\(k=\pm 3\)

p=-5, q=-6

k=7

x²-6x+7

-2 and -3

Let α and \(\frac { 1 }{ \alpha } \) be two zeroes of the given polynomial, which are reciprocal of each other.
On comparing the given polynomial with Ax²+Bx+C, we get
A=a²+9, B=13 and C=6a
Now, product of zeroes,
\(\alpha \times \frac { 1 }{ \alpha } =\frac { Constant \ term }{ Coefficient \ of \ x^{ 2 } } \) 
\(\\ \Rightarrow \ 1=\frac { 6\alpha }{ a^{ 2 }+9 }\) 
\( \\ \Rightarrow \ a^{ 2 }+9=6a\) 
\(\\ \Rightarrow \ a^{ 2 }-6a+9=0\) 
\(\\ \Rightarrow \ (a-3)^{ 2 }=0\ \ \left[ \because \quad (x-y)^{ 2 }=x^{ 2 }+y^{ 2 }-2xy \right] \) 
\(\\ \therefore \ a=3\)

\(a=-\frac { 3 }{ 2 } \)

α+β+αβ=\(-\frac { 7 }{ 2 } +\frac { 5 }{ 2 } =-1\)

α³+β³=(α³+β³)-3αβ(α+β) \(\frac { 215 }{ 27 } \)

α+β=7  ...(i)
α-β=1 ....(ii)
On solving Eqs. (i) and (ii), we get α=4 and β=3
Now, k=αβ=12

Area of path = (Area of the field including the path) - (Area of the field encluding the path)
1.5 m.

k < 4

\(x^{2} + x - 342 = 0 ; 18 , 19\)

Either p < - 6 or p > 6

k = 5

36th term

a_n = 1 - 4n
\(\Rightarrow\)  a₁ = 1 - 4 x 1 = -3
a₂ = 1 - 4 x 2 = -7
d = a₂ - a₁
= - 7 - ( - 3 ) = - 4
a₂₅ = a + 24d
- 3 = 24 x ( - 4 ) = - 99
Now,    S₂₅ = \(\frac{25}{2}({a}_{1}+{a}_{25})\)
\(=\frac{25}{2}(-3-99)\)
= 25 x ( - 51 ) = - 1275

Here a=, d=5-2=3
n=10
Now, S_n=\({n\over}[2a+(n-1)d]\)
\(\Rightarrow\ \ S_{10}={10\over2}[2\times2+9\times3]\)
=155

\(\therefore \alpha ,\beta \) are roots of 2x²-10x+5=0
\(\therefore \alpha +\beta =-\frac { -10 }{ 2 } =5\)   ....(i)
and  \(\alpha \beta =\frac { 5 }{ 2 } \) ...(ii)
Now \((\alpha +2)(\beta +2)=\alpha \beta +2(\alpha +\beta )+4\)
\(=\frac { 5 }{ 2 } +2(5)+4=\frac { 33 }{ 2 } \)

Let one roots be \(\alpha \) 
Other root be =\(\frac { 1 }{ \alpha } \)  
Now, product of roots =\(\frac { c }{ a } \) 
\(\Rightarrow \alpha \times \frac { 1 }{ \alpha } =\frac { c }{ 2 } \Rightarrow c=2\) 

Here,   t₈ = \(\frac{{t}_{2}}{2}\Rightarrow\) a + 7d = \(\frac{a+d}{2}\)
\(\Rightarrow\)  2a + 14d = a + d \(\Rightarrow\) a = - 13d (i)
and  t₁₁-\(\frac{{t}_{4}}{3}=1\)
\(\Rightarrow\)  a + 10d - \(\frac{a+3d}{3}=1\)
\(\Rightarrow\frac{3a+30d-a-3d}{3}=1\)
\(\Rightarrow\)  2a + 27d = 3
\(\Rightarrow\)   2 ( - 13d ) + 27d = 3  [ using (i) ]
\(\Rightarrow\)     -26d + 27d = 3
\(\Rightarrow\)   d = 3,
\(\therefore\)  from (i)  a = -13 x 3 = - 39
Therefore,,  t₁₅ = a + 14d
= - 39 + 14 x 3
= - 39 + 42 = 3

\(\because\) a, 7, b, 23, c are in AP
\(\therefore\)    7 - a = b - 7 and 23 - b = c - 23
\(\Rightarrow\)    \(\frac{89}{5}-\frac{4n}{5}<0\)
\(\Rightarrow\)     \(-\frac{4}{5}n<-\frac{89}{5}\)
\(\Rightarrow\)    \(n>\frac{89}{4}\Rightarrow n>22\frac{1}{4}\)
\(\therefore\)   n = 23.
That is, 23rd term is the Ist negative term.

