Mix Examples - Quadratic Equations Questions in English

(D) Comparing the given equation $3x^{2} + x - 2 = 0$ with the standard form $ax^{2} + bx + c = 0$,we get:
$a = 3, b = 1, c = -2$.
The discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Substituting the values:
$D = (1)^{2} - 4(3)(-2)$
$D = 1 + 24 = 25$.
Since $D = 25 > 0$,the discriminant is positive and a perfect square.
Therefore,the quadratic equation has two distinct real and rational roots.

(N/A) Comparing the given equation $x^{2}+5x+5=0$ with the standard quadratic form $ax^{2}+bx+c=0$,we get:
$a=1, b=5, c=5$
The discriminant $D$ is given by the formula $D = b^{2}-4ac$.
Substituting the values:
$D = (5)^{2} - 4(1)(5)$
$D = 25 - 20$
$D = 5$
Since $D > 0$,the quadratic equation has two distinct real roots. Furthermore,since $5$ is not a perfect square,the roots are irrational.

(B) Comparing the given equation $4x^{2} + x - 3 = 0$ with the standard form $ax^{2} + bx + c = 0$:
Here,$a = 4$,$b = 1$,and $c = -3$.
The discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Substituting the values,we get $D = (1)^{2} - 4(4)(-3)$.
$D = 1 + 48 = 49$.
Since $D = 49 > 0$,the quadratic equation has two distinct real roots.
Furthermore,since $49$ is a perfect square,the roots are rational.

(D) Comparing the given equation $4x^{2} + 11x + 10 = 0$ with the standard form $ax^{2} + bx + c = 0$,we get:
$a = 4, b = 11, c = 10$
The discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Substituting the values:
$D = (11)^{2} - 4(4)(10)$
$D = 121 - 160$
$D = -39$
Since $D < 0$,the quadratic equation has no real roots.

(B) Comparing the given equation $4x^{2} + 12x + 9 = 0$ with the standard form $ax^{2} + bx + c = 0$,we get:
$a = 4, b = 12, c = 9$.
The discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Substituting the values:
$D = (12)^{2} - 4(4)(9)$
$D = 144 - 144$
$D = 0$.
Since the discriminant $D = 0$,the quadratic equation has real and equal (repeated) roots.

(D) Comparing the given equation $x^{2}-2 \sqrt{2} x+1=0$ with the standard form $ax^{2}+bx+c=0$,we get:
$a=1, b=-2 \sqrt{2}, c=1$
The discriminant $D$ is given by the formula $D=b^{2}-4ac$.
Substituting the values:
$D=(-2 \sqrt{2})^{2}-4(1)(1)$
$D=(4 \times 2)-4$
$D=8-4=4$
Since $D > 0$,the quadratic equation has two distinct real roots.

(C) Comparing the equation $(k+1) x^{2}-2(k+3) x+(2 k+3)=0$ with the standard form $ax^{2}+bx+c=0$,we get:
$a = k+1, b = -2(k+3), c = 2k+3$
Since the quadratic equation has two equal and real roots,the discriminant $D$ must be zero.
$D = b^{2} - 4ac = 0$
$(-2(k+3))^{2} - 4(k+1)(2k+3) = 0$
$4(k^{2} + 6k + 9) - 4(2k^{2} + 3k + 2k + 3) = 0$
Divide by $4$:
$(k^{2} + 6k + 9) - (2k^{2} + 5k + 3) = 0$
$k^{2} + 6k + 9 - 2k^{2} - 5k - 3 = 0$
$-k^{2} + k + 6 = 0$
$k^{2} - k - 6 = 0$
$(k-3)(k+2) = 0$
Therefore,$k = 3$ or $k = -2$.

(D) Comparing the given quadratic equation $x^{2}-(k+10)x+9(k+1)=0$ with the standard form $ax^{2}+bx+c=0$,we get:
$a=1, b=-(k+10), c=9(k+1)$.
Since the quadratic equation has two equal and real roots,the discriminant $D$ must be equal to $0$.
$D = b^{2}-4ac = 0$
Substituting the values:
$( -(k+10) )^{2} - 4(1)(9(k+1)) = 0$
$(k+10)^{2} - 36(k+1) = 0$
$k^{2} + 20k + 100 - 36k - 36 = 0$
$k^{2} - 16k + 64 = 0$
This is a perfect square trinomial:
$(k-8)^{2} = 0$
Taking the square root on both sides:
$k-8 = 0$
$k = 8$
Therefore,the value of $k$ is $8$.

