Solution of a Quadratic Equation by Factorisation Questions in English

(A) To find the roots of the quadratic equation $2x^{2}-5x+3=0$ by factorisation,we split the middle term $-5x$ into $-2x$ and $-3x$,because $(-2x) \times (-3x) = 6x^{2}$,which is the product of the coefficient of $x^{2}$ $(2)$ and the constant term $(3)$.
Now,rewrite the equation as:
$2x^{2}-2x-3x+3=0$
Factor by grouping:
$2x(x-1)-3(x-1)=0$
$(2x-3)(x-1)=0$
Setting each factor to zero:
$2x-3=0 \implies 2x=3 \implies x=\frac{3}{2}$
$x-1=0 \implies x=1$
Thus,the roots of the equation are $x=1$ and $x=\frac{3}{2}$.

(A) We have the quadratic equation $6x^{2}-x-2=0$.
To find the roots,we factorize the quadratic expression by splitting the middle term:
$6x^{2}-x-2 = 6x^{2}+3x-4x-2$
Grouping the terms,we get:
$= 3x(2x+1)-2(2x+1)$
$= (3x-2)(2x+1)$
The roots of $6x^{2}-x-2=0$ are the values of $x$ for which $(3x-2)(2x+1)=0$.
Setting each factor to zero:
$3x-2=0$ or $2x+1=0$
Solving for $x$:
$3x=2 \implies x=\frac{2}{3}$
$2x=-1 \implies x=-\frac{1}{2}$
Therefore,the roots of the quadratic equation are $\frac{2}{3}$ and $-\frac{1}{2}$.

(N/A) The given quadratic equation is $3x^{2}-2\sqrt{6}x+2=0$.
We can factorize the middle term $-2\sqrt{6}x$ as $-\sqrt{6}x - \sqrt{6}x$.
So,$3x^{2}-\sqrt{6}x-\sqrt{6}x+2=0$.
Taking common terms from the first two and last two terms:
$\sqrt{3}x(\sqrt{3}x-\sqrt{2})-\sqrt{2}(\sqrt{3}x-\sqrt{2})=0$.
This simplifies to $(\sqrt{3}x-\sqrt{2})(\sqrt{3}x-\sqrt{2})=0$.
Setting each factor to zero,we get $\sqrt{3}x-\sqrt{2}=0$,which implies $x = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}}$.
Since both factors are identical,the roots are equal.
Therefore,the roots of the equation are $\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}$.

(A) To find the roots of the quadratic equation $x^{2}-3x-10=0$ by factorisation,we split the middle term.
We need two numbers whose product is $-10$ and whose sum is $-3$. These numbers are $-5$ and $2$.
$x^{2}-5x+2x-10=0$
Factor by grouping:
$x(x-5)+2(x-5)=0$
$(x-5)(x+2)=0$
Setting each factor to zero:
$x-5=0$ or $x+2=0$
Therefore,the roots are $x=5$ or $x=-2$.

(A) To find the roots of the quadratic equation $2x^{2} + x - 6 = 0$ by factorisation,we split the middle term.
We need two numbers whose product is $2 \times (-6) = -12$ and whose sum is $1$.
These numbers are $4$ and $-3$.
So,$2x^{2} + 4x - 3x - 6 = 0$.
Grouping the terms,we get $2x(x + 2) - 3(x + 2) = 0$.
Factoring out $(x + 2)$,we have $(x + 2)(2x - 3) = 0$.
Setting each factor to zero:
$x + 2 = 0 \implies x = -2$
$2x - 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2}$.
Thus,the roots are $x = -2$ and $x = \frac{3}{2}$.

(A) To find the roots of the quadratic equation $\sqrt{2} x^{2}+7 x+5 \sqrt{2}=0$ by factorisation:
$1$. Split the middle term $7x$ into two parts such that their sum is $7x$ and their product is $(\sqrt{2} x^{2}) \times (5 \sqrt{2}) = 10x^{2}$.
$2$. The two parts are $5x$ and $2x$ because $5x + 2x = 7x$ and $5x \times 2x = 10x^{2}$.
$3$. Rewrite the equation: $\sqrt{2} x^{2} + 2x + 5x + 5 \sqrt{2} = 0$.
$4$. Factor by grouping: $\sqrt{2} x(x + \sqrt{2}) + 5(x + \sqrt{2}) = 0$.
$5$. Take out the common factor $(x + \sqrt{2})$: $(x + \sqrt{2})(\sqrt{2} x + 5) = 0$.
$6$. Set each factor to zero:
- $x + \sqrt{2} = 0 \implies x = -\sqrt{2}$
- $\sqrt{2} x + 5 = 0 \implies \sqrt{2} x = -5 \implies x = -\frac{5}{\sqrt{2}}$
Thus,the roots are $x = -\sqrt{2}$ and $x = -\frac{5}{\sqrt{2}}$.

