As 2 is a root of the equation x2 + 3kx + 2 = 0
\(\Rightarrow\)   (2)²+ 3k(2) + 2 = 0
\(\Rightarrow\)   4 + 6k + 2 = 0
\(\Rightarrow\)   6K = - 6
\(\Rightarrow\)   K = - 1

24, 21 ; 19,26

Original price =  Rs 50 ; Honesty

\(- {\sqrt {2}\over 6} , \sqrt {2}\)

\(x^{2}+2x - 168 = 0\)

Here \(\sqrt { 2x+9 } +x=13\)
\(\Rightarrow \sqrt { 2x+9 } =13-x\)
On squaring both side, we get
\(\left( \sqrt { 2x+9 } \right) ^{ 2 }=(13-x)^{ 2 }\)
\(\Rightarrow 2x+9=169+x^{ 2 }-26x\)
\(\Rightarrow x^{ 2 }-28x+160=0\)
\(\Rightarrow x^{ 2 }-20x-8x+160=0\)
\(\Rightarrow (x-20)(x-8)=0\)
\(\Rightarrow \) x = 20 or x = 8

Here a= \(\alpha -12,b=2(\alpha -12),c=2\)
For equal roots D=0 \(\Rightarrow \) b²-4ac=0
\(\Rightarrow \) [2( \(\alpha -12)]^{ 2 }\) -4X[2 \(\alpha -12)]=0\)
\(2(\alpha -12)[2\alpha -12)]=0\)
\(\Rightarrow \left( \alpha -12 \right) [(2\alpha -12)-4]=0\)
\(\Rightarrow \alpha =12,14\)
\(\alpha =12\) is not possible take \(\alpha =14\)

The given equation is 4ax²+4bx+3=0. Let \(\alpha \) and \(\beta \) are the roots of equation 
Here A=4a,B= 4b, C=3
Sum of roots = \(-\frac { B }{ A } =-\frac { 4b }{ 4a } =\frac { -b }{ a } \)
\(\therefore -\frac { b }{ a } =\frac { 1 }{ 2 } \) [Given]  \(\Rightarrow \) -2b=a  ....(i)
Product of roots = \(\frac { C }{ A } \Rightarrow \frac { 3 }{ 16 } =\frac { 3 }{ 4a } \)
\(\Rightarrow 4a=16\Rightarrow a=4\)
Putting in equation  (i) ,we get b = -2
Hence a=4 and b=-2

4x²-2(a²+b²)x+a²b²=0
4x²-2a²x-2b²x+a²b²=0
2x(2x-a2)-b²(2x-a²)=0
(2x-a²)(2x-b²)=0
\(x={a^2\over2}, x={b^2\over 2}\)
\(x={-B\pm\sqrt D\over 2A}={4a^2\pm\sqrt{16b^4}\over2X4}\)
\(={4a^2+4b^2\over8},{4a^2-4b^2\over 8}\)
\(={a^2+b^2\over 2},{a^2-b^2\over 2}\)

Given equation is x²+5x+5=0
Here a=1, b=5, c=5.
D=b²-4ac \(\Rightarrow\) D=(5)²-4 x 1 x 5 \(\Rightarrow\) D=5>0
\(\therefore\) solution is given by
\(x=\frac { -b+\sqrt { D } }{ 2a } ,​​\frac { -b-\sqrt { D } }{ 2a } \quad \Rightarrow \quad \frac { -5+\sqrt { 5 } }{ 2\times 1 } ,\frac { -5-\sqrt { 5 } }{ 2\times 1 } \)  are solutions

This is of the form ax²+bx+c=0, where a=2, b=1 and c=-4
Discriminant(D)=b^2-4ac=(1)₂-4x2(-4)=1+32=33>0
Let roots be \(\alpha\) and \(\beta\)
\(\alpha =\frac { -b+\sqrt { D } }{ 2a } =\frac { -1+\sqrt { 33 } }{ 4 } \)
\(\beta =\frac { -b-\sqrt { D } }{ 2a } =\frac { -1-\sqrt { 33 } }{ 4 } \)
Hence, the roots are \(\frac { -1+\sqrt { 33 } }{ 4 } ,\frac { -1-\sqrt { 33 } }{ 4 } \)

\(2x^2-x+{1\over 8}=0\)
\({16x^2-8x+1\over 8}=0\)
16x²-4x-4x+1=0
4x(4x+1)-1(4x-1)=0
(4x-1)(4x-1)=0
4x-1=0 or  4x-1=0
\(x={1\over 4}\ or \ x={1\over 4}\)
Hence the roots are, \({1\over 4},{1\over 4}\)

LHS=2x²+5x-12
When x=-4
LHS =2(-4)²+5(-4)-12
=32-20-12=0=RHS
x=-4 is a solution of the given equation.

(x-1)(x+2)=x+3
x²+x-2=x+3
x²-5=0
Here degree of the equation is 2.
It is a quadratic equation.

x²-2x-8=0
When x=-2, LHS=(-2)²-2(-2)+8=4+4+8=16≠0
x=-2 is not a solution of the given equation.