(B) The given quadratic equation is $\frac{1}{4} x^{2} - 2x + 1 = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = \frac{1}{4}$,$b = -2$,and $c = 1$.
The discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Substituting the values: $D = (-2)^{2} - 4(\frac{1}{4})(1)$.
$D = 4 - 1 = 3$.
Since $D > 0$,the roots of the equation are real and distinct.

(D) The given equation is $x(x - 5) = 36$.
Expanding the equation: $x^2 - 5x = 36$.
Rearranging into the standard form $ax^2 + bx + c = 0$: $x^2 - 5x - 36 = 0$.
Here,$a = 1$,$b = -5$,and $c = -36$.
The discriminant $D$ is given by the formula $D = b^2 - 4ac$.
Substituting the values: $D = (-5)^2 - 4(1)(-36)$.
$D = 25 + 144 = 169$.
Since $D > 0$ and $169$ is a perfect square,the roots are real,rational,and distinct.

(D) For a quadratic equation of the form $ax^{2} + bx + c = 0$,the discriminant $D$ is given by $D = b^{2} - 4ac$.
Here,$a = 5$,$b = -6$,and $c = 2$.
Substituting these values into the formula: $D = (-6)^{2} - 4(5)(2)$.
$D = 36 - 40 = -4$.
Since the discriminant $D < 0$,the equation has no real roots.

(B) The given quadratic equation is $5x^{2} - 4\sqrt{5}x + 4 = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = 5$,$b = -4\sqrt{5}$,and $c = 4$.
The discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Substituting the values,we get $D = (-4\sqrt{5})^{2} - 4(5)(4)$.
$D = (16 \times 5) - 80 = 80 - 80 = 0$.
Since the discriminant $D = 0$,the roots of the quadratic equation are real and equal.

(A) The given quadratic equation is $3x^{2} - 18x + 27 = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = 3$,$b = -18$,and $c = 27$.
The discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Substituting the values: $D = (-18)^{2} - 4(3)(27)$.
$D = 324 - 324 = 0$.
Since the discriminant $D = 0$,the roots of the quadratic equation are real,rational,and equal.

(A) For the quadratic equation $ax^{2} + bx + c = 0$,the discriminant $D$ is given by $D = b^{2} - 4ac$.
Comparing $x^{2} + x - 3 = 0$ with the standard form,we have $a = 1, b = 1, c = -3$.
Substituting these values into the formula:
$D = (1)^{2} - 4(1)(-3)$
$D = 1 + 12 = 13$.
Since $D > 0$,the roots of the equation are real and distinct.

(A) The given quadratic equation is $x^{2} = 9$,which can be written in the standard form $ax^{2} + bx + c = 0$ as $x^{2} + 0x - 9 = 0$.
Here,$a = 1$,$b = 0$,and $c = -9$.
The discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Substituting the values,we get $D = (0)^{2} - 4(1)(-9) = 0 + 36 = 36$.
Since $D > 0$ and $D$ is a perfect square $(6^{2} = 36)$,the roots of the equation are real,rational,and distinct.

(D) The given quadratic equation is $x^{2}-2x-15=0$.
Comparing this with the standard form $ax^{2}+bx+c=0$,we get $a=1$,$b=-2$,and $c=-15$.
The discriminant $D$ is given by the formula $D = b^{2}-4ac$.
Substituting the values: $D = (-2)^{2} - 4(1)(-15) = 4 + 60 = 64$.
Since $D > 0$ and $D$ is a perfect square $(8^{2} = 64)$,the roots of the equation are real,rational,and distinct.

(A) For the quadratic equation $ax^{2}+bx+c=0$,the discriminant $D$ is given by $D = b^{2}-4ac$.
Comparing $4x^{2}-6x+2=0$ with the standard form,we get $a=4$,$b=-6$,and $c=2$.
Substituting these values into the formula: $D = (-6)^{2} - 4(4)(2) = 36 - 32 = 4$.
Since $D > 0$ and $D$ is a perfect square,the roots are real,rational,and distinct.