(A) Given equation: $2x^{2} - x + \frac{1}{8} = 0$
Multiply the entire equation by $8$ to remove the fraction:
$16x^{2} - 8x + 1 = 0$
Now,factorise the quadratic expression by splitting the middle term:
$16x^{2} - 4x - 4x + 1 = 0$
Group the terms:
$4x(4x - 1) - 1(4x - 1) = 0$
$(4x - 1)(4x - 1) = 0$
$(4x - 1)^{2} = 0$
Setting the factor to zero:
$4x - 1 = 0$
$4x = 1$
$x = \frac{1}{4}$
Thus,the roots of the equation are $x = \frac{1}{4}, \frac{1}{4}$.

(A) Given quadratic equation: $100 x^{2}-20 x+1=0$
To factorise,we split the middle term $-20x$ into $-10x - 10x$:
$100 x^{2}-10 x-10 x+1=0$
Group the terms:
$10 x(10 x-1)-1(10 x-1)=0$
Factor out the common term $(10 x-1)$:
$(10 x-1)(10 x-1)=0$
$(10 x-1)^{2}=0$
For the roots,set the factor to zero:
$10 x-1=0$
$10 x=1$
$x=\frac{1}{10}$
Since the factor is squared,the roots are equal: $x = \frac{1}{10}, \frac{1}{10}$.

(A) Let the number of John's marbles be $x$.
Therefore,the number of Jivanti's marbles $= 45 - x$.
After losing $5$ marbles each:
Number of John's marbles $= x - 5$.
Number of Jivanti's marbles $= 45 - x - 5 = 40 - x$.
It is given that the product of their remaining marbles is $124$.
So,$(x - 5)(40 - x) = 124$.
Expanding the equation: $40x - x^2 - 200 + 5x = 124$.
Rearranging terms: $-x^2 + 45x - 200 = 124$.
$x^2 - 45x + 324 = 0$.
Factoring the quadratic equation: $x^2 - 36x - 9x + 324 = 0$.
$x(x - 36) - 9(x - 36) = 0$.
$(x - 36)(x - 9) = 0$.
Thus,$x = 36$ or $x = 9$.
If John had $36$ marbles,Jivanti had $45 - 36 = 9$ marbles.
If John had $9$ marbles,Jivanti had $45 - 9 = 36$ marbles.

(A) Let the number of toys produced in a day be $x$.
The cost of production of each toy (in rupees) is $(55 - x)$.
The total cost of production is given by the product of the number of toys and the cost per toy:
$x(55 - x) = 750$
Expanding the equation:
$55x - x^2 = 750$
Rearranging into the standard quadratic form $ax^2 + bx + c = 0$:
$x^2 - 55x + 750 = 0$
To solve this,we factorize the quadratic equation by splitting the middle term:
$x^2 - 25x - 30x + 750 = 0$
Grouping the terms:
$x(x - 25) - 30(x - 25) = 0$
$(x - 25)(x - 30) = 0$
Setting each factor to zero:
$x - 25 = 0 \Rightarrow x = 25$
$x - 30 = 0 \Rightarrow x = 30$
Therefore,the number of toys produced on that day was either $25$ or $30$.

(N/A) Let the first number be $x$. Then the second number is $27-x$.
Given that their product is $182$,we have:
$x(27-x) = 182$
Expanding the equation:
$27x - x^2 = 182$
Rearranging into the standard quadratic form $ax^2 + bx + c = 0$:
$x^2 - 27x + 182 = 0$
Factoring the quadratic equation:
$x^2 - 13x - 14x + 182 = 0$
$x(x - 13) - 14(x - 13) = 0$
$(x - 13)(x - 14) = 0$
Setting each factor to zero:
$x - 13 = 0 \Rightarrow x = 13$
$x - 14 = 0 \Rightarrow x = 14$
If the first number is $13$,the second number is $27 - 13 = 14$.
If the first number is $14$,the second number is $27 - 14 = 13$.
Therefore,the two numbers are $13$ and $14$.

(A) Let the consecutive positive integers be $x$ and $x+1$.
Given that $x^{2} + (x+1)^{2} = 365$.
Expanding the equation: $x^{2} + x^{2} + 2x + 1 = 365$.
Simplifying: $2x^{2} + 2x + 1 = 365$.
$2x^{2} + 2x - 364 = 0$.
Dividing by $2$: $x^{2} + x - 182 = 0$.
Factoring the quadratic equation: $x^{2} + 14x - 13x - 182 = 0$.
$x(x + 14) - 13(x + 14) = 0$.
$(x + 14)(x - 13) = 0$.
This gives $x = -14$ or $x = 13$.
Since the integers must be positive,we take $x = 13$.
Therefore,the next integer is $x + 1 = 13 + 1 = 14$.
The two consecutive positive integers are $13$ and $14$.

