Let P be the required location of the pole. Let the distance of the pole from the gate B be x m, i.e., BP = x m. Now the difference of the distances of the pole from the two gates = AP - BP (or, BP - AP) = 7 m.

Therefore, AP = (x + 7) m.
Now, AB = 13m, and since AB is a diameter,
∠APB = 90°
Therefore, AP² + PB² = AB² (By Pythagoras theorem)
i.e., (x + 7)² + x² = 13²
i.e., x² + 14x + 49 + x² = 169
i.e., 2x² + 14x - 120 = 0
So, the distance ‘x’ of the pole from gate B satisfies the equation
x² + 7x - 60 = 0
So, it would be possible to place the pole if this equation has real roots. To see if this is so or not, let us consider its discriminant. The discriminant is
b² - 4ac = 7² - 4 x 1 x (-60) = 289 > 0.
So, the given quadratic equation has two real roots, and it is possible to erect the pole on the boundary of the park.
Solving the quadratic equation x² + 7x - 60 = 0, by the quadratic formula, we get
\(x=\frac{-7 \pm \sqrt{289}}{2}=\frac{-7 \pm 17}{2}\)
Therefore, x = 5 or -12.
Since x is the distance between the pole and the gate B, it must be positive.
Therefore, x = -12 will have to be ignored. So, x = 5.
Thus, the pole has to be erected on the boundary of the park at a distance of 5m from the gate B and 12m from the gate A.

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.

2x² + kx + 3 = 0
Comparing equation with ax² + bx + c = 0, we get
a = 2, b = k and c = 3
Discriminant = b² - 4ac = (k)² - 4(2) (3)
= k² - 24
For equal roots,
Discriminant = 0
k² - 24 = 0
k² = 24
\(k=\pm \sqrt{24}=\pm 2 \sqrt{6}\)

Let the number of marbles John had be x
Then the number of marbles Jivanti had be = 45 – x (Why?).
The number of marbles left with john, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marble = 45 – x – 5 = 40 – x
Therefore, their product = (x – 5) (40 – x)
= 40x – x² – 200 + 5x
= – x² + 45x – 200
So, – x² + 45x – 200 = 124 (Given that product = 124)
i.e., – x² + 45x – 324 = 0
i.e., x² – 45x + 324 = 0
Therefore, the number of marbles John had, satisfies the quadratic equation
x² – 45x + 324 = 0
which is the required representation of the problem mathematically

(i) LHS = (x – 2)² + 1 = x² – 4x + 4 + 1 = x2 – 4x + 5
Therefore, (x – 2)² + 1 = 2x – 3 can be rewritten as
x² – 4x + 5 = 2x – 3
i.e., x² – 6x + 8 = 0
It is of the form ax² + bx + c = 0.
Therefore, the given equation is a quadratic equation
(ii) Since x(x + 1) + 8 = x² + x + 8 and (x + 2)(x – 2) = x² – 4
Therefore, x² + x + 8 = x² – 4
i.e., x + 12 = 0
It is not of the form ax² + bx + c = 0.
Therefore, the given equation is not a quadratic equation
(iii) Here, LHS = x (2x + 3) = 2x² + 3x
So, x (2x + 3) = x² + 1 can be rewritten as
2x² + 3x = x² + 1
Therefore, we get x² + 3x – 1 = 0
It is of the form ax² + bx + c = 0.
So, the given equation is a quadratic equation
(iv) Here, LHS = (x + 2)³ = x3 + 6x² + 12x + 8
Therefore, (x + 2)³ = x³ – 4 can be rewritten as
x³ + 6x² + 12x + 8 = x3 – 4
i.e., 6x² + 12x + 12 = 0 or, x² + 2x + 2 = 0
It is of the form ax² + bx + c = 0.
So, the given equation is a quadratic equation

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 \(\frac{-b}{2 a}, \frac{-b}{2 a}, \text { i.e., } \frac{2}{6}, \frac{2}{6}, \text { i.e., } \frac{1}{3}, \frac{1}{3} .\)

Let the breadth of the park be x m.
Given, perimeter of a rectangular park = 80 m
\(\Rightarrow\) 2(Length + Breadth) = 80 m
\(\Rightarrow\) Length + Breadth = 40 m
\(\Rightarrow\) Length = (40 - x)m
Now, area of a rectangular park = Length \(\times\)Breadth
=(40 - x)x m²
But according to the question, area of the rectangular park is 400 m².
\(\therefore\) (40 - x) x = 400
\(\Rightarrow\) x² - 40x + 400 = 0
\(\Rightarrow\) x² - 2x \(\times\) 20 + (20)² = 0
\(\Rightarrow\) (x - 20)² = 0 [\(\because\) a² - 2ab + b² = (a - b)²]
\(\Rightarrow\) x = 20
thus, breadth of the park = 20 m
and length of the park = 40 - 20 = 20 m
Hence, it is possible to design the rectangular park having perimeter 80 m and area 400 m with equal length and breadth, i.e. 20 m each.