(D) For the quadratic equation $ax^{2} + bx + c = 0$,the discriminant $D$ is given by $D = b^{2} - 4ac$.
Comparing $x^{2} + x + \frac{1}{4} = 0$ with the standard form,we have $a = 1$,$b = 1$,and $c = \frac{1}{4}$.
Substituting these values into the formula: $D = (1)^{2} - 4(1)(\frac{1}{4}) = 1 - 1 = 0$.
Since the discriminant $D = 0$,the roots of the quadratic equation are real,rational,and equal.

(C) For a quadratic equation of the form $ax^{2} + bx + c = 0$,the roots are equal and real if the discriminant $D = b^{2} - 4ac = 0$.
Given the equation $k x^{2} - 24 x + 16 = 0$,we have $a = k$,$b = -24$,and $c = 16$.
Substituting these values into the discriminant formula:
$(-24)^{2} - 4(k)(16) = 0$
$576 - 64k = 0$
$64k = 576$
$k = \frac{576}{64}$
$k = 9$.
Therefore,the value of $k$ is $9$.

(D) For a quadratic equation of the form $ax^{2} + bx + c = 0$ to have two equal and real roots,its discriminant $D$ must be equal to $0$.
Here,$a = 1$,$b = 2k$,and $c = 4$.
The discriminant $D = b^{2} - 4ac$.
Setting $D = 0$,we get $(2k)^{2} - 4(1)(4) = 0$.
$4k^{2} - 16 = 0$.
$4k^{2} = 16$.
$k^{2} = 4$.
Taking the square root on both sides,we get $k = \pm 2$.

(A) For a quadratic equation of the form $ax^{2} + bx + c = 0$,the roots are equal and real if the discriminant $D = b^{2} - 4ac = 0$.
Given the equation $3x^{2} - 18x + k = 0$,we identify the coefficients: $a = 3$,$b = -18$,and $c = k$.
Substituting these values into the discriminant formula:
$D = (-18)^{2} - 4(3)(k) = 0$
$324 - 12k = 0$
$12k = 324$
$k = \frac{324}{12} = 27$.
Therefore,the value of $k$ is $27$.

(B) The given equation is $x(4 - kx) = 3 - 2x$.
Expanding the equation: $4x - kx^2 = 3 - 2x$.
Rearranging the terms to form the standard quadratic equation $ax^2 + bx + c = 0$: $kx^2 - 6x + 3 = 0$.
For a quadratic equation to have two equal and real roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$.
Here,$a = k$,$b = -6$,and $c = 3$.
Substituting these values into the discriminant formula: $(-6)^2 - 4(k)(3) = 0$.
$36 - 12k = 0$.
$12k = 36$.
$k = 3$.

(C) For a quadratic equation of the form $ax^{2} + bx + c = 0$,the roots are equal and real if the discriminant $D = b^{2} - 4ac = 0$.
Given the equation: $16x^{2} - 40x + \frac{k-1}{2} = 0$.
Here,$a = 16$,$b = -40$,and $c = \frac{k-1}{2}$.
Substituting these values into the discriminant formula:
$(-40)^{2} - 4(16)\left(\frac{k-1}{2}\right) = 0$
$1600 - 64\left(\frac{k-1}{2}\right) = 0$
$1600 - 32(k-1) = 0$
$1600 - 32k + 32 = 0$
$1632 = 32k$
$k = \frac{1632}{32} = 51$.
Thus,the value of $k$ is $51$.

(C) For a quadratic equation of the form $ax^{2} + bx + c = 0$,the roots are equal and real if the discriminant $D = b^{2} - 4ac = 0$.
Given equation: $x^{2} - kx + 25 = 0$.
Here,$a = 1$,$b = -k$,and $c = 25$.
Substituting these values into the discriminant formula:
$(-k)^{2} - 4(1)(25) = 0$
$k^{2} - 100 = 0$
$k^{2} = 100$
$k = \pm \sqrt{100}$
$k = 10$ or $k = -10$.
Thus,the values of $k$ are $10$ and $-10$.