Let Ist term of the AP = a and common
difference = d
Now   a₃ = p \(\Rightarrow\) a + 2d = p     ...(i)
and    a₄ = q \(\Rightarrow\) a + 3d = q     ...(ii)
Subtracting equation (i) from equation (ii),
a + 3d = q
a + 2d = p
-  -          -
d = q -  p
When d = ( q - p ), equation (i) becomes
a + 2 (q - p ) = p
a = 3p - 2q
\(\therefore\)   a₁₀ = a + 9d
= ( 3p - 2q ) + 9 ( q - p )
= 7q - 6p

Fare for 1st km = Rs.15
Fare for 2 km = 15 + 8 = Rs.23
Fare foe 3 km = 15 + 2 x 8 = Rs.31
AP is 15, 23, 31,...........

 \(\sqrt{2},\sqrt{8},\sqrt{18}......\) =  \(\sqrt{2},2\sqrt{2},3\sqrt{2}.....\)
Next term is \(4\sqrt{2}\)

10th term from end of AP 4, 9, 14, ....., 254 is 10th term of the AP 254, 249, 244, .... 14, 9, 4
Here  a = 254
d = 249 - 254 = -5
a₁₀ = a + 9d
\(\Rightarrow\) a₁₀ = 254 + 9 x (-5) = 209

Let list price of the book be Rs.x
Total cost=Rs.1200
\(\therefore\) Number of books=\(​​\frac { 1200 }{ x } \)
If list price of the bood=Rs.(x+10)
Then number of books=\(\frac { 1200 }{ x+10 } \)
ATQ  \(\frac { 1200 }{ x } -\frac { 1200 }{ x+10 } =10\)
\(\frac { 1200 }{ x } -\frac { 1200 }{ x+10 } =10\)
\(\Rightarrow\) 12000=10x(x+10)
\(\Rightarrow\) x²+10x-1200=0
\(\Rightarrow\) (x+40)(x-30)=0
\(\Rightarrow\) x=-40 (rejecting) or x=30
\(\therefore\) List price of the book is Rs.30

Let one number be x then other number be x+5
Also \(​​\frac { 1 }{ x } -\frac { 1 }{ x+5 } =\frac { 1 }{ 10 } \)
\(\Rightarrow \frac { x+5-x }{ x(x+5) } =\frac { 1 }{ 10 } \)
\(\Rightarrow\) x(x+5)=50
\(\Rightarrow\) x²+4x-50=0
\(\Rightarrow\) (x+10)(x-3)=0
\(\Rightarrow\) x=-10, x=5
When x=-10, then other number
=-10+5=-5
When x=5, then other number
=5+5=10
Hence, numbers are -10, -5 or 5, 10.

Let digit at unit's place be x and digit at ten's place by y
Number=10y+x    ....(i)
According to given condition,
10y+x=6(x+y)
\(\Rightarrow\) 10y+x=6x+6y
\(\Rightarrow\)   4y=5x y=\(\frac {3}{4}\)x   .....(ii)
Also 10y+x=3xy-6
\(\Rightarrow \quad 10\times \frac { 5 }{ 4 } x+x=3x\times \frac { 5 }{ 4 } x-6\quad (using\quad (ii))\)
\(\Rightarrow \quad \frac { 27 }{ 2 } =\frac { 15{ x }^{ 2 }-24 }{ 4 } \)
\(\Rightarrow\) 54x=15x2-24
\(\Rightarrow\) 15x²-54x-24=0
\(\Rightarrow\) 5x²-18x-8=0
\(\Rightarrow\) 5x²-20x+2x-8=0
\(\Rightarrow\) 5x(x-1)+2(x-4)=0
\(\Rightarrow\) (x-4)(5x+2)=0
\(\Rightarrow\) x=4 or x=\(-\frac{2}{5}\)
Rejecting x=\(-\frac{2}{5}\), we have x=4 when
x=4, y=\(-\frac{2}{5}\) x r=4  [Using (i)]
\(\therefore\)  Number=10 x 5+4=54

For consecutive terms of AP
2(2k² + 3k + 6) = (3k² + 4k + 4) + (4k + 8)
\(\Rightarrow\)  4k² + 6k + 12 = 3k² + 8k + 12
\(\Rightarrow\)  k² - 2k = 0
\(\Rightarrow\)  k(k - 2) = 0
\(\Rightarrow\)  k = 0  or  k = 2

