{tex}\begin{array}{l}\sqrt2=2^{1/2}\\(\sqrt2)^2=(2^\frac12)^2\\=2^{\frac12\times2}\\=2^1\\=2\\Hence\;the\;root\;is\;removed\;\end{array}{/tex}

f(t)=t^2-4t+3

{tex}\begin{array}{l}\alpha+\beta=4,\alpha\beta=3\\\alpha^4\beta^3+\alpha^3\beta^4\\=\alpha^3\beta^3(\alpha+\beta)\\=(3)^3(4)\\=27\times4\\=108\\\end{array}{/tex}

I think you didn't have a clear image of this question in your mind.
Lets see this,

We can see here that to be an AP the common difference "d" must be the same between each successive terms.

But here, 5500, 6050, 6655, 7320.5 are not making an AP which can be seen clearly.

0.2 x + 0.3 y = 1.3 ;  0.4 x + 0.5 y = 2.3
The given system of linear equations is:
0.2 x + 0.3 y = 1.3..............(1)
0.4 x + 0.5 y = 2.3...................(2)
From equation (1), 
0.3 y = 1.3 - 0.2 x
{tex}\Rightarrow \quad y = \frac { 1.3 - 0.2 x } { 0.3 }{/tex}.........................(3)
Substituting this value of y in equation(2), we get
{tex}0.4 x + 0.5 \left( \frac { 1.3 - 0.2 x } { 0.3 } \right) = 2.3{/tex}
{tex}\Rightarrow{/tex}0.12 x + 0.65 - 0.1 x = 0.69
{tex}\Rightarrow{/tex}0.12 x - 0.1 x = 0.69 - 0.65
{tex}\Rightarrow{/tex}0.02 x = 0.04
{tex}\Rightarrow{/tex}{tex}\mathrm { x } = \frac { 0.04 } { 0.02 } = 2{/tex}
Substituting this value of x in equation(3), we get
{tex}y = \frac { 1.3 - 0.2 ( 2 ) } { 0.3 } = \frac { 1.3 - 0.4 } { 0.3 } = \frac { 0.9 } { 0.3 } = 3{/tex}
Therefore, the solution is x = 2, y = 3, we find that both equation (1) and (2) are satisfied as shown below:
0.2 x + 0.3 y = ( 0.2 )( 2 )+( 0.3)( 3 ) = 0.4 + 0.9 = 1.3
0.4 x + 0.5 y= ( 0.4 )( 2 ) + ( 0.5 )( 3 ) } = 0.8 + 1.5 = 2.3
This verifies the solution.​​​​​​​

0.2x+0.3y = 1.3 ....... (i)
eq (i) ×10
2x+3y=13
0.4x+0.5y=2.3 ...... (ii)
eq(ii) ×10
4x+5y=23
from eq (i)
x =13-3y/2

putting value of x in eq (ii)
4(13-3y/2)+5y=23
26-6y+5y=23
26-y=23
y= 3
putting value of y in eq (i)
2x+3(3)=13
2x=13-9
x=4/2
x=2

x² - 6x + 12
Here, a = 1, b = -6 and c = 12
D = b² - 4ac
D = (-6)² - 4(1)(12)
D = 36 - 48
D = -12
Hence, D < 0
Therefore, there exists no real root of the equation.

{tex}\begin{array}{l}\frac{7\sin\;\theta+\;5\cos\theta}{\;5\sin\theta\;+\;7\;\cos\;\theta}=\frac{7\sin\;\theta+7\sin\;\theta}{7\sin\;\theta+{\displaystyle\frac{\;7\times5}5}\cos\theta}\\=\frac{\displaystyle14\sin\;\theta}{7\sin\;\theta+{\displaystyle\frac75}\times7\sin\;\theta}=\frac{70\sin\;\theta}{7\sin\;\theta+49\sin\;\theta}\\=\frac{\displaystyle70\sin\;\theta}{\displaystyle56\sin\;\theta}=\frac{70}{56}=\frac54\end{array}{/tex}

On a graph paper, draw a horizontal line XOX' and a vertical line YOY' as the x-axis and they-axis respectively.
{tex}3x + y - 11 = 0{/tex} {tex}\Rightarrow{/tex}{tex}y = (11 -3x){/tex}. ...(i)
Putting {tex}x = 2{/tex} in (i), we get {tex}y = 5{/tex}.
Putting {tex}x = 3{/tex} in (i), we get {tex}y = 2{/tex}.
Putting {tex}x = 5{/tex} in (i), we get {tex}y = -4{/tex}.

On the graph paper, plot the points {tex}A (2, 5), B(3, 2)\ and\ C(5, -4).{/tex}
Join AB and BC to get the graph line ABC.
Thus, the line ABC is the graph of the equation {tex}3x + y - 11 = 0{/tex}.
{tex}x - y - 1 = 0{/tex} {tex}\Rightarrow{/tex}{tex}y = (x - 1){/tex} . ...(ii)
Putting {tex}x = -3{/tex} in (ii), we get {tex}y = -4{/tex}.
Putting {tex}x = 0{/tex} in (ii), we get {tex}y = -1{/tex}.
Putting {tex}x = 3{/tex} in (ii), we get {tex}y = 2{/tex}.

