The given equation is of the form ax² + bx + c = 0, where a = 2, b = - 4 and c = 3. Therefore, the discriminant
b² - 4ac = (-4)² - (4 x 2 x 3) = 16 - 24 = -8 < 0
So, the given equation has no real roots.

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.

Multiplying Equation (2) by \(\frac{5}{3},\) we get
5x – 8y + 1 = 0
But, this is the same as Equation (1). Hence the lines represented by Equations (1) and (2) are coincident. Therefore, Equations (1) and (2) have infinitely many solutions Plot few points on the graph and verify it yourself.

1000 = 2 x 2 x 2 x 5 x 5 x 5
56 = 2 x 2 x 2 x 7
HCF of 1000 and 56 = 8
\(\therefore\) Maximum number of columns = 8

Given equation is, x³-3x²+5x=(x-2)³
x³-3x²+5x=x³-6x²+12x-8
3x²-7x+8=0
Which is of the form ax2+bx+c=0
Hence, given equation is a quadratic equation.

x/2 + 2y/3 = - 1 and x – y/3 = 3
By elimination method
x/2 + 2y/3 = -1 ... (i)
x – y/3 = 3 ... (ii)
Multiplying equation (i) by 2, we get
x + 4y/3 = - 2 ... (iii)
x – y/3 = 3 ... (ii)
Subtracting equation (ii) from equation (iii), we get
5y/3 = -5
Dividing by 5 and multiplying by 3, we get
y = -15/5
y = - 3
Putting this value in equation (ii), we get
x – y/3 = 3 ... (ii)
x – (-3)/3 = 3
x + 1 = 3
x = 2
Hence our answer is x = 2 and y = −3.
By substitution method
x – y/3 = 3 ... (ii)
Add y/3 both side, we get
x = 3 + y/3 ... (iv)
Putting this value in equation (i) we get
x/2 + 2y/3 = - 1 ... (i)
(3+ y/3)/2 + 2y/3 = -1
3/2 + y/6 + 2y/3 = - 1
Multiplying by 6, we get
9 + y + 4y = - 6
5y = -15
y = - 3
Hence our answer is x = 2 and y = −3.

b=-3, c=3

Geometrically, the zeroes of a polynomial f(x) are the x-coordinates of the points, where the graph of y=f(x) intersects the X-axis.

Let a be any positive integer.
We know that, any odd positive integer is of the form 2q + 1, where q is a whole number.
\(\therefore \)  a = 2q + 1
\(\Rightarrow \)  a² = (2q + 1)²           [squaring both sides]
\(\Rightarrow \)   a² = 4q(q + 1) + 1            ...(i)
Note that q(q + 1) is either '0' or even, for any whole number q.
So, let q(q + 1) = 2m where m is a whole number.
From Eq.(i), we get  a² = 4(2m) + 1 = 8m + 1

Let the speed of the passenger train be x km/hr. Then,
Speed of the express train = (x + 11)km/hr
Time taken by the passenger train to cover 132 km between Mysore to Bangalore \(\frac{132}{x} \mathrm{hr}\)
Time taken by the express train to cover 132 km between Mysore to Bangalore = \(=\frac{132}{x+11} \mathrm{hr}\)
Therefore,
\( \frac{132}{x}-\frac{132}{x+11}=1 \)
\( \frac{132(x+11)-132 x}{x(x+11)}=1 \)
\(\frac{132 x+1452-132 x}{x^{2}+11}=1 \)
\(\frac{1452}{x^{2}+11}=1 \)
1452 = x² + 11
x² + 11 - 1452 = 0
x2 - 33x + 44x - 1452 = 0
x(x - 33) + 44(x - 33) = 0
(x - 33)(x + 44) = 0
So, either 
x - 33 = 0
x = 33
Or
x + 44 = 0
x = -44
But, the speed of the train can never be negative.
Thus, when x = 33 then speed of express train
= x + 11
= 33 + 11
= 44
e speed of the passenger train is x = 33 km/hr
and the speed of the express train is x = 44 km/hr respectively.