(A) For a quadratic equation of the form $ax^{2} + bx + c = 0$,the roots are equal and real if the discriminant $D = b^{2} - 4ac = 0$.
Given equation: $5x^{2} - 2kx + 20 = 0$.
Here,$a = 5$,$b = -2k$,and $c = 20$.
Substituting these values into the discriminant formula:
$(-2k)^{2} - 4(5)(20) = 0$
$4k^{2} - 400 = 0$
$4k^{2} = 400$
$k^{2} = 100$
$k = \pm 10$.
Thus,the values of $k$ are $10$ and $-10$.

(A) For a quadratic equation $ax^2 + bx + c = 0$ to have two equal and real roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$.
Here,$a = (k+1)$,$b = -2(k-1)$,and $c = 1$.
Substituting these values into the discriminant formula:
$D = [-2(k-1)]^2 - 4(k+1)(1) = 0$
$4(k-1)^2 - 4(k+1) = 0$
Dividing by $4$:
$(k-1)^2 - (k+1) = 0$
$k^2 - 2k + 1 - k - 1 = 0$
$k^2 - 3k = 0$
$k(k-3) = 0$
Therefore,$k = 0$ or $k = 3$.

(C) For a quadratic equation $ax^2 + bx + c = 0$ to have two equal and real roots,the discriminant $D$ must be equal to $0$.
$D = b^2 - 4ac = 0$
Here,$a = 2k$,$b = -8$,and $c = k$.
Substituting these values into the discriminant formula:
$(-8)^2 - 4(2k)(k) = 0$
$64 - 8k^2 = 0$
$8k^2 = 64$
$k^2 = 8$
$k = \pm\sqrt{8}$
$k = \pm 2\sqrt{2}$
Thus,the values of $k$ are $-2\sqrt{2}$ and $2\sqrt{2}$.

(D) For a quadratic equation of the form $ax^{2} + bx + c = 0$,the roots are equal and real if the discriminant $D = b^{2} - 4ac = 0$.
Given the equation $4x^{2} - 5x + k = 0$,we have $a = 4$,$b = -5$,and $c = k$.
Substituting these values into the discriminant formula:
$D = (-5)^{2} - 4(4)(k) = 0$
$25 - 16k = 0$
$16k = 25$
$k = \frac{25}{16}$

(A) For a quadratic equation of the form $ax^{2} + bx + c = 0$,the roots are equal and real if the discriminant $D = b^{2} - 4ac = 0$.
Given equation: $k x^{2} - 2 \sqrt{5} x + 4 = 0$.
Here,$a = k$,$b = -2 \sqrt{5}$,and $c = 4$.
Substituting these values into the discriminant formula:
$(-2 \sqrt{5})^{2} - 4(k)(4) = 0$
$(4 \times 5) - 16k = 0$
$20 - 16k = 0$
$16k = 20$
$k = \frac{20}{16} = \frac{5}{4}$.

(B) In an isosceles right-angled triangle,the two sides containing the right angle are equal in length.
Let the length of each of these sides be $x \, cm$.
By the Pythagoras theorem,in a right-angled triangle,the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Therefore,$(x)^2 + (x)^2 = (8)^2$.
$2x^2 = 64$.
$x^2 = 32$.
$x = \sqrt{32} = \sqrt{16 \times 2} = 4 \sqrt{2} \, cm$.
Since the length cannot be negative,the length of each of the other two sides is $4 \sqrt{2} \, cm$.

(C) Let the breadth of the rectangle be $x\,cm$.
Then,by the given condition,the length of the rectangle is $(3x - 2)\,cm$.
Now,by the data,the area of the rectangle is $280\,cm^2$.
$\therefore$ Area of rectangle $= 280$
$\therefore$ Length $\times$ Breadth $= 280$
$\therefore (3x - 2)(x) = 280$
$\therefore 3x^2 - 2x = 280$
$\therefore 3x^2 - 2x - 280 = 0$
$\therefore (3x + 28)(x - 10) = 0$
$\therefore 3x + 28 = 0$ or $x - 10 = 0$
$\therefore x = -\frac{28}{3}$ or $x = 10$
As $x$ is the breadth of a rectangle,it cannot be negative.
$\therefore x = -\frac{28}{3}$ is not possible.
$\therefore x = 10$
Now,the length of the rectangle $= 3x - 2 = 3(10) - 2 = 28\,cm$.
Hence,the length of the given rectangle is $28\,cm$.