(A) Let the base of the right triangle be $x \, cm$.
Its altitude is $(x - 7) \, cm$.
According to the Pythagoras theorem:
$\text{Base}^2 + \text{Altitude}^2 = \text{Hypotenuse}^2$
$x^2 + (x - 7)^2 = 13^2$
$x^2 + x^2 - 14x + 49 = 169$
$2x^2 - 14x - 120 = 0$
Dividing by $2$:
$x^2 - 7x - 60 = 0$
Factoring the quadratic equation:
$x^2 - 12x + 5x - 60 = 0$
$x(x - 12) + 5(x - 12) = 0$
$(x - 12)(x + 5) = 0$
So,$x = 12$ or $x = -5$.
Since the length of a side cannot be negative,we take $x = 12$.
Therefore,the base is $12 \, cm$ and the altitude is $12 - 7 = 5 \, cm$.

(N/A) Let the number of articles produced be $x$.
Therefore,the cost of production of each article $= (2x + 3)$ rupees.
It is given that the total cost of production is ₹ $90$.
Thus,$x(2x + 3) = 90$.
Expanding the equation: $2x^2 + 3x - 90 = 0$.
Factoring the quadratic equation: $2x^2 + 15x - 12x - 90 = 0$.
$x(2x + 15) - 6(2x + 15) = 0$.
$(2x + 15)(x - 6) = 0$.
This gives $x = -15/2$ or $x = 6$.
Since the number of articles produced must be a positive integer,we have $x = 6$.
Therefore,the number of articles produced is $6$,and the cost of each article is $2(6) + 3 = 15$ rupees.

(A) The given quadratic equation is $9x^{2} + \frac{3}{4}x - \sqrt{2} = 0$.
To complete the square,we first express the equation in the form $(ax)^{2} + 2(ax)(b) + b^{2}$.
We can rewrite the equation as $(3x)^{2} + 2(3x)(\frac{1}{8}) - \sqrt{2} = 0$.
Here,the term $b$ is $\frac{1}{8}$.
To complete the square,we need to add and subtract $b^{2} = (\frac{1}{8})^{2} = \frac{1}{64}$.
Thus,the constant that must be added and subtracted is $\frac{1}{64}$.

(C) Yes,such a quadratic equation exists.
Consider the quadratic equation $\sqrt{3}x^2 - 7\sqrt{3}x + 12\sqrt{3} = 0$.
Here,the coefficients are $a = \sqrt{3}$,$b = -7\sqrt{3}$,and $c = 12\sqrt{3}$,which are all distinct irrational numbers.
Dividing the entire equation by $\sqrt{3}$,we get $x^2 - 7x + 12 = 0$.
Factoring the quadratic,we get $(x - 3)(x - 4) = 0$.
Thus,the roots are $x = 3$ and $x = 4$,which are both rational numbers.

(D) To find the roots of $6x^{2} - \sqrt{2}x - 2 = 0$,we factorize the quadratic polynomial.
We need two numbers whose product is $6 \times (-2) = -12$ and whose sum is $-\sqrt{2}$.
These numbers are $-3\sqrt{2}$ and $2\sqrt{2}$.
$6x^{2} - 3\sqrt{2}x + 2\sqrt{2}x - 2 = 0$
$3x(2x - \sqrt{2}) + \sqrt{2}(2x - \sqrt{2}) = 0$
$(3x + \sqrt{2})(2x - \sqrt{2}) = 0$
Setting each factor to zero:
$3x + \sqrt{2} = 0 \implies x = -\frac{\sqrt{2}}{3}$
$2x - \sqrt{2} = 0 \implies x = \frac{\sqrt{2}}{2}$
Thus,the roots are $-\frac{\sqrt{2}}{3}$ and $\frac{\sqrt{2}}{2}$.

(A) Let one side be $x \, cm$. Then the other side will be $(x+5) \, cm$.
According to the Pythagoras Theorem:
$x^{2} + (x+5)^{2} = (25)^{2}$
Expanding the equation:
$x^{2} + x^{2} + 10x + 25 = 625$
$2x^{2} + 10x - 600 = 0$
Dividing by $2$:
$x^{2} + 5x - 300 = 0$
Factoring the quadratic equation:
$x^{2} + 20x - 15x - 300 = 0$
$x(x+20) - 15(x+20) = 0$
$(x-15)(x+20) = 0$
Thus,$x = 15$ or $x = -20$.
Since the length cannot be negative,we reject $x = -20$.
Therefore,one side is $15 \, cm$ and the other side is $15 + 5 = 20 \, cm$.

(A) Given equation: $\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}$
Taking the common denominator on the left side:
$\frac{(x-7)-(x+4)}{(x+4)(x-7)}=\frac{11}{30}$
Simplifying the numerator:
$\frac{x-7-x-4}{x^2-7x+4x-28}=\frac{11}{30}$
$\frac{-11}{x^2-3x-28}=\frac{11}{30}$
Dividing both sides by $11$:
$\frac{-1}{x^2-3x-28}=\frac{1}{30}$
Cross-multiplying:
$x^2-3x-28 = -30$
$x^2-3x-28+30 = 0$
$x^2-3x+2 = 0$
Factoring the quadratic equation:
$x^2-2x-x+2 = 0$
$x(x-2)-1(x-2) = 0$
$(x-2)(x-1) = 0$
Therefore,the roots are $x = 1$ or $x = 2$.