(i) (a):
Speed of the boat in downtream = x + y km/hr
(ii) (b):
 Speed of the boat in upstream = x - y km/hr
According to this statement \(\frac{42}{x+y}+\frac{42}{x-y}=20 \Rightarrow \frac{21}{x+y}+\frac{21}{x-y}=10\) and \(\frac{14}{x+y}=\frac{6}{x-y} \Rightarrow \frac{7}{x+y}-\frac{3}{x-y}=0\)
Let \(\frac{1}{x+y}=\mathrm{p}\) and \(\frac{1}{x-y}=\mathrm{q}\) then we have 21p + 21q = 0 and 7p - 3q = 0
Solving 21p + 21q = 0 and 7p - 3q = 0 
we get \(\mathrm{p}=\frac{1}{7}=\frac{1}{x+y} \Rightarrow \mathrm{x}+\mathrm{y}=7\) and \(\mathrm{q}=\frac{1}{3}=\frac{1}{x-y} \Rightarrow \mathrm{x}-\mathrm{y}=3\) 
Again solving x + y = 7 and x - y = 3, we get x = 5 and y = 2
(iii) (a):
 Speed of the boat = 5 km/hr  
(iv) (b): 
 Speed of the river = 2 km/hr 
(v) (c):
Speed of the boat in upstream = 5 – 2 = 3 km/hr

(i) (a): To have no real roots, discriminant (D = b² - 4ac) should be < 0.
(a) D = 7² - 4(-4)(-4) = 49 - 64 = -15 < 0
(b) D=7²-4(-4)(-2)=49-32=17>0
(c) D = 5² - 4(-2)(-2) = 25 - 16 = 9 > 0
(d) D = 6² - 4(3)(2) = 36 - 24 = 12> 0
(ii) (b): To have rational roots, discriminant (D = b² - 4ac) should be> 0 and also a perfect square
(a) D = 1²- 4(1)( -1) = 1 + 4 = 5, which is not a perfect square.
(b) D = (-5)² - 4(1)(6) = 25 - 24 = I, which is a perfect square.
(c) D = (-3)² - 4(4)(-2) = 9 + 32 = 41, which is not a perfect square.
(d) D = (-1)² - 4(6)(11) = 1 - 264 = -263, which is not a perfect square.
(iii) (c) : To have irrational roots, discriminant (D = b² - 4ac) should be > 0 but not a perfect square.
(a) D = 2² - 4(3)(2) = 4 - 24 = -20 < 0
(b) D = (-7)² - 4(4)(3) = 49 - 48 = 1 > 0 and also a perfect square.
(c) D = (-3)² - 4(6)(-5) = 9 + 120 = 129> 0 and not a perfect square.
(d) D = 3² - 4(2)(-2) = 9 + 16 = 25 > 0 and also a perfect square.
(iv) (d): To have equal roots, discriminant (D = b² - 4ac) should be = 0.
(a) D=(-3)²-4(1)(4)=9-16=-7<0
(b) D = (-2)² - 4(2)(1) = 4 - 8 = -4 < 0
(c) D = (-10)² - 4(5)(1) = 100 - 20 = 80 > 0
(d) D = 6² - 4(9)(1) = 36 - 36 = 0
(v) (a): To have two distinct real roots, discriminant (D = b² - 4ac) should be > 0.
(a) D = 3² - 4(1)(1) = 9 - 4 = 5 > 0
(b) D = 3² - 4(-1)( -3) = 9 - 12 = -3 < 0
(c) D=8²- 4(4)(4) = 64-64 = 0
(d) D = 6² - 4(3)(4) = 36 - 48 = -12 < 0

(i) (a) : Suppose two cars meet at point Q. Then,
Distance travelled by car X = A Q,
Distance travelled by car Y = BQ.
It is given that two cars meet in 9 hours.
\(\therefore\) Distance travelled by car X in 9 hours = 9x km
\(\Rightarrow\) AQ=9x
Distance travelled by car Y in 9 hours = 9y km
\(\Rightarrow\) BQ=9y

