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.

we found that if the breadth of the hall is x m, then x satisfies the equation 2x² + x - 300 = 0. Applying the factorisation method, we write this equation as
2x² - 24x + 25x - 300 = 0
2x (x - 12) + 25 (x - 12) = 0
i.e., (x - 12)(2x + 25) = 0
So, the roots of the given equation are x = 12 or x = - 12.5. Since x is the breadth of the hall, it cannot be negative.
Thus, the breadth of the hall is 12 m. Its length = 2x + 1 = 25 m.

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.

Let the breadth of the plot be x m.
Then, the length of the plot be (2x+1)m.
[by given condition]
\(\therefore\) Area of the rectangular plot=Length x breadth 
\(\therefore\) 528 = (2x + 1)x    [given]
\(\Rightarrow\) 528=2x²+x
\(\Rightarrow\) 2x² + x - 528 = 0,
which is the required quadratic equation.

kx(x - 2) + 6 = 0
or kx² - 2kx + 6 = 0
Comparing this equation with ax² + bx + c = 0, we get
a = k, b = - 2k and c = 6
= ( - 2k)² - 4 (k) (6)
= 4k² - 24k
For equal roots,
b² - 4ac = 0
4k² - 24k = 0
4k (k - 6) = 0
Either 4k = 0 or
k = 6 = 0
k = 0 or k = 6
However, if k = 0, then the equation will not have the terms 'x²' and 'x'.
Therefore, if this equation has two equal roots, k should be 6 only.

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
\(k=\pm \sqrt{24}=\pm 2 \sqrt{6}\)

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 
\(x^{ 2 }-3x-10=0\quad are-2\quad and\quad 5\)

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

b² - 4ac = (2)² - 4 x 1 x 3 = - 8 < 0
\(\Rightarrow\) Given equation has no real roots.

For equal roots
b² - 4ac = 0 ⇒ (-2k)² - 4 x 4 x k =0
\(\Rightarrow\) 4k² - 16k = 0 \(\Rightarrow\) k = 0, 4