(C) Let the shorter side of the rectangle be $x\, m$.
Then,the longer side of the rectangle is $(x+30)\, m$.
According to the Pythagorean theorem,the diagonal of the rectangle is $\sqrt{x^2 + (x+30)^2}$.
It is given that the diagonal is $60\, m$ more than the shorter side,so $\sqrt{x^2 + (x+30)^2} = x + 60$.
Squaring both sides,we get $x^2 + (x+30)^2 = (x+60)^2$.
Expanding the terms: $x^2 + x^2 + 60x + 900 = x^2 + 120x + 3600$.
Simplifying the equation: $x^2 - 60x - 2700 = 0$.
Factoring the quadratic equation: $x^2 - 90x + 30x - 2700 = 0$,which gives $x(x-90) + 30(x-90) = 0$.
So,$(x-90)(x+30) = 0$.
This gives $x = 90$ or $x = -30$.
Since the length cannot be negative,the shorter side is $90\, m$.
The longer side is $x + 30 = 90 + 30 = 120\, m$.

(A) Let the uniform speed of the train be $x\, km/h$.
Time taken to cover $360\, km$ at speed $x$ is $t_1 = \frac{360}{x}\, h$.
If the speed is increased by $5\, km/h$,the new speed is $(x + 5)\, km/h$.
Time taken at the new speed is $t_2 = \frac{360}{x + 5}\, h$.
According to the problem,the difference in time is $1\, hour$:
$\frac{360}{x} - \frac{360}{x + 5} = 1$
$360 \left( \frac{1}{x} - \frac{1}{x + 5} \right) = 1$
$360 \left( \frac{x + 5 - x}{x(x + 5)} \right) = 1$
$360 \left( \frac{5}{x^2 + 5x} \right) = 1$
$1800 = x^2 + 5x$
$x^2 + 5x - 1800 = 0$
Factoring the quadratic equation:
$x^2 + 45x - 40x - 1800 = 0$
$x(x + 45) - 40(x + 45) = 0$
$(x + 45)(x - 40) = 0$
This gives $x = -45$ or $x = 40$.
Since speed cannot be negative,the speed of the train is $40\, km/h$.

(A) Let the average speed of the passenger train be $x\, km/h$.
Average speed of the express train $= (x + 11)\, km/h$.
Time taken by the passenger train to cover $132\, km = \frac{132}{x}\, hours$.
Time taken by the express train to cover $132\, km = \frac{132}{x + 11}\, hours$.
According to the problem,the express train takes $1\, hour$ less than the passenger train:
$\frac{132}{x} - \frac{132}{x + 11} = 1$
$132 \left[ \frac{x + 11 - x}{x(x + 11)} \right] = 1$
$\frac{132 \times 11}{x^2 + 11x} = 1$
$x^2 + 11x = 1452$
$x^2 + 11x - 1452 = 0$
Solving the quadratic equation by splitting the middle term:
$x^2 + 44x - 33x - 1452 = 0$
$x(x + 44) - 33(x + 44) = 0$
$(x + 44)(x - 33) = 0$
$x = -44$ or $x = 33$.
Since speed cannot be negative,$x = 33\, km/h$.
Therefore,the speed of the passenger train is $33\, km/h$ and the speed of the express train is $33 + 11 = 44\, km/h$.

(A) Let the sides of the two squares be $x \, m$ and $y \, m$. Therefore,their perimeters are $4x$ and $4y$ respectively,and their areas are $x^2$ and $y^2$ respectively.
It is given that the difference of their perimeters is $24 \, m$:
$4x - 4y = 24$
$x - y = 6$
$x = y + 6$
It is also given that the sum of their areas is $468 \, m^2$:
$x^2 + y^2 = 468$
Substituting $x = y + 6$ into the area equation:
$(y + 6)^2 + y^2 = 468$
$y^2 + 12y + 36 + y^2 = 468$
$2y^2 + 12y - 432 = 0$
Dividing by $2$:
$y^2 + 6y - 216 = 0$
Factoring the quadratic equation:
$y^2 + 18y - 12y - 216 = 0$
$y(y + 18) - 12(y + 18) = 0$
$(y + 18)(y - 12) = 0$
This gives $y = -18$ or $y = 12$. Since the side of a square cannot be negative,we take $y = 12 \, m$.
Then,$x = 12 + 6 = 18 \, m$.
Thus,the sides of the two squares are $18 \, m$ and $12 \, m$.

