(x - 1) = (2x - 3)(x - 1)
Now, 2x² - 5x + 3 = 0 can be rewritten as (2x - 3)(x - 1) = 0.
So, the values of x for which 2x² - 5x + 3 = 0 are the same for which (2x - 3)(x - 1) = 0,
i.e., either 2x - 3 = 0 or x - 1 = 0.
Now, 2x - 3 = 0 gives x = \(\frac{3}{2}\) and x - 1 = 0 gives x = 1.
So, x = \(\frac{3}{2}\) and x = 1 are the solutions of the equation.
In other words, 1 and \(\frac{3}{2}\) are the roots of the equation 2x² - 5x + 3 = 0.
Verify that these are the roots of the given equation.
Note that we have found the roots of 2x² - 5x + 3 = 0 by factorising 2x² - 5x + 3 into two linear factors and equating each factor to zero.

Given, pair of linear equations is
2x + y - 6 = 0             ....(i)
and 4x - 2y - 4 = 0     ....(ii)
On Comparing with standard form of pair of linear equations, we get a₁ = 2, b₁ = 1, c₁ = -6
and a₂=4,b₂=-2, c₂= -4
Here, 
\(\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2}, \frac{b_{1}}{b_{2}}=-\frac{1}{2}\) and \(\frac{c_{1}}{c_{2}}=\frac{-6}{-4}=\frac{3}{2}\) 
Thus,  \(\frac{1}{2}\neq -\frac{1}{2}, i.e.\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}\)
 Hence, the given pair of linear equations is consistent. Now, on plotting the graph similar to Q. 1, we get the following graph.

The lines AB and DE intersect at point C(2, 2).
Hence, the unique solution is x = 2, y =2.

Let p(x) = x² - 2x - 8
= x² - 4x + 2x - 8 [by splitting the middle term]
= x(x - 4) + 2(x - 4) = (x + 2) (x - 4)
To find zeroes, put p(x) = 0
\(\Rightarrow\) x - 4 = 0 or x + 2 = 0 \(\Rightarrow\) x = 4 or x = -2
Hence, zeroes of the given polynomial are -2 and 4.
Verification
Here, sum of the zeroes = -2 + 4 = 2 = -(-2/1)
\(=-\frac{Coefficient \quad of \quad x}{Coefficient \quad of \quad x^{2}}\)
and product of the zeroes = -2 \(\times\) 4 = -8 = -8/1
\(=\frac{Constant \quad term}{Coefficient \quad of \quad x^{2}}\)
So, the relationship between the zeroes and its coefficients is verified.

Given, a pair of linear equations is :
0.2x + 0.3y = 1.3        .....(i)
and 0.4 x + 0.5y= 2.3  ....(ii)
Multiply both sides of Eq. (i) and Eq. (ii) by 10, we get
2x + 3y = 13                ....(iii)
and 4x + 5y = 23          ....(iv)
On substituting y from eqn. (iii) in eqn. (ii),
\(\frac { 4 }{ 10 } x+\frac { 5 }{ 10 } \times \frac { (13-2x) }{ 3 } =\frac { 23 }{ 10 } \)
\(\Rightarrow \quad 4x+\frac { 5 }{ 3 } (13-2x)=23\)
\(\Rightarrow\) 12x + 5 (13 - 2x) = 3 \(\times\) 23
\(\Rightarrow\) 12x + 65 - 10x = 69
\(\Rightarrow\) 2x = 69 - 65 = 4
\(\therefore\) x = 2
On substituting x = 2 in eqn. (iii), we get
\(\\ y=\frac { 13-2\times 2 }{ 3 } =\frac { 9 }{ 3 } \)
i.e., y = 3
Hence, x = 2,y = 3

Let us assume, to the contrary, that 5 - \( \sqrt{3}\) is rational.
That is, we can find coprime a and b (b \(\neq\)0) such that 5 - \( \sqrt{3}\) = \(\frac{a}{b}\)
Therefore, 5 - \(\frac{a}{b}=\sqrt{3}\)
Rearranging this equation, we get \(\sqrt{3}\) = 5 - \(\frac{a}{b}\)=\(\frac{5b-a}{b}\)
Since a and b are integers, we get 5 - \(\frac{a}{b}\) is rational, and so \( \sqrt{3}\) is rational.
But this contradicts the fact that \(\sqrt{3}\) is irrational.
This contradiction has arisen because of our incorrect assumption that 5 - \( \sqrt{3}\) is rational.
So, we conclude that 5 - \( \sqrt{3}\) is irrational.