Let Ist term = a
Common difference, d = \(\frac{5}{4}\) (Given)
Also,  a₉ = - 6
\(\Rightarrow\) a + 8d = -6
\(\Rightarrow\)  a + 8 x \(\frac{5}{4}\) = - 6
\(\Rightarrow\)  a = -16
Now a₂₅ = a + 24d
 = -16 + 24 x \(\frac{5}{4}\) = 14

a_n = a + ( n - 1 ) d; a_k = a + ( k - 1 )d
Now,     a_n - a_k = [ a + ( n - 1 ) d ] - [ a + ( k - 1 ) d ] = ( n - 1 ) d - ( k - 1 ) d = ( n - 1 - k + 1 )d
\(\Rightarrow\)  a_n - a_k = ( n - k ) d  
(i) Here a₁₁ = 5, a₁₃ = 79
Taking n = 13, k = 11, eq. (i) becomes
a₁₃ - a₁₁ = ( 13 - 11 )d \(\Rightarrow\) 79 - 5 = 2d \(\Rightarrow\) d = 37
(ii) Let a₁₈ = x
\(\therefore\)      a₂₀ = x + 10
Taking n = 20 and k = 18, equation (i) becomes
a₂₀ - a₁₈ = (  20 - 18 ) d \(\Rightarrow\) ( x + 10 ) -x = 2d \(\Rightarrow\) d = 5

a = \(5\frac{1}{2}=\frac{11}{2},\) d = a₂ - a₁ = 11 - \(\frac{11}{2}=\frac{11}{2}\) and x = 550
A.T.Q., a_k = x
\(\Rightarrow\)   a + ( k - d ) = 550   \(\Rightarrow\) \(\frac{11}{2}+(k-1)\frac{11}{2}=550\)
\(\Rightarrow\)    \(\frac{11}{2}+\frac{11}{2}k-\frac{11}{2}=550\)  \(\Rightarrow\)   \(\frac{11}{2}\) k = 550
\(\Rightarrow\)    k = \(\frac{550\times2}{11}=100\)

Here a = 21, common difference d = 42 - 21 = 21.
Let       a_n = 420
Now,     a_n = a + (n - 1)d
\(\Rightarrow\)       420 = 21 + (n - 1)21
 n = 20

Let Ist term of AP = a and common difference = d.
Now,    a₆ = -10 \(\Rightarrow\)  a +  5d  = -10                 ..(i)
Also,     a₁₀ = -26  \(\Rightarrow\)    a + 9d = -26           ...(ii)
Subtract (i) from (ii),    
a + 9d = - 26
a + 5d = -10
-  -          +
 4d = -16   \(\Rightarrow\) d = - 4
Substituting in (i), we get
a + 5 x ( -4 ) = - 10  \(\Rightarrow\) a = 10
Now,  a₁₅ = a + 14d = 10 + 14 X  - 4 = - 46

The given quadratic equation is
3x²-2kx+m=0
when x=2 and m=3 eq.(i)becomes
3(2)²-2k(2)+3=0
12-4k+3=0
k=\(15\over 4\).

Since 2 is a root of the given equation.
(2)²+b(2)+12=0
16+2b=0
b=-8

\(\sqrt{x^2+4}=(x^2+1)^2\)
\(=x^2+1+2x^2\)
\(\sqrt{x^2+4}-1=x^4+2x^2\)
\(x^2+4+1-2\sqrt{x^2+4}=(x^4+2x^2)^2\)
It is not a quadratic equation.

(x+1)(x+3)=(x-1)(x-4)
x²+4x+3=x²-5x+4
9x-1=0
Degree of equation is 1.
It is not a quadratic equation.

Let speed of the train be x km/h
Total distance to be covered=480 km
Time=\(\frac {distance}{speed}=\frac{480}{x}\)hours
Decreased speed of the train=(x-8) km/h
Now, time=\(\frac {480}{x-8}\)hours
ATQ  \(\frac {480}{x-8}-\frac{480}{x}=3\)
\(\Rightarrow 480\left[ \frac { 1 }{ x-8 } -\frac { 1 }{ x } \right] =3\quad \Rightarrow \quad 480\left[ \frac { x-x+8 }{ x(x-8) } \right] =3\)
\(\Rightarrow\) 480 x 8=3x(x-8) \(\Rightarrow\) 3840=3x²-24x
\(\Rightarrow\)  3x²-24x-3840=0 x²-8x-1280=0
Which is the required quadratic equation.