(C) Given equation is $2x^{2} + \frac{5}{3}x - 2 = 0$.
On multiplying both sides by $3$,we get:
$6x^{2} + 5x - 6 = 0$.
To factorise,we split the middle term such that the product of the two numbers is $6 \times (-6) = -36$ and their sum is $5$. The numbers are $9$ and $-4$.
$6x^{2} + 9x - 4x - 6 = 0$
Taking common terms:
$3x(2x + 3) - 2(2x + 3) = 0$
$(2x + 3)(3x - 2) = 0$
Now,setting each factor to zero:
$2x + 3 = 0 \Rightarrow x = -\frac{3}{2}$
$3x - 2 = 0 \Rightarrow x = \frac{2}{3}$
Thus,the roots of the equation are $-\frac{3}{2}$ and $\frac{2}{3}$.

(B) Given equation is $\frac{2}{5} x^{2}-x-\frac{3}{5}=0$.
On multiplying by $5$ on both sides,we get:
$2x^{2}-5x-3=0$
To factorise,we split the middle term $-5x$ into $-6x + x$:
$2x^{2}-6x+x-3=0$
Grouping the terms:
$2x(x-3)+1(x-3)=0$
Taking $(x-3)$ as a common factor:
$(x-3)(2x+1)=0$
Setting each factor to zero:
$x-3=0 \Rightarrow x=3$
$2x+1=0 \Rightarrow x=-\frac{1}{2}$
Hence,the roots of the equation are $-\frac{1}{2}$ and $3$.

(A) Given equation is $3 \sqrt{2} x^{2}-5 x-\sqrt{2}=0$.
To factorise,we split the middle term $-5x$ into $-6x + x$ such that their product is $(3 \sqrt{2}) \times (-\sqrt{2}) = -6$.
$3 \sqrt{2} x^{2}-6 x+x-\sqrt{2}=0$
$3 \sqrt{2} x(x-\sqrt{2})+1(x-\sqrt{2})=0$
$(x-\sqrt{2})(3 \sqrt{2} x+1)=0$
Setting each factor to zero:
$x-\sqrt{2}=0 \Rightarrow x=\sqrt{2}$
$3 \sqrt{2} x+1=0 \Rightarrow x=-\frac{1}{3 \sqrt{2}}$.
Rationalizing the denominator: $x = -\frac{1 \times \sqrt{2}}{3 \sqrt{2} \times \sqrt{2}} = -\frac{\sqrt{2}}{3 \times 2} = -\frac{\sqrt{2}}{6}$.
Thus,the roots are $-\frac{\sqrt{2}}{6}$ and $\sqrt{2}$.

(B) Given equation is $3x^{2} + 5\sqrt{5}x - 10 = 0$.
To factorise,we split the middle term such that the product is $3 \times (-10) = -30$ and the sum is $5\sqrt{5}$.
The factors are $6\sqrt{5}$ and $-\sqrt{5}$,since $(6\sqrt{5}) \times (-\sqrt{5}) = -6 \times 5 = -30$ and $6\sqrt{5} - \sqrt{5} = 5\sqrt{5}$.
$3x^{2} + 6\sqrt{5}x - \sqrt{5}x - 10 = 0$
$3x(x + 2\sqrt{5}) - \sqrt{5}(x + 2\sqrt{5}) = 0$
$(x + 2\sqrt{5})(3x - \sqrt{5}) = 0$
Setting each factor to zero:
$x + 2\sqrt{5} = 0 \Rightarrow x = -2\sqrt{5}$
$3x - \sqrt{5} = 0 \Rightarrow x = \frac{\sqrt{5}}{3}$
Thus,the roots are $-2\sqrt{5}$ and $\frac{\sqrt{5}}{3}$.

(C) Given equation is $21 x^{2}-2 x+\frac{1}{21}=0$.
To simplify,multiply the entire equation by $21$ on both sides:
$21(21 x^{2}-2 x+\frac{1}{21}) = 21(0)$
$441 x^{2}-42 x+1=0$
Now,factorise the quadratic equation by splitting the middle term:
$441 x^{2}-21 x-21 x+1=0$
Group the terms:
$(441 x^{2}-21 x)-(21 x-1)=0$
$21 x(21 x-1)-1(21 x-1)=0$
Factor out $(21 x-1)$:
$(21 x-1)(21 x-1)=0$
Setting each factor to zero:
$21 x-1=0 \Rightarrow x=\frac{1}{21}$
$21 x-1=0 \Rightarrow x=\frac{1}{21}$
Thus,the roots of the equation are $\frac{1}{21}$ and $\frac{1}{21}$.

(A) Let her actual marks be $x$.
According to the problem,if she scored $10$ more marks,the new score would be $(x + 10)$.
The condition states that $9$ times these marks would be the square of her actual marks:
$9(x + 10) = x^2$
$9x + 90 = x^2$
Rearranging the terms to form a quadratic equation:
$x^2 - 9x - 90 = 0$
Factoring the quadratic equation:
$x^2 - 15x + 6x - 90 = 0$
$x(x - 15) + 6(x - 15) = 0$
$(x + 6)(x - 15) = 0$
This gives $x = -6$ or $x = 15$.
Since marks cannot be negative,we discard $x = -6$.
Therefore,the actual marks obtained by Ajita is $15$.