Let the number of marbles John had be x
Then the number of marbles Jivanti had be = 45 – x (Why?).
The number of marbles left with john, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marble = 45 – x – 5 = 40 – x
Therefore, their product = (x – 5) (40 – x)
= 40x – x² – 200 + 5x
= – x² + 45x – 200
So, – x² + 45x – 200 = 124 (Given that product = 124)
i.e., – x² + 45x – 324 = 0
i.e., x² – 45x + 324 = 0
Therefore, the number of marbles John had, satisfies the quadratic equation
x² – 45x + 324 = 0
which is the required representation of the problem mathematically

Let s and t be the ages (in years) of Aftab and his daughter, respectively.
Then, the pair of linear equations that represent the situation is
s – 7 = 7 (t – 7), i.e., s – 7t + 42 = 0  ...(1)
and s + 3 = 3 (t + 3), i.e., s – 3t = 6   ...(2)
Using Equation (2), we get s = 3t + 6.
Putting this value of s in Equation (1), we get
(3t + 6) – 7t + 42 = 0,
i.e., 4t = 48, which gives t = 12.
Putting this value of t in Equation (2), we get
s = 3 (12) + 6 = 42
So, Aftab and his daughter are 42 and 12 years old, respectively.
Verify this answer by checking if it satisfies the conditions of the given problems.

3x² – x – 4 = 3x² – 4x + 3x – 4
= x(3x - 4) +1(3x - 4)
=(3x - 4)(x +1)
The zeroes of the polynomial are {4/3, -1}
Relationship between the zeroes and the coefficient of the polynomial:
Also sum of the zeroes = \(\frac{4}{3}-1=\frac{4-3}{3}=\frac{1}{3}\) 
Also product of the zeroes = \(\frac{4}{3} x-1=-\frac{4}{3}\) 
Hence verified.

Let p(u) = 4u² + 8u = 4u(u+2)
To find zeroes, put p(u) = 0
\(\Rightarrow\) 4u(u+2) = 0 \(\Rightarrow\) u = 0 or u + 2 = 0 [\(\because\) 4 \(\neq\)0]
\(\Rightarrow\) u = 0 or u = -2
Hence, zeroes of the given polynonial are 0 and -2.
Verification
Here, sum of zeroes = 0 - 2 = -2 = -(8/4)
=-\(\frac{Coefficient \quad of \quad u}{Coefficient \quad of \quad u^{2}}\)
and product of zeroes
=0 \(\times\)-2 = 0 = (0/4) = \(\frac{Constant \quad term}{Coefficient \quad of \quad u^{2}}\)
so, the relationship between the zeroes and its coefficients is verified.

Let us assume to the contrary that \(\frac{1}{\sqrt{2}}\) is rational, i.e. we can find coprime integers p and q (q \(\neq\)0), such that
\(\frac{1}{\sqrt{2}}=\frac{p}{q}\Rightarrow \frac{1\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}}=\frac{p}{q}\)
\(\Rightarrow \frac{\sqrt{2}}{2}=\frac{p}{q}\Rightarrow \sqrt{2}=\frac{2p}{q}\)
Since, p, q are integers and q \(\neq\) 0, therefore \(\frac{2p}{q}\) is rational.
This implies that \( \sqrt{2}\) is rational.
But this contradicts the fact that \( \sqrt{2}\) is irrational.
This shows that our assumption is wrong.
So, we conclude that \(\frac{1}{\sqrt{2}}\) is an irrational. Hence proved.

We have: 6 = 2 \(\times\) 3, 72 = 2³ \(\times\) 3², 120 = 2³ \(\times\) 3 \(\times\) 5
Here, 2¹ and 3¹ are the smallest powers of the common factors 2 and 3, respectively.
So, HCF (6, 72, 120) = 2¹ \(\times\) 3¹ = 2 \(\times\) 3 = 6
2³, 3² and 5¹ are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the three numbers.
So,    LCM (6, 72, 120) = 2³ \(\times\) 3² \(\times\) 5¹ = 360
Remark : Notice, 6 × 72 × 120 ≠ HCF (6, 72, 120) × LCM (6, 72, 120). So, the product of three numbers is not equal to the product of their HCF and LCM.

