Since 1 is a root of equation ay²+ay+3=0,
\(\therefore\) a(1)²+a(1)+3=0
\(\Rightarrow\) 2a+3=0 \(\Rightarrow\) a=\(\frac {-3}{2}\)
Also, 1 is a root y²+y+b=0
\(\therefore\) (1)²+(1)+b=0
\(\Rightarrow\) 2+b=0 \(\Rightarrow\) b=-2
Now, ab=\(​​\left( \frac { -3 }{ 2 } \right) \)(-2)=3
Hence, the value of ab is 3.

Since p and q are the roots of x²-px+q=0
\(\therefore \) Sum of roots=p+q=p is q=0
Also, product of roots=pq=q
\(\Rightarrow\)  pq-q=0
\(\Rightarrow\)  q(p-1)=0
 \(\Rightarrow\) q=0 or p=1
Hence, the value of p and q are 1 and 0 respectively.

Here, \(\sqrt { x+\sqrt { x+\sqrt { x+....\infty } } } =6\)
\(\Rightarrow \quad \sqrt { x+6 } =6\)
Squaring both sides, we have
x+6=36
\(\Rightarrow\) x=36-6=30
Hence, the value of x is 30.

Given, equation is (x+2)³=2x(x²-1)
\(\Rightarrow\) x³ + 8 + 3x²(2) + 3x (2)² = 2x³ - 2x
[\(\because\) (a + b)³ = a³ + b³ + 3a²b + 3ab²]
\(\Rightarrow\) x³ + 8 + 6x² + 12x = 2x³ - 2x
\(\Rightarrow\) x³ + 8 + 6x² + 12x - 2x³ + 2x = 0
\(\Rightarrow\) -x³ + 6x² + 14x + 8 = 0
which is not of the form ax² + bx + c = 0,
because it has cubic term, i.e. x³.
\(\therefore\) It is not a quadratic equation.

(2x-1)(x-3)=(x+5)(x-1)
2x²-6x-x+3=x²-x+5x-5
2x²-6x-x+3-x²+x-5x+5=0
x²-11x+8=0
Which is of the form ax²+bx+c=0. Hence the given equation is a quadratic equation.

Let Integers be x,x+1 and x+2
ATQ x(x+1) (x+2) =16(x+x+1+x+2)
\(\Rightarrow (x^{ 2 }+x)(x+2)=16(3x+3)\)
\(\Rightarrow x^{ 3 }+2x^{ 2 }+x^{ 2 }+2x=48+48 \)
\(\Rightarrow x^{ 3 }+3x^{ 2 }-46x-48=0\)
LHS when x=-1, we have
(x+1) is a factor
For others factor

\(\therefore x^{ 3 }+3x^{ 2 }-46x-48=0\)
\(\Rightarrow (x+1)(x^{ 2 }+2x-48=0)\)
\(\Rightarrow (x+1)(x+8)(x-6)=0\)
\(\Rightarrow x=-1,x=-8,x=6\)
Rejecting -ve values x = 6 \(\therefore \) positive integers are 6,

Let number of days taken by B to finish a work = x
Number of days taken by A to finish a work = (x — 6) 
Number of days taken by A and B together = 4
Now
As one day work + B's one day work = one day's work of A and B together 
\(\Rightarrow \frac { 1 }{ x-6 } +\frac { 1 }{ x } =\frac { 1 }{ 4 } \Rightarrow \frac { x+x-6 }{ (x-6)x } =\frac { 1 }{ 4 } \)
\(\Rightarrow 8x-24=x^{ 2 }-6x\Rightarrow x^{ 2 }-14x+24=0\)
\(\Rightarrow (x-12)(x-2)=0\Rightarrow x=12,x=2\) (not possible)
Number of days taken by B to finish the work = 12 

Let consecutive positive integers be n, + I and n + 2
According to the question, 
n²+(n+1)(n+2) =46 \(\Rightarrow \) 2n²+3n-44=0
\(\Rightarrow 2n^{ 2 }+11n-8n-44=0\Rightarrow n(2n-11)-4(2n+11)=0\)
\(\Rightarrow (n-4)(2n+11)=0\Rightarrow n-4=0\) or 2n+11=0
\(\Rightarrow n=4\) or \(n=\frac { 11 }{ 2 } \)
Integers are 4,5,6 

Let digit at ten's place be x
and digit at unit's place be y
Also xy=15  \(\Rightarrow x=\frac { 15 }{ y } \) ....(i)
ATQ 
10x+y+18=10y+x
\(\Rightarrow \) 9x-9y+18=0
\(\Rightarrow \) x-y+2=0
\(\Rightarrow \frac { 15 }{ y } -y+2=0\) [From (i)]
\(\Rightarrow \) 15-y²+2y=0
\(\Rightarrow \) y²-2y-15=0
\(\Rightarrow \) (y-5) (y+3) =0
\(\Rightarrow \) y=5, y=3 [y=-3 rejected]
On putting the value of y= 5 in equation
 (i). we get 
\(x=\frac { 15 }{ 5 } =3\)
\(\therefore \) Number = 3X10+5 = 35

x²+5x-(a²+a-6)=0
D=(5)²-4x1x[-(a²+a-6)]
=25+4a²+4a-24
=4a²+4a+1=(2a+1)²
\(\therefore x=\frac { -b\pm \sqrt { D } }{ 2a } =\frac { -5\pm \sqrt { (2a+1) } ^{ 2 } }{ 2\times 1 } \)
\(=\frac { -5\pm (2a+1) }{ 2 } =\frac { -4+2a }{ 2 } ,\frac { -6-2a }{ 2 } \)
=-2+a,-3-a