(D) Let $n$ be the required natural number.
According to the problem,the square of the number diminished by $84$ is $n^2 - 84$.
The expression for thrice of $8$ more than the number is $3(n + 8)$.
Setting these equal,we get the quadratic equation:
$n^2 - 84 = 3(n + 8)$
$n^2 - 84 = 3n + 24$
$n^2 - 3n - 108 = 0$
To solve this,we factor the quadratic equation:
$n^2 - 12n + 9n - 108 = 0$
$n(n - 12) + 9(n - 12) = 0$
$(n - 12)(n + 9) = 0$
This gives $n = 12$ or $n = -9$.
Since $n$ must be a natural number,we discard $n = -9$.
Therefore,the required natural number is $12$.

(A) Let the natural number be $x$.
According to the question,
$x + 12 = \frac{160}{x}$
On multiplying both sides by $x$,we get:
$x^2 + 12x = 160$
$x^2 + 12x - 160 = 0$
Now,factorizing the quadratic equation:
$x^2 + 20x - 8x - 160 = 0$
$x(x + 20) - 8(x + 20) = 0$
$(x + 20)(x - 8) = 0$
This gives two possible values for $x$:
$x = -20$ or $x = 8$.
Since the question specifies a natural number,$x$ must be positive.
Therefore,$x = 8$ is the required natural number.

(C) Let the actual age of Zeba be $x\, \text{years}$.
Her age when she was $5\, \text{years}$ younger is $(x - 5)\, \text{years}$.
According to the given condition:
$(x - 5)^2 = 5x + 11$
Expanding the left side:
$x^2 - 10x + 25 = 5x + 11$
Rearranging the terms to form a quadratic equation:
$x^2 - 10x - 5x + 25 - 11 = 0$
$x^2 - 15x + 14 = 0$
Factoring the quadratic equation by splitting the middle term:
$x^2 - 14x - x + 14 = 0$
$x(x - 14) - 1(x - 14) = 0$
$(x - 1)(x - 14) = 0$
This gives two possible values for $x$: $x = 1$ or $x = 14$.
If $x = 1$, then her age $5\, \text{years}$ ago would be $1 - 5 = -4$, which is impossible as age cannot be negative.
Therefore, the actual age of Zeba is $14\, \text{years}$.

(D) To factorize the quadratic polynomial $p(x) = x^{2} + 5x - 24$,we need to find two numbers whose product is $-24$ and whose sum is $5$.
These two numbers are $8$ and $-3$,because $8 \times (-3) = -24$ and $8 + (-3) = 5$.
Now,split the middle term $5x$ as $8x - 3x$:
$p(x) = x^{2} + 8x - 3x - 24$
Group the terms:
$p(x) = (x^{2} + 8x) - (3x + 24)$
Factor out the common terms:
$p(x) = x(x + 8) - 3(x + 8)$
Finally,factor out the common binomial $(x + 8)$:
$p(x) = (x + 8)(x - 3)$

(A) Given equation: $\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{5}{6}$
Taking the common denominator on the left side:
$\frac{(x+1)^{2}-(x-1)^{2}}{(x-1)(x+1)}=\frac{5}{6}$
Expanding the squares and simplifying the denominator:
$\frac{(x^{2}+2x+1)-(x^{2}-2x+1)}{x^{2}-1}=\frac{5}{6}$
$\frac{x^{2}+2x+1-x^{2}+2x-1}{x^{2}-1}=\frac{5}{6}$
$\frac{4x}{x^{2}-1}=\frac{5}{6}$
Cross-multiplying:
$24x = 5(x^{2}-1)$
$24x = 5x^{2}-5$
Rearranging into standard quadratic form $ax^{2}+bx+c=0$:
$5x^{2}-24x-5=0$
Factorizing the quadratic equation:
$5x^{2}-25x+x-5=0$
$5x(x-5)+1(x-5)=0$
$(x-5)(5x+1)=0$
Setting each factor to zero:
$x-5=0 \implies x=5$
$5x+1=0 \implies x=-\frac{1}{5}$
Thus,the roots are $5$ and $-\frac{1}{5}$.

(B) Given equation: $\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6}$
Taking the common denominator on the left side:
$\frac{(x+5)-(x-3)}{(x-3)(x+5)}=\frac{1}{6}$
Simplifying the numerator and denominator:
$\frac{x+5-x+3}{x^{2}+5x-3x-15}=\frac{1}{6}$
$\frac{8}{x^{2}+2x-15}=\frac{1}{6}$
Cross-multiplying:
$x^{2}+2x-15 = 48$
$x^{2}+2x-15-48 = 0$
$x^{2}+2x-63 = 0$
Factoring the quadratic equation:
$x^{2}+9x-7x-63 = 0$
$x(x+9)-7(x+9) = 0$
$(x+9)(x-7) = 0$
Setting each factor to zero:
$x+9 = 0 \implies x = -9$
$x-7 = 0 \implies x = 7$
Thus,the roots of the equation are $-9$ and $7$.

