We have
6x² – x – 2 = 6x² + 3x – 4x – 2
= 3x (2x + 1) – 2 (2x + 1)
= (3x – 2)(2x + 1)
The roots of 6x² - x - 2 = 0 are the values of x for which (3x - 2)(2x + 1) = 0
Therefore, 3x - 2 = 0 or 2x + 1 = 0,
ie., x = \(\frac{2}{3}\) or x = \(-\frac{1}{2}\)
Therefore, the roots of 6x² - x - 2 = 0 are \(\frac{2}{3}\) and \(-\frac{1}{2}\)
We verify the roots, by checking that \(\frac{2}{3}\) and \(-\frac{1}{2}\) satisfy 6x² - x - 2 = 0.

Given quadratic equation  is 
\((k-12)x^{ 2 }+2(k-12)x+2=0\)
On comparing with \(ax^{ 2 }+bx+x=0,\) we get
a=k-12,b=2(k-12) and c=2
Now, \(D=b^{ 2 }-4ac\)
=\([2(k-12)]^{ 2 }-4(k-12)(2)\)
\(=4(k-12)^{ 2 }-8(k-12)\)
=\((k-12)[4(k-12)-8]\)
= \((k-12)(4k-48-8)\)
=(k-12)(4k-56)

Let two successive integral multiples of 5
be  x and x + 5.
ATQ x (x + 5) = 300 
\(\Rightarrow x^{ 2 }+5x-300=0\)
\(\Rightarrow (x+20)(x-15)=0\)
\(\Rightarrow \) x=-20 or x=15
When x=-20, multiples are -20 and -20+5= -15
When x=15 , multiples are 15 and 15+5 =20

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

Let the two consecutive integers be x and x+1
ATQ x²+(x+1)2=365
\(\Rightarrow\) x²+x2+2x+1=365 \(\Rightarrow\) 2x²+2x-364=0
\(\Rightarrow\) x²+x-182=0 \(\Rightarrow\) x²+14x-13x-182=0
\(\Rightarrow\) x(x+14)-13(x+14)=0 \(\Rightarrow\) (x-13)(x+14)=0
\(\Rightarrow\) x=13, -14 (-14 is rejected because it is a negative integer)
Hence, the two consecutive positive integers are 13 and 13+1=14

Let the number of toys produced on that day be x.
Therefore, the cost of production (in rupees) of each toy that day = 55 – x
So, the total cost of production (in rupees) that day = x (55 – x)
Therefore, x (55 – x) = 750
i.e., 55x – x² = 750
i.e., – x² + 55x – 750 = 0
i.e., x² – 55x + 750 = 0
Therefore, the number of toys produced that day satisfies the quadratic equation
x² – 55x + 750 = 0
which is the required representation of the problem mathematically.

Here a = 3, b = - 2 and \(c=\frac{1}{3} \text { . }\)
Therefore, discriminant b² - 4ac = (-2)² - 4 x 3 x \(\frac{1}{3}\) = 4 - 4 = 0.
Hence, the given quadratic equation has two equal real roots.
The roots are \(\frac{-b}{2 a}, \frac{-b}{2 a}, \text { i.e., } \frac{2}{6}, \frac{2}{6}, \text { i.e., } \frac{1}{3}, \frac{1}{3} .\)

k = 3

(1+p)x²+2(1+2p)x+(1+p)=0
Here a=1+p,b=2(1+2p),c=(1+p)
D=b²-4ac=[2(1+2p)²-4(1+p)(1+p)
=4(1+4p²+4p)-4(1+p²+2p)
=4+16p²+16p-4-4p²-8p
=12p²+8p
For coincident roots i.e. equal roots
D=0
\(\Rightarrow \) 12p²+8p=0 \(\Rightarrow \) 4p(3p+2)=0
\(\Rightarrow \) p=0 or p=\(\frac { -2 }{ 3 } \)

Let breadth of a rectangular mango grove be x m
Then, length of a rectangular mango grove = 2x m
According to the question,
Area of rectangular mango grove = 800 m²
\(\Rightarrow\) 2x(x) = 800  [\(\because\) Area = length \(\times\) breadth]
\(\Rightarrow\) 2x² = 800 \(\Rightarrow\) x² = 400
\(\Rightarrow\) x = \(\pm\)20
But, x = -20 is not possible because breadth can never be negative. so, x = 20.
Thus, length = 2x = 40 m and breadth = 20 m.