(D) Let the speed of the motorboat in still water be $x \, km/hr$.
Therefore,the speed of the motorboat going downstream is $(x+3) \, km/hr$ and its speed going upstream is $(x-3) \, km/hr$.
Hence,the time taken to cover $12 \, km$ downstream is $\frac{12}{x+3}$ hours and the time taken to cover the same distance,i.e.,$12 \, km$ to come back upstream is $\frac{12}{x-3}$ hours.
According to the problem,the total time taken is $3$ hours.
$\frac{12}{x+3} + \frac{12}{x-3} = 3$
$12(x-3) + 12(x+3) = 3(x+3)(x-3)$
$12x - 36 + 12x + 36 = 3(x^2 - 9)$
$24x = 3x^2 - 27$
$3x^2 - 24x - 27 = 0$
Dividing by $3$,we get $x^2 - 8x - 9 = 0$.
$(x-9)(x+1) = 0$
$x = 9$ or $x = -1$.
Since the speed cannot be negative,$x = 9 \, km/hr$.

(A) Let the smaller of two consecutive even positive integers be $x$.
Then,the greater even positive integer is $x+2$.
According to the problem,the sum of their squares is $244$.
$\therefore x^{2} + (x+2)^{2} = 244$
$\therefore x^{2} + x^{2} + 4x + 4 = 244$
$\therefore 2x^{2} + 4x + 4 - 244 = 0$
$\therefore 2x^{2} + 4x - 240 = 0$
Dividing the equation by $2$,we get:
$\therefore x^{2} + 2x - 120 = 0$
Factoring the quadratic equation:
$\therefore (x+12)(x-10) = 0$
$\therefore x+12 = 0$ or $x-10 = 0$
$\therefore x = -12$ or $x = 10$
Since the required numbers are positive integers,$x = -12$ is not possible.
Therefore,$x = 10$.
The smaller number is $10$ and the greater number is $10+2 = 12$.
Hence,the required consecutive even positive integers are $10$ and $12$.

(B) Let the required non-zero number be $x$.
Then,its reciprocal is $\frac{1}{x}$.
According to the problem,their sum is $\frac{10}{3}$.
So,$x + \frac{1}{x} = \frac{10}{3}$.
Multiplying the entire equation by $3x$,we get $3x^2 + 3 = 10x$.
Rearranging the terms,we get the quadratic equation $3x^2 - 10x + 3 = 0$.
Factoring the quadratic equation: $3x^2 - 9x - x + 3 = 0$.
$3x(x - 3) - 1(x - 3) = 0$.
$(3x - 1)(x - 3) = 0$.
This gives $x = \frac{1}{3}$ or $x = 3$.
If the number is $3$,its reciprocal is $\frac{1}{3}$. If the number is $\frac{1}{3}$,its reciprocal is $3$.
Thus,the required numbers are $3$ and $\frac{1}{3}$.

(C) Let the digit at the ten's place be $x$.
Since the product of the digits is $6$,the digit at the unit's place is $\frac{6}{x}$.
The original number is $10x + \frac{6}{x}$.
When the digits are interchanged,the new number is $10(\frac{6}{x}) + x = \frac{60}{x} + x$.
According to the problem,subtracting $9$ from the original number gives the new number:
$10x + \frac{6}{x} - 9 = \frac{60}{x} + x$
Multiply the entire equation by $x$:
$10x^2 + 6 - 9x = 60 + x^2$
$9x^2 - 9x - 54 = 0$
Divide by $9$:
$x^2 - x - 6 = 0$
$(x - 3)(x + 2) = 0$
Since a digit cannot be negative,$x = 3$.
The original number is $10(3) + \frac{6}{3} = 30 + 2 = 32$.