\(3 x^{2}-4 \sqrt{3} x+4=0\)
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 \(\frac{2}{\sqrt{3}} \text { and } \frac{2}{\sqrt{3}}\)

Given equation is 2x² + x - 6 = 0
\(\Rightarrow\) 2x² + 4x - 3x - 6 = 0
[\(\because\) 4 \(\times\)(-3) = -12 and 4 - 3 = 1]
\(\Rightarrow\) 2x (x + 2) -3 (x + 2) = 0
\(\Rightarrow\) (x + 2)(2x - 3) = 0
\(\Rightarrow\) x + 2 = 0 or 2x - 3 = 0
\(\Rightarrow\)   x = -2  or \(x=\frac{3}{2}\)
Hence, the roots of the equation
2x² + x - 6 = 0 are -2 and \(\frac{3}{2}\)

Given equation (x-3)(2x+1)=x(x+5)
2x² + x - 6x - 3 = x² + 5x
\(\Rightarrow\) 2x² + x - 6x - 3 - x² - 5x = 0
\(\Rightarrow\) x² - 10x - 3 = 0 has its highest power 2
Hence it is a quadratic equation.

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.

\(3 x^{2}-2 \sqrt{6} x+2=3 x^{2}-\sqrt{6} x-\sqrt{6} x+2\)
\(=\sqrt{3} x(\sqrt{3} x-\sqrt{2})-\sqrt{2}(\sqrt{3} x-\sqrt{2})\)
\(=(\sqrt{3} x-\sqrt{2})(\sqrt{3} x-\sqrt{2})\)
So, the roots of the equation are the values of x for which
\((\sqrt{3} x-\sqrt{2})(\sqrt{3} x-\sqrt{2})=0\)
Now, \(\sqrt{3} x-\sqrt{2}=0 \text { for } x=\sqrt{\frac{2}{3}}\)
So, this root is repeated twice, one for each repeated factor \(\sqrt{3} x-\sqrt{2}\)
Therefore, the roots of 3x² - 2\(\sqrt6\)x + 2 = 0 are \(\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}\)

Let the number of pottery articles produced on a particular day be x.
Then, cost of production of each article = Rs (2x + 3)
[by given condition]
So, the total cost of production = Number of pottery articles \(\times\) Cost of production of each article
= Rsx(2x + 3)
According to the question,
x(2x + 3) = 90
\(\Rightarrow\) 2x² + 3x = 90
\(\Rightarrow\) 2x² + 3x - 90 = 0
\(\Rightarrow\) 2x²  +15x - 12x - 90 = 0
[\(\because\) 15 \(\times\)(-12) = -180 and 15 - 12 = 3]
\(\Rightarrow\) x(2x + 15) - 6(2x + 15) = 0
\(\Rightarrow\) (2x + 15) (x - 6) = 0
\(\Rightarrow\) 2x + 15 = 0 or x - 6 = 0
\(\Rightarrow\) \(x=-\frac{15}{2}\) or x = 6
But x cannot be negative, as number of pottery articles should be positive.
\(\therefore\) x = 6
Hence, the number of articles produced = 6
and the cost of each article = 2 \(\times\) 6 + 3 = Rs 15

Let base be x cm, then height be (x-7) cm
By Pythagoras Theorem, (base)² + (height)²=(hypotenuse)²
\(\Rightarrow\) x²+(x-7)2=(13)² \(\Rightarrow\) x²+x²+49-14x=169
\(\Rightarrow\) 2x²-14x-120=0 \(\Rightarrow\) x²-7x-60=0
\(\Rightarrow\) x²-12x+5x-60=0 \(\Rightarrow\) x(x-12)+5(x-12)=0
\(\Rightarrow\) (x+5)(x-12)=0 \(\Rightarrow\) x-12=0 or x+5=0
\(\Rightarrow\) x=12 or x=-5 (-5 is rejected as sides can never be negative)
\(\Rightarrow\) Base=12 cm and altitude=12-7=5 cm

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.