\(3 x^{2}-4 \sqrt{3} x+4=0\)
Comparing it with ax² + bx + c = 0, we get
a = 3, b = \(-4 \sqrt{3}\)  and c = 4
Discriminant = b² - 4ac
\(=(-4 \sqrt{3})^{2}-4(3)(4)\)
= 48 - 48 = 0
As b² - 4ac = 0,
Therefore, real roots exist for the given equation and they are equal to each other.
And the roots will be \(\frac{-b}{2 a}\)
Therefore, the roots are \(\frac{2}{\sqrt{3}} \text { and } \frac{2}{\sqrt{3}}\)

Let us first split the middle term - 5x as -2x -3x [because (-2x) x (-3x) = 6x² = (2x² ) x 3].
So, 2x² - 5x + 3 = 2x² - 2x - 3x + 3 = 2x (x - 1) -3(x - 1) = (2x - 3)(x - 1)
Now, 2x² - 5x + 3 = 0 can be rewritten as (2x - 3)(x - 1) = 0.
So, the values of x for which 2x² - 5x + 3 = 0 are the same for which (2x - 3)(x - 1) = 0,
i.e., either 2x - 3 = 0 or x - 1 = 0.
Now, 2x - 3 = 0 gives x = \(\frac{3}{2}\) and x - 1 = 0 gives x = 1.
So, x = \(\frac{3}{2}\) and x = 1 are the solutions of the equation.
In other words, 1 and \(\frac{3}{2}\) are the roots of the equation 2x² - 5x + 3 = 0.
Verify that these are the roots of the given equation.
Note that we have found the roots of 2x² - 5x + 3 = 0 by factorising 2x² - 5x + 3 into two linear factors and equating each factor to zero.

Given equations is \(x^{ 2 }-3x-10=0\)
\(\Rightarrow x^{ 2 }-5x+2x-10=0\)   
\(\left[ \because -5x(2)=-10\\ and-5+2=-3 \right] \)
\(\Rightarrow x(x-5)+2(x-5)=0\)
\(\Rightarrow x-5=0\quad \Rightarrow x=5\)
\(\Rightarrow x+2=0\Rightarrow x=-2\)
Hence, the roots of the equation 
\(x^{ 2 }-3x-10=0\quad are-2\quad and\quad 5\)

Given equation is 
\((x+2)^{ 2 }=2x(x^{ 2 }-1)\)
\(\Rightarrow x^{ 3 }+8+3x^{ 2 }(2)+3x(2)^{ 2 }=2x^{ 3 }-2x\)
\(\left[ \because (a+b)^{ 3 }=a^{ 3 }+b^{ 3 }+3a^{ 2 }b+3ab^{ 2 } \right] \)
\(\Rightarrow x^{ 2 }+8+6x^{ 2 }+12x-2x^{ 2 }-2x\)
\(\Rightarrow -x^{ 2 }+8+6x^{ 2 }+12x-2x^{ 3 }+2x=0\)
Which is not of the form \(ax^{ 2 }+bx+c=0\)
because it has cubic term i.e.\(x^{ 3 }\)
Hence it is not a quadratic equation.