(D) Let the shorter side among the sides containing the right angle (i.e.,base) be $x$ metres.
Then,the other side containing the right angle (i.e.,altitude) is $(x+7)$ metres and the hypotenuse is $(2x+1)$ metres.
By Pythagoras' theorem,$(\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Altitude})^2$.
Therefore,$(2x+1)^2 = x^2 + (x+7)^2$.
$4x^2 + 4x + 1 = x^2 + x^2 + 14x + 49$.
$4x^2 + 4x + 1 = 2x^2 + 14x + 49$.
$2x^2 - 10x - 48 = 0$.
Dividing by $2$,we get $x^2 - 5x - 24 = 0$.
$(x-8)(x+3) = 0$.
So,$x=8$ or $x=-3$.
Since the side of a triangle cannot be negative,$x=8$.
Thus,the base is $8\,m$,the altitude is $8+7=15\,m$,and the hypotenuse is $2(8)+1=17\,m$.
The sides of the garden are $8\,m, 15\,m$ and $17\,m$.

(A) Let the present age of Virat Kohli be $x$ years.
Age $8$ years ago was $(x - 8)$ years.
Age $6$ years later will be $(x + 6)$ years.
According to the problem,the product of these ages is $680$:
$(x - 8)(x + 6) = 680$
$x^2 + 6x - 8x - 48 = 680$
$x^2 - 2x - 48 - 680 = 0$
$x^2 - 2x - 728 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a = 1, b = -2, c = -728$:
$x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-728)}}{2(1)}$
$x = \frac{2 \pm \sqrt{4 + 2912}}{2}$
$x = \frac{2 \pm \sqrt{2916}}{2}$
$x = \frac{2 \pm 54}{2}$
Since age cannot be negative,we take the positive value:
$x = \frac{2 + 54}{2} = \frac{56}{2} = 28$
Therefore,his present age is $28$ years.

(B) Let the original speed of the car be $x\,km/hr$.
Time taken at original speed,$T_1 = \frac{150}{x}$ hours.
New speed = $(x + 5)\,km/hr$.
Time taken at new speed,$T_2 = \frac{150}{x + 5}$ hours.
Given that the car takes $60$ minutes ($1$ hour) less,so $T_1 - T_2 = 1$.
$\frac{150}{x} - \frac{150}{x + 5} = 1$.
$150(x + 5 - x) = x(x + 5)$.
$150(5) = x^2 + 5x$.
$x^2 + 5x - 750 = 0$.
$(x + 30)(x - 25) = 0$.
Since speed cannot be negative,$x = 25\,km/hr$.

(A) Let the speed of the current be $x \text{ km/hr}$.
Given, speed of motorboat in still water = $25 \text{ km/hr}$.
Speed downstream = $(25 + x) \text{ km/hr}$.
Speed upstream = $(25 - x) \text{ km/hr}$.
Time taken to travel $60 \text{ km}$ downstream = $\frac{60}{25 + x} \text{ hours}$.
Time taken to travel $60 \text{ km}$ upstream = $\frac{60}{25 - x} \text{ hours}$.
Total time = $5 \text{ hours}$.
So, $\frac{60}{25 + x} + \frac{60}{25 - x} = 5$.
Divide by $5$: $\frac{12}{25 + x} + \frac{12}{25 - x} = 1$.
$12(25 - x + 25 + x) = (25 + x)(25 - x)$.
$12(50) = 625 - x^2$.
$600 = 625 - x^2$.
$x^2 = 25$.
$x = 5 \text{ km/hr}$ (since speed cannot be negative).
Thus, the speed of the current is $5 \text{ km/hr}$.

(D) Let the number of children be $x$.
Each child receives $\frac{120}{x}$ kites.
If there are $4$ children fewer,the number of children is $(x - 4)$.
In this case,each child receives $\frac{120}{x - 4}$ kites.
According to the problem,$\frac{120}{x - 4} - \frac{120}{x} = 1$.
Multiplying by $x(x - 4)$,we get $120x - 120(x - 4) = x(x - 4)$.
$120x - 120x + 480 = x^2 - 4x$.
$x^2 - 4x - 480 = 0$.
Factoring the quadratic equation: $(x - 24)(x + 20) = 0$.
Since the number of children cannot be negative,$x = 24$.