(C) Given equation: $x - \frac{1}{x} = \frac{45}{14}$
Multiply by $x$ to clear the denominator: $\frac{x^2 - 1}{x} = \frac{45}{14}$
Cross-multiply: $14(x^2 - 1) = 45x$
Rearrange into standard quadratic form $ax^2 + bx + c = 0$: $14x^2 - 45x - 14 = 0$
Factorize by splitting the middle term: $14x^2 - 49x + 4x - 14 = 0$
Group the terms: $7x(2x - 7) + 2(2x - 7) = 0$
Factor out the common binomial: $(2x - 7)(7x + 2) = 0$
Set each factor to zero: $2x - 7 = 0$ or $7x + 2 = 0$
Solve for $x$: $x = \frac{7}{2}$ or $x = -\frac{2}{7}$
Thus,the roots of the quadratic equation are $\frac{7}{2}$ and $-\frac{2}{7}$.

(D) Given equation: $\frac{x^{2}-1}{x^{2}+1}=\frac{7}{9}$
Cross-multiplying,we get:
$9(x^{2}-1) = 7(x^{2}+1)$
Expanding the terms:
$9x^{2} - 9 = 7x^{2} + 7$
Rearranging the terms to one side:
$9x^{2} - 7x^{2} - 9 - 7 = 0$
Simplifying the equation:
$2x^{2} - 16 = 0$
Dividing by $2$:
$x^{2} - 8 = 0$
Using the difference of squares identity $a^{2}-b^{2}=(a-b)(a+b)$,where $8 = (\sqrt{8})^{2} = (2\sqrt{2})^{2}$:
$(x - 2\sqrt{2})(x + 2\sqrt{2}) = 0$
Setting each factor to zero:
$x - 2\sqrt{2} = 0$ or $x + 2\sqrt{2} = 0$
Therefore,the roots are:
$x = 2\sqrt{2}$ or $x = -2\sqrt{2}$

(A) Given equation: $6(2x+1)^2 - (2x+1) - 5 = 0$
Let $m = 2x+1$.
Substituting $m$ into the equation,we get: $6m^2 - m - 5 = 0$.
Factorizing the quadratic equation: $6m^2 - 6m + 5m - 5 = 0$.
$6m(m-1) + 5(m-1) = 0$.
$(6m+5)(m-1) = 0$.
Therefore,$m = 1$ or $m = -\frac{5}{6}$.
Case $1$: If $m = 1$,then $2x+1 = 1 \implies 2x = 0 \implies x = 0$.
Case $2$: If $m = -\frac{5}{6}$,then $2x+1 = -\frac{5}{6} \implies 2x = -\frac{5}{6} - 1 \implies 2x = -\frac{11}{6} \implies x = -\frac{11}{12}$.
Thus,the roots of the equation are $0$ and $-\frac{11}{12}$.

(B) Given equation: $4x^{2} + 4x = 15$
Rearranging the equation to standard form $ax^{2} + bx + c = 0$:
$4x^{2} + 4x - 15 = 0$
To factorize,we need two numbers whose product is $4 \times (-15) = -60$ and whose sum is $4$. These numbers are $10$ and $-6$.
Splitting the middle term:
$4x^{2} + 10x - 6x - 15 = 0$
Grouping the terms:
$2x(2x + 5) - 3(2x + 5) = 0$
Taking out the common factor:
$(2x + 5)(2x - 3) = 0$
Setting each factor to zero:
$2x + 5 = 0$ or $2x - 3 = 0$
Solving for $x$:
$2x = -5 \implies x = -\frac{5}{2}$
$2x = 3 \implies x = \frac{3}{2}$
Thus,the solution set is $\left\{-\frac{5}{2}, \frac{3}{2}\right\}$.

(C) Given equation: $x - \frac{1}{x} - \frac{45}{14} = 0$.
Multiplying the entire equation by $14x$ (the least common multiple of the denominators) to clear the fractions:
$14x^2 - 14 - 45x = 0$.
Rearranging into standard quadratic form $ax^2 + bx + c = 0$:
$14x^2 - 45x - 14 = 0$.
To factorize,we look for two numbers whose product is $14 \times (-14) = -196$ and whose sum is $-45$. These numbers are $-49$ and $4$.
$14x^2 - 49x + 4x - 14 = 0$.
Grouping the terms:
$7x(2x - 7) + 2(2x - 7) = 0$.
$(2x - 7)(7x + 2) = 0$.
Setting each factor to zero:
$2x - 7 = 0 \implies x = \frac{7}{2}$.
$7x + 2 = 0 \implies x = -\frac{2}{7}$.
Thus,the solution set is $\{ \frac{7}{2}, -\frac{2}{7} \}$.