On the same graph paper as above, plot the points {tex}P(-3, -4)\ and\ Q(0, -1){/tex}. The third point {tex}B(3,2){/tex} is already plotted.
Join PQ and QB to get the graph line PQB.
Thus, line PQB is the graph of the equation {tex}x - y - 1 = 0{/tex}.
The two graph lines intersect at the point {tex}B(3,2){/tex}. {tex}x = 3, y = 2{/tex} is the solution of the given system of equations.
The region bounded by these lines and the y-axis has been shaded.

On extending the graph lines on both sides, we find that these graph lines intersect the y-axis at the points Q(0, -1) and R(0,11).

We know that, the ratio of the areas of two similar triangles is equal to the square of their corresponding sides.
Also,   
{tex}\Delta ADE{/tex} {tex}\sim{/tex} {tex}\triangle{/tex}ABC [given]
Now AD = 1, DB = 2 
{tex}\therefore{/tex} AB = AD + DB = 1 + 2 = 3
{tex}\therefore{/tex} {tex}\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle A D E)}{/tex} = {tex}\frac{A B^{2}}{A D^{2}}{/tex} = {tex}\left(\frac{3}{1}\right)^{2}{/tex} = {tex}\frac{9}{1}{/tex} 

Let the ten's and unit's digits of the required number be x and y respectively.
Then, the number = (10x + y).
The number obtained on reversing the digits = (10y + x).
As per given condition
The sum of a two-digit number and the number obtained by reversing the order of its digits is 99.
{tex}\therefore{/tex} (10y + x) + (10x + y) = 99
{tex}\Rightarrow{/tex} 11(x + y) = 99
{tex}\Rightarrow{/tex} x + y = 9.
The digits differ by 3
So, (x - y) = ±3.
Thus, we have
x + y = 9 ........ (i)
x - y = 3 .......... (ii)
or x + y = 9 ......... (iii)
x - y = -3 ............ (iv)
From (i) and (ii), we get x = 6, y = 3.
From (iii) and (iv), we get x = 3, y = 6.
Hence, the required number is 63 or 36.

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Let the numerator and denominator of the fraction be x and y respectively.
Then,
Fraction= {tex} \frac { x } { y }{/tex}
It  is given that
Denominator = 2 (Numerator) + 4
{tex}\Rightarrow{/tex} {tex}y = 2x + 4 {/tex}
{tex}\Rightarrow{/tex}{tex}2x - y + 4 =0{/tex}
According to the given condition, we have
{tex}y - 6 = 12(x - 6){/tex}
{tex}\Rightarrow{/tex} {tex}y - 6 =12x -72{/tex}
{tex}\Rightarrow{/tex} {tex}12x - y - 66 =0{/tex}
Thus, we have the following system of equations
2x - y + 4 =0 ............(i)
12x - y -66 =0 .............(ii)
Subtracting equation (i) from equation (ii), we get
10x - 70 =0 {tex}\Rightarrow{/tex} x = 7
Putting x = 7 in equation (i), we get
14 - y + 4 =0 {tex}\Rightarrow{/tex} y = 18
Hence, required fraction ={tex}\frac { 7 } { 18 }.{/tex}

Let the ten's and unit digit be y and x respectively.
So the number is 10y + x
The number when digits are reversed becomes 10x + y
So, 7(10x + y) = 4(10y + x)
or, 70x+ 7y= 40y + 4x
or, 70x - 4x = 40y - 7y
or 66x = 33y
{tex} \Rightarrow {/tex} 2x = y .....(i)
The difference of the digits is 3
y - x= 3
2x - x = 3
x = 3
x = 3 and y = 6
Hence the number is 63

We have equation 
{tex}a + b p ^ { \frac { 1 } { 3 } } + c p ^ { \frac { 2 } { 3 } } = 0{/tex} ...(i)
Multiplying both sides by {tex}p ^ { \frac { 1 } { 3 } }{/tex}in eq(i), we get
{tex}a p ^ { \frac { 1 } { 3 } } + b p ^ { \frac { 2 } { 3 } } + c p = 0{/tex} ...(ii)
Multiplying (i) by b and (ii) by c, we get
{tex}ab+b^2p^\frac13+bcp^\frac23=0{/tex} .......(iii) 

{tex}acp^\frac13+bcp^\frac23+c^2p=0{/tex} .......(iv).