(A) Let the two sides forming the right angle be $x \, cm$ and $(x + 7) \, cm$.
By the Pythagorean theorem,the hypotenuse is $\sqrt{x^2 + (x + 7)^2} = \sqrt{x^2 + x^2 + 14x + 49} = \sqrt{2x^2 + 14x + 49}$.
The perimeter is given as $30 \, cm$,so $x + (x + 7) + \sqrt{2x^2 + 14x + 49} = 30$.
$2x + 7 + \sqrt{2x^2 + 14x + 49} = 30$.
$\sqrt{2x^2 + 14x + 49} = 23 - 2x$.
Squaring both sides: $2x^2 + 14x + 49 = (23 - 2x)^2$.
$2x^2 + 14x + 49 = 529 - 92x + 4x^2$.
$2x^2 - 106x + 480 = 0$.
$x^2 - 53x + 240 = 0$.
$(x - 5)(x - 48) = 0$.
Since $x$ cannot be $48$ (as the perimeter is $30$),$x = 5$.
The sides are $5 \, cm$,$5 + 7 = 12 \, cm$,and the hypotenuse is $\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, cm$.

(B) Let the breadth of the rectangular garden be $x \, m$.
Then,the length of the garden is $(x + 6) \, m$.
The area of the rectangle is given by $\text{length} \times \text{breadth} = 216 \, m^2$.
So,$x(x + 6) = 216$.
$x^2 + 6x - 216 = 0$.
To solve this quadratic equation,we factorize it: $x^2 + 18x - 12x - 216 = 0$.
$x(x + 18) - 12(x + 18) = 0$.
$(x - 12)(x + 18) = 0$.
Since the breadth cannot be negative,we take $x = 12$.
Thus,the breadth is $12 \, m$ and the length is $12 + 6 = 18 \, m$.

(D) Let the cost price of the flowerpot be $x$ rupees.
Given that the profit percentage is equal to the cost price,the profit percentage is $x\%$.
Profit = $\frac{x}{100} \times x = \frac{x^2}{100}$.
Selling Price = Cost Price + Profit = $x + \frac{x^2}{100}$.
Given that the selling price is Rs. $96$,we have the equation: $x + \frac{x^2}{100} = 96$.
Multiplying the entire equation by $100$,we get $100x + x^2 = 9600$.
Rearranging into a standard quadratic equation: $x^2 + 100x - 9600 = 0$.
Factoring the quadratic equation: $x^2 + 160x - 60x - 9600 = 0$.
$x(x + 160) - 60(x + 160) = 0$.
$(x - 60)(x + 160) = 0$.
This gives $x = 60$ or $x = -160$.
Since the cost price cannot be negative,the cost price is Rs. $60$.

(D) Let the cost price of the pen be $x$ rupees.
Given that the loss percentage is equal to the cost price, the loss percentage $= x\%$.
Loss $= \text{Cost Price} - \text{Selling Price} = x - 24$.
Also, $\text{Loss} = \text{Cost Price} \times \frac{\text{Loss Percentage}}{100} = x \times \frac{x}{100} = \frac{x^2}{100}$.
Equating the two expressions for loss: $x - 24 = \frac{x^2}{100}$.
$100x - 2400 = x^2$.
$x^2 - 100x + 2400 = 0$.
Factoring the quadratic equation: $(x - 60)(x - 40) = 0$.
Therefore, $x = 60$ or $x = 40$.
Thus, the cost price of the pen is Rs. $40$ or Rs. $60$.

(A) Given the quadratic equation: $x^{2} - 7 = 0$.
We can rewrite this as: $x^{2} - (\sqrt{7})^{2} = 0$.
Using the algebraic identity $a^{2} - b^{2} = (a - b)(a + b)$,we get:
$(x - \sqrt{7})(x + \sqrt{7}) = 0$.
Setting each factor to zero:
$x - \sqrt{7} = 0 \implies x = \sqrt{7}$.
$x + \sqrt{7} = 0 \implies x = -\sqrt{7}$.
Thus,the solutions are $x = \sqrt{7}$ and $x = -\sqrt{7}$.