Given, a pair of linear equations is :
s - t = 3          
\(\Rightarrow\) s = t + 3        ....(i)
and \(\frac { s }{ 3 } +\frac { t }{ 2 } =6\) ...(ii)
On substituting s = t + 3, from egn. (i) in eqn. (ii),
we get
\(\\ \frac { t+3 }{ 3 } +\frac { t }{ 2 } =6\)
\(\Rightarrow\) 2(t + 3) + 3t = 36
\(\Rightarrow\) 5t + 6 = 36
\(\Rightarrow\) 5t = 30
\(\Rightarrow\) t = 6
From eqn., (i), s = 6 + 3 = 9
Hence, s = 9, t = 6

\(3 x^{2}-2 \sqrt{6} x+2=3 x^{2}-\sqrt{6} x-\sqrt{6} x+2\)
\(=\sqrt{3} x(\sqrt{3} x-\sqrt{2})-\sqrt{2}(\sqrt{3} x-\sqrt{2})\)
\(=(\sqrt{3} x-\sqrt{2})(\sqrt{3} x-\sqrt{2})\)
So, the roots of the equation are the values of x for which
\((\sqrt{3} x-\sqrt{2})(\sqrt{3} x-\sqrt{2})=0\)
Now, \(\sqrt{3} x-\sqrt{2}=0 \text { for } x=\sqrt{\frac{2}{3}}\)
So, this root is repeated twice, one for each repeated factor \(\sqrt{3} x-\sqrt{2}\)
Therefore, the roots of 3x² - 2\(\sqrt6\)x + 2 = 0 are \(\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}\)

Let the two consecutive integers be x and x+1
ATQ x²+(x+1)2=365
\(\Rightarrow\) x²+x2+2x+1=365 \(\Rightarrow\) 2x²+2x-364=0
\(\Rightarrow\) x²+x-182=0 \(\Rightarrow\) x²+14x-13x-182=0
\(\Rightarrow\) x(x+14)-13(x+14)=0 \(\Rightarrow\) (x-13)(x+14)=0
\(\Rightarrow\) x=13, -14 (-14 is rejected because it is a negative integer)
Hence, the two consecutive positive integers are 13 and 13+1=14

(x+1)²=2(x-3)
x²+2x+1=2x-6
x²+2x-2x+1+6=0
x²+7=0
Which is of the form ax²+bx+c=0. Hence the given equation is a quadratic equation.

\(\frac{3}{2} x+\frac{5}{3} y=7,9 x-10 y=14\) 
Here,\(\frac{a_{1}}{a_{2}}=\frac{3}{2 \times 9}=\frac{1}{6}, \frac{b_{1}}{b_{2}}=-\frac{5}{3 \times 10}=\frac{-1}{6}, \frac{c_{1}}{c_{2}}=\frac{-7}{-14}=\frac{1}{2}\) 
\(\because \quad \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\) 
\(\therefore\) Pair of equations is consistent with unique solution
 

336 and 54
336 = 2 x 168
= 2 x 168
= 2 x 2 x 84
= 2 x 2 x 2 x 42
= 2 x 2 x 2 x 2 x 21
= 2 x 2 x 2 x 2 x 7 x 3 x 1
Therefore 336 = 2 x 2 x 2 x 2 x 7 x 3 .......(A)
54 = 2 x 27
= 2 x 3 x 9
= 2 x 3 x 3 x 3
Therefore 54 = 2 x 3 x 3 x 3 ......(B)
From (A) and (B) HCF of 336 and 54 = 2 x 3 = 6
LCM of 336 and 54 = 2 x 3 x 2 x 2 x 2 x 7 x 3 x 3 = 2⁴ x 3³ x 7 = 3024
Product of 336 and 54 = 18144
Product of LCM and HCF = 6 x 3024 = 18144
Therefore it is proved that LCM X HCF = Product of the two numbers.

17, 23 and 29
17 = 1 x 17
23 = 1 x 23
29 = 1 x 29
From the above, HCF (17, 23, 29) = 1 and LCM (17, 23, 29) = 11339

\(0, \sqrt{5}\) 
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
Given \(\alpha+B=0=-\frac{b}{a}\) 
\(\alpha \beta=\sqrt{5}=\frac{c}{a}\) 
If \(a=1, b=0, c=\sqrt{5}\) 
Therefore, the quadratic polynomial is \(x^{2}+\sqrt{5}\)

Let the polynomial be ax² + bx + c, and its zeroes be α and ß
Given \(\alpha+\beta=\sqrt{2}=\frac{-b}{a}\) 
\(\alpha \beta=\frac{1}{3}=\frac{c}{a}\) 
If a = 3, then \(b=\sqrt{2}\) and  c = 1/3
Therefore, the quadratic polynomial is \(3 x^{2}-3 \sqrt{2} x+1\).