Clearly, AQ - BQ = AB
\(\Rightarrow\) 9x - 9y = 90
\(\Rightarrow\) x-y =10
(ii) (c): Suppose two cars meet at point P. Then
Distance travelled by car X = AP and
Distance travelled by car Y = BP.
In this case, two cars meet in 917 hours.
\(\therefore\) Distance travelled by car X in 9/7 hours = \(\frac {9}{7}\) x km
\(\Rightarrow A P=\frac{9}{7} x\)
Distance travelled by car Y in 9/7 hours \(\frac {9}{7}\) y km
\(\Rightarrow B P=\frac{9}{7} y\)
Clearly, AP + BP = AB
\(\Rightarrow \quad \frac{9}{7} x+\frac{9}{7} y=90 \Rightarrow \frac{9}{7}(x+y)=90 \Rightarrow x+y=70\)
(iii) (b): We have x - y = 10
\(\Rightarrow x+y=70\)
Adding equations (i) and (ii), we get
2x = 80 \(\Rightarrow\) x = 40
Hence, speed of car X is 40 km/hr.
(iv) (c): We have x - y = 10
\(\Rightarrow\) 40 - Y = 10 \(\Rightarrow\) Y = 30
Hence, speed of car y is 30 km/hr.
(v) (d)

(i) (b): Since, the given graph is parabolic is shape, therefore it will represent a quadratic polynomial.
[\(\therefore\) Graph of quadratic polynomial is parabolic in shape 1
(ii) (c): Since, the graph cuts the x-axis at -1, 5. So the polynomial has 2 zeroes i.e., -1 and 5.
(iii) (a) : Sum of zeroes = -1 + 5 = 4
(iv) (c): Since a and β are zeroes of the given polynomial and β > a
\(\therefore\)a = - 1 and β = 5.
\(\therefore|8 \alpha+\beta|=|8(-1)+5|=|-8+5|=|-3|=3 .\)
(v) (d): Since the zeroes of the given polynomial are - 1 and 5.
\(\therefore\) Required polynomial p(x)
= k{ x² -(-1 + 5)x + (-1)(5)} = k(.x² - 4x - 5)
For k = -1, we get
p(x) = -.x² + 4x + 5, which is the required polynomial.

(i) (d) : For a number to end in zero it must be divisible by 5, but 4ⁿ = 22ⁿ is never divisible by 5. So, 4ⁿ never ends in zero for any value of n.
(ii) (c) : We know that product of two rational numbers is also a rational number.
So, a² = a x a = rational number
a³ = a² x a = rational number
a⁴ = a³ x a = rational number
................................................
...............................................
aⁿ = a^n-1 x a = rational number.
(iii) (d): Let x = 2m + 1 and y = 2k + 1
Then x² + y² = (2m + 1)² + (2k + 1)²
= 4m² + 4m + 1 + 4k² + 4k + 1 = 4(m² + k² + m + k) + 2 So, it is even but not divisible by 4.
(iv) (a): Let three consecutive positive integers be n, n + 1 and n + 2.
We know that when a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
So, n = 3p or 3p + lor 3p + 2, where p is some integer. If n = 3p, then n is divisible by 3.
If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.
If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.
So, we can say that one of the numbers among n, n + 1 and n + 2 Wi always divisible by 3.
(v) (d): Any odd number is of the form of (2k +1), where k is any integer.
So, n² - 1 = (2k + 1)2 -1 = 4k² + 4k
For k = 1, 4k² + 4k = 8, which is divisible by 8.
Similarly, for k = 2, 4k² + 4k = 24, which is divisible by 8.
And for k = 3, 4k² + 4k = 48, which is also divisible by 8.
So, 4k² + 4k is divisible by 8 for all integers k, i.e., n² - 1 is divisible by 8 for all odd values of n.