6, 11

(i) (a) Since, Ajay's car travels a distance in one hour is (x + 5) km. Therefore, Ajay's car travels a distance two hours is 2(x + 5) km.
(ii) (c) \(\because \text { Time }=\frac{\text { Distance }}{\text { Speed }}\)
Time taken by Ajay and Raj to complete the 400 km journey,
\(t_1=\frac{400}{x+5} \mathrm{~h} \)
and \(t_2=\frac{400}{x} \mathrm{~h}\)
According to the question,
t₂ = t₁ + 4
\( \therefore \frac{400}{x}=\frac{400}{x+5}+4\)
\(\Rightarrow \frac{100}{x}=\frac{100}{x+5}+1\) [dividing by 4]
\(\Rightarrow\) 100(x + 5) = 100x + x(x + 5)
\(\Rightarrow\) 100x + 500 = 100x + x² + 5x
\(\Rightarrow\) x²  +5x - 500 = 0
(iii) (a) Consider the quadratic equation x² + 5x - 500 = 0
On comparing with ax² + bx + c =0, we get
a = 1, b = 5 and c = -500
\(\begin{aligned} \because \quad x & =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{-5 \pm \sqrt{(5)^2-4 \times(1)(-500)}}{2 \times 1} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{-5 \pm \sqrt{25+2000}}{2}=\frac{-5 \pm \sqrt{2025}}{2} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{-5 \pm 45}{2}=\frac{-50}{2}, \frac{40}{2}=-25,20 \end{aligned}\)
Since, speed cannot be negative, so we consider only x = 20.
Hence, the speed of Raj's car is 20 km/h.
(iv) (d) To travel 400 km, time taken by Ajay
\(\begin{aligned} t_1 & =\frac{400}{(x+5)} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{400}{20+5}=\frac{400}{25}=16 \mathrm{~h} \end{aligned}\)

(i) (b):We have \(6 x^{2}+x-2=0\)
\(\Rightarrow \quad 6 x^{2}-3 x+4 x-2=0 \)
\(\Rightarrow \quad(3 x+2)(2 x-1)=0 \)
\(\Rightarrow \quad x=\frac{1}{2}, \frac{-2}{3}\)
(ii) (c): \(2 x^{2}+x-300=0\)
\(\Rightarrow \quad 2 x^{2}-24 x+25 x-300=0 \)
\(\Rightarrow \quad(x-12)(2 x+25)=0 \)
\(\Rightarrow \quad x=12, \frac{-25}{2}\)
(iii) (d): \(x^{2}-8 x+16=0\)
\(\Rightarrow(x-4)^{2}=0 \Rightarrow(x-4)(x-4)=0 \Rightarrow x=4,4\)
(iv) (d): \(6 x^{2}-13 x+5=0\)
\(\Rightarrow \quad 6 x^{2}-3 x-10 x+5=0 \)
\(\Rightarrow \quad(2 x-1)(3 x-5)=0 \)
\(\Rightarrow \quad x=\frac{1}{2}, \frac{5}{3}\)
(v) (a): \(100 x^{2}-20 x+1=0\)
\(\Rightarrow(10 x-1)^{2}=0 \Rightarrow x=\frac{1}{10}, \frac{1}{10}\)
 

(i) (b): Let two consecutive integers be x, x + 1.
Given, x² + (x + 1)² = 650
\(\begin{array}{l} \Rightarrow 2 x^{2}+2 x+1-650=0 \\ \Rightarrow 2 x^{2}+2 x-649=0 \end{array}\)
(ii) (c): Let the two numbers be x and 15 - x.
Given, \(\frac{1}{x}+\frac{1}{15-x}=\frac{3}{10}\)
\(\begin{array}{l} \Rightarrow 10(15-x+x)=3 x(15-x) \\ \Rightarrow 50=15 x-x^{2} \Rightarrow x^{2}-15 x+50=0 \end{array}\)
(iii) (d): Let the numbers be x and x + 3.
Given, x(x + 3) = 504
\(\Rightarrow\) x² + 3x - 504 = 0
(iv) (c): Let the number be x.
According to question, x² - 84 = 3(x + 8)
\(\Rightarrow x^{2}-84=3 x+24 \Rightarrow x^{2}-3 x-108=0\)
(v) (d): Let the number be x.
According to question, x + 12 = \(\frac {160}{x}\)
\(\Rightarrow x^{2}+12 x-160=0\)