(B) To solve the quadratic equation $x^{2} + 5x - 66 = 0$ by factorization,we need to find two numbers whose product is $-66$ and whose sum is $5$.
These two numbers are $11$ and $-6$,since $11 \times (-6) = -66$ and $11 + (-6) = 5$.
Now,rewrite the middle term $5x$ as $11x - 6x$:
$x^{2} + 11x - 6x - 66 = 0$
Group the terms to factorize:
$x(x + 11) - 6(x + 11) = 0$
$(x - 6)(x + 11) = 0$
Setting each factor to zero:
$x - 6 = 0 \implies x = 6$
$x + 11 = 0 \implies x = -11$
Thus,the solutions are $x = 6$ and $x = -11$.

(C) Given equation: $\frac{x-1}{x-2} + \frac{x-3}{x-4} = \frac{10}{3}$.
Taking $LCM$ on the left side: $\frac{(x-1)(x-4) + (x-3)(x-2)}{(x-2)(x-4)} = \frac{10}{3}$.
Expanding the terms: $\frac{(x^2 - 5x + 4) + (x^2 - 5x + 6)}{x^2 - 6x + 8} = \frac{10}{3}$.
Simplifying: $\frac{2x^2 - 10x + 10}{x^2 - 6x + 8} = \frac{10}{3}$.
Dividing numerator by $2$: $\frac{2(x^2 - 5x + 5)}{x^2 - 6x + 8} = \frac{10}{3} \implies \frac{x^2 - 5x + 5}{x^2 - 6x + 8} = \frac{5}{3}$.
Cross-multiplying: $3(x^2 - 5x + 5) = 5(x^2 - 6x + 8)$.
$3x^2 - 15x + 15 = 5x^2 - 30x + 40$.
Rearranging terms: $2x^2 - 15x + 25 = 0$.
Factorizing: $2x^2 - 10x - 5x + 25 = 0 \implies 2x(x - 5) - 5(x - 5) = 0$.
$(2x - 5)(x - 5) = 0$.
Thus,$x = 5$ or $x = \frac{5}{2}$.

(D) Given equation: $3x^{2} = -11x - 10$.
Rearranging the terms to standard form $ax^{2} + bx + c = 0$:
$3x^{2} + 11x + 10 = 0$.
To factorize,we need two numbers whose product is $3 \times 10 = 30$ and whose sum is $11$.
These numbers are $6$ and $5$.
Splitting the middle term:
$3x^{2} + 6x + 5x + 10 = 0$.
Grouping the terms:
$3x(x + 2) + 5(x + 2) = 0$.
$(3x + 5)(x + 2) = 0$.
Setting each factor to zero:
$3x + 5 = 0 \implies x = -\frac{5}{3}$.
$x + 2 = 0 \implies x = -2$.
Therefore,the solutions are $x = -\frac{5}{3}, -2$.

(A) Given equation: $x^{2} + (x + 5)^{2} = 625$
Expand the square: $x^{2} + (x^{2} + 10x + 25) = 625$
Combine like terms: $2x^{2} + 10x + 25 = 625$
Subtract $625$ from both sides: $2x^{2} + 10x - 600 = 0$
Divide the entire equation by $2$: $x^{2} + 5x - 300 = 0$
Factorize the quadratic equation by splitting the middle term: $x^{2} + 20x - 15x - 300 = 0$
Group the terms: $x(x + 20) - 15(x + 20) = 0$
Factor out $(x + 20)$: $(x + 20)(x - 15) = 0$
Set each factor to zero: $x + 20 = 0$ or $x - 15 = 0$
Therefore,the solutions are $x = -20$ and $x = 15$.

(B) To solve the quadratic equation $2x^2 + ax - a^2 = 0$ by factorization,we need to split the middle term $ax$ into two parts such that their product is $(2x^2) \times (-a^2) = -2a^2x^2$ and their sum is $ax$.
We can write $ax$ as $2ax - ax$.
Substituting this into the equation: $2x^2 + 2ax - ax - a^2 = 0$.
Grouping the terms: $2x(x + a) - a(x + a) = 0$.
Factoring out the common term $(x + a)$: $(2x - a)(x + a) = 0$.
Setting each factor to zero:
$1$) $2x - a = 0 \implies 2x = a \implies x = \frac{a}{2}$.
$2$) $x + a = 0 \implies x = -a$.
Therefore,the solutions are $x = -a$ and $x = \frac{a}{2}$.