If the number 4ⁿ , for any n, were to end with the digit zero, then it would be divisible by 5. That is, the prime factorisation of 4ⁿ would contain the prime 5. This is not possible because 4ⁿ = (2)²ⁿ ; so the only prime in the factorisation of 4ⁿ is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 4ⁿ . So, there is no natural number n for which 4ⁿ ends with the digit zero.
You have already learnt how to find the HCF and LCM of two positive integers using the Fundamental Theorem of Arithmetic in earlier classes, without realising it. This method is also called the prime factorisation method. Let us recall this method through an example.

The given equations can be rewritten as
3x + 2y - 5 = 0 and 2x - 3y - 7 = 0
On comparing with standard form of pair of linear equations, we get a₁ = 3, b₁ = 2, c₁ = -5
and a₂ = 2, b₂ = -3, c₂ = -7
Now, \(\frac{a_{1}}{a_{2}}=\frac{3}{2}, \frac{b_{1}}{b_{2}}=-\frac{2}{3}\) and \(\frac{c_{1}}{c_{2}}=\frac{5}{7}\)
Thus, \(\frac{3}{2}\neq -\frac{2}{3},i.e.\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}\)
Hence, the pair of linear equations is consistent.

2x² - 6x + 3 = 0
Comparing this equation with ax² + bx + c = 0, we get
a = 2, b = -6, c = 3
Discriminant = b2 - 4ac
= (-6)² - 4 (2) (3)
= 36 - 24 = 12
As b2 - 4ac > 0,
Therefore, distinct real roots exist for this equation
\( x=\frac{-b \pm b^{2}-4 a c}{2 a} \)
\(=\frac{-(-6) \pm \sqrt{(-6)^{2}-4(2)(3)}}{2(2)} \)
\(=\frac{6 \pm \sqrt{12}}{4}=\frac{6 \pm 2 \sqrt{3}}{4} \)
\(=\frac{3 \pm \sqrt{3}}{2} \)
Therefore the root are \(=\frac{3 \pm \sqrt{3}}{2} \)

(i) (b): Let two consecutive integers be x, x + 1.
Given, x² + (x + 1)² = 650
\(\begin{array}{l} \Rightarrow 2 x^{2}+2 x+1-650=0 \\ \Rightarrow 2 x^{2}+2 x-649=0 \end{array}\)
(ii) (c): Let the two numbers be x and 15 - x.
Given, \(\frac{1}{x}+\frac{1}{15-x}=\frac{3}{10}\)
\(\begin{array}{l} \Rightarrow 10(15-x+x)=3 x(15-x) \\ \Rightarrow 50=15 x-x^{2} \Rightarrow x^{2}-15 x+50=0 \end{array}\)
(iii) (d): Let the numbers be x and x + 3.
Given, x(x + 3) = 504
\(\Rightarrow\) x² + 3x - 504 = 0
(iv) (c): Let the number be x.
According to question, x² - 84 = 3(x + 8)
\(\Rightarrow x^{2}-84=3 x+24 \Rightarrow x^{2}-3 x-108=0\)
(v) (d): Let the number be x.
According to question, x + 12 = \(\frac {160}{x}\)
\(\Rightarrow x^{2}+12 x-160=0\)

(i) (a): 1^st situation can be represented as x + 7y = 650 ...(i) and
2^nd situation can be represented as x + 11y = 970 ...(ii)
(ii) (b): Subtracting equations (i) from (ii), we get 
\(4 y=320 \Rightarrow y=80\)
\(\therefore\) Proportional expense for each person is Rs 80.
(iii) (c): Puttingy = 80 in equation (i), we get
x + 7 x 80 = 650 \(\Rightarrow\) x = 650 - 560 = 90
\(\therefore\) Fixed expense for the party is Rs 90
(iv) (d): If there will be 15 guests, then amount that Mr Jindal has to pay = Rs (90 + 15 x 80) = Rs 1290
(v) (a): We have a₁ = 1, b₁ = 7, c₁ = -650 and 
\(a_{2}=1, b_{2}=11, c_{2}=-970 \)
\(\therefore \frac{a_{1}}{a_{2}}=1, \frac{b_{1}}{b_{2}}=\frac{7}{11}, \frac{c_{1}}{c_{2}}=\frac{-650}{-970}=\frac{65}{97}\)
\(\text { Here, } \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Thus, system of linear equations has unique solution.