subtracting equation (iv) from equation (iii) we get 

{tex}ab+b^2p^\frac13+bcp^\frac23-acp^\frac13-bcp^\frac23-c^2p=0{/tex}  

{tex}\Rightarrow\left(b^2p^\frac13-acp^\frac13\right)+\left(ab-c^2p\right)=0{/tex}
{tex}\Rightarrow \quad \left( b ^ { 2 } - a c \right) p ^ \frac{1}{3} + a b - c ^ { 2 } p = 0{/tex}
{tex}\Rightarrow \quad b ^ { 2 } - a c = 0 \text { and } a b - c ^ { 2 } p = 0 \quad \left[ \because p ^ { 1 / 3 } \text { is irrational } \right]{/tex}
{tex}\Rightarrow \quad b ^ { 2 } = a c \text { and } a b = c ^ { 2 } p{/tex}
{tex}\Rightarrow \quad b ^ { 2 } = a c \text { and } a ^ { 2 } b ^ { 2 } = c ^ { 4 } p ^ { 2 }{/tex}
{tex}\Rightarrow \quad a ^ { 2 } ( a c ) = c ^ { 4 } p ^ { 2 }{/tex} [Putting b² = ac in a²b² = c⁴p²]
{tex}\Rightarrow \quad a ^ { 3 } c - p ^ { 2 } c ^ { 4 } = 0{/tex}
{tex}\Rightarrow \quad \left( a ^ { 3 } - p ^ { 2 } c ^ { 3 } \right) c = 0{/tex}
{tex}\Rightarrow \quad a ^ { 3 } - p ^ { 2 } c ^ { 3 } = 0 , \text { or } c = 0{/tex}
Now, {tex}a ^ { 3 } - p ^ { 2 } c ^ { 3 } = 0{/tex}
 {tex}\Rightarrow p^2=\left(\frac ab\right)^3{/tex} 

cube root both side we get 

{tex}\left(p^2\right)^\frac13=\frac{a}{b}{/tex}  

{tex}\left(p^\frac13\right)^2=\frac{a}{b}{/tex}

this is not possible as {tex}p ^ { 1 / 3 }{/tex} is irrational and {tex}\frac { a } { b }{/tex} is rational.
{tex}\therefore \quad a ^ { 3 } - p ^ { 2 } c ^ { 3 } \neq 0{/tex}
Hence, c = 0
Putting c = 0 in b² = ac = 0, we get b = 0
Putting b = 0 and c = 0 in equation (i) {tex}a + b p ^ { 1 / 3 } + c p ^ { 2 / 3 } = 0{/tex}   

a + 0 +0 = 0, we get a = 0
Hence, a = b = c = 0.

By Euclid’s division lemma, we have
a = bq + r; 0 ≤ r < b
For a = n and b = 5, we have
n = 5q + r …(i)
Where q is an integer
and 0 ≤ r < 5, i.e. r = 0, 1, 2, 3, 4.
Putting r = 0 in (i), we get
n = 5q
⇒ n is divisible by 5.
n + 4 = 5q + 4
⇒ n + 4 is not divisible by 5.
n + 8 = 5q + 8
⇒ n + 8 is not divisible by 5.
n + 12 = 5q + 12
⇒ n + 12 is not divisible by 5.
n + 16 = 5q + 16
⇒ n + 16 is not divisible by 5.
Putting r = 1 in (i), we get
n = 5q + 1
⇒ n is not divisible by 5.
n + 4 = 5q + 5 = 5(q + 1)
⇒ n + 4 is divisible by 5.
n + 8 = 5q + 9
⇒ n + 8 is not divisible by 5.
n + 12 = 5q + 13
⇒ n + 12 is not divisible by 5.
n + 16 = 5q + 17
⇒ n + 16 is not divisible by 5.
Putting r = 2 in (i), we get
n = 5q + 2
⇒ n is not divisible by 5.
n + 4 = 5q + 9
⇒ n + 4 is not divisible by 5.
n + 8 = 5q + 10 = 5(q + 2)
⇒ n + 8 is divisible by 5.
n + 12 = 5q + 14
⇒ n + 12 is not divisible by 5.
n + 16 = 5q + 18
⇒ n + 16 is not divisible by 5.
Putting r = 3 in (i), we get
n = 5q + 3
⇒ n is not divisible by 5.
n + 4 = 5q + 7
⇒ n + 4 is not divisible by 5.

n + 8 = 5q + 11
⇒ n + 8 is not divisible by 5.
n + 12 = 5q + 15 = 5(q + 3)
⇒ n + 12 is divisible by 5.
n + 16 = 5q + 19
⇒ n + 16 is not divisible by 5.
Putting r = 4 in (i), we get
n = 5q + 4
⇒ n is not divisible by 5.
n + 4 = 5q + 8
⇒ n + 4 is not divisible by 5.
n + 8 = 5q + 12
⇒ n + 8 is not divisible by 5.
n + 12 = 5q + 16
⇒ n + 12 is not divisible by 5.
n + 16 = 5q + 20 = 5(q + 4)
⇒ n + 16 is divisible by 5.
Thus for each value of r such that 0 ≤ r < 5 only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5.

{tex}\begin{array}{l}\alpha+\beta+\gamma=-\frac ba\\\alpha\beta\gamma=-\frac ca\\\alpha\beta+\beta\gamma+\gamma\alpha=\frac da\\\\\end{array}{/tex}