(D) Given equation: $\frac{1}{x-2}+\frac{1}{x+3}=\frac{7}{2x}$
Multiplying both sides by the least common multiple $(LCM)$ of the denominators,which is $2x(x-2)(x+3)$:
$2x(x+3) + 2x(x-2) = 7(x-2)(x+3)$
Expanding the terms:
$2x^2 + 6x + 2x^2 - 4x = 7(x^2 + x - 6)$
$4x^2 + 2x = 7x^2 + 7x - 42$
Rearranging the terms to form a standard quadratic equation $ax^2 + bx + c = 0$:
$7x^2 - 4x^2 + 7x - 2x - 42 = 0$
$3x^2 + 5x - 42 = 0$
Factorizing the quadratic equation by splitting the middle term:
$3x^2 + 14x - 9x - 42 = 0$
$x(3x + 14) - 3(3x + 14) = 0$
$(3x + 14)(x - 3) = 0$
Setting each factor to zero:
$3x + 14 = 0 \implies x = -\frac{14}{3}$
$x - 3 = 0 \implies x = 3$
Thus,the solution set is $\left\{-\frac{14}{3}, 3\right\}$.

(A) Let $(x+2) = y$. Substituting this into the equation,we get:
$y^{2} - 14y + 45 = 0$
To factorize,we look for two numbers whose product is $45$ and sum is $-14$. These numbers are $-9$ and $-5$.
$y^{2} - 9y - 5y + 45 = 0$
$y(y - 9) - 5(y - 9) = 0$
$(y - 9)(y - 5) = 0$
So,$y = 9$ or $y = 5$.
Now,substitute $y = x + 2$ back:
Case $1$: $x + 2 = 9 \implies x = 7$
Case $2$: $x + 2 = 5 \implies x = 3$
Thus,the solutions are $x = 3, 7$.

(B) Let $y = 2x + 1$. The equation becomes $6y^2 = y + 5$.
Rearranging the terms,we get $6y^2 - y - 5 = 0$.
To factorize,we look for two numbers that multiply to $(6 \times -5) = -30$ and add to $-1$.
These numbers are $-6$ and $5$.
So,$6y^2 - 6y + 5y - 5 = 0$.
$6y(y - 1) + 5(y - 1) = 0$.
$(6y + 5)(y - 1) = 0$.
This gives $y = 1$ or $y = -\frac{5}{6}$.
Substituting $y = 2x + 1$ back:
Case $1$: $2x + 1 = 1 \implies 2x = 0 \implies x = 0$.
Case $2$: $2x + 1 = -\frac{5}{6} \implies 2x = -\frac{5}{6} - 1 = -\frac{11}{6} \implies x = -\frac{11}{12}$.
Thus,the solutions are $x = 0$ and $x = -\frac{11}{12}$.

(C) Given equation: $x + 3 + \frac{2}{x} = 0$.
Multiply the entire equation by $x$ (where $x \neq 0$):
$x^2 + 3x + 2 = 0$.
Now,factorize the quadratic equation by splitting the middle term:
$x^2 + 2x + x + 2 = 0$.
$x(x + 2) + 1(x + 2) = 0$.
$(x + 1)(x + 2) = 0$.
Setting each factor to zero:
$x + 1 = 0 \implies x = -1$.
$x + 2 = 0 \implies x = -2$.
Thus,the solutions are $x = -1, -2$.

(A) Given equation: $(x+1)^{2} + x^{2} = 221$
Expand the square: $(x^{2} + 2x + 1) + x^{2} = 221$
Combine like terms: $2x^{2} + 2x + 1 = 221$
Subtract $221$ from both sides: $2x^{2} + 2x - 220 = 0$
Divide the entire equation by $2$: $x^{2} + x - 110 = 0$
Factorize the quadratic equation: $x^{2} + 11x - 10x - 110 = 0$
Group the terms: $x(x + 11) - 10(x + 11) = 0$
$(x - 10)(x + 11) = 0$
Therefore,$x - 10 = 0$ or $x + 11 = 0$
$x = 10$ or $x = -11$
Thus,the solutions are $10, -11$.

(A) Given the equation: $x - \frac{20}{x} = 1$
Multiply the entire equation by $x$ to eliminate the fraction: $x^2 - 20 = x$
Rearrange the equation into the standard quadratic form $ax^2 + bx + c = 0$: $x^2 - x - 20 = 0$
To factorize,find two numbers that multiply to $-20$ and add to $-1$: These numbers are $-5$ and $4$.
Rewrite the middle term: $x^2 - 5x + 4x - 20 = 0$
Group the terms: $x(x - 5) + 4(x - 5) = 0$
Factor out the common binomial: $(x - 5)(x + 4) = 0$
Set each factor to zero: $x - 5 = 0$ or $x + 4 = 0$
Thus,the solutions are $x = 5$ or $x = -4$.

(B) Given equation: $x^{2} + (x + 5)^{2} = 125$
Expand the term $(x + 5)^{2}$ using the identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$:
$x^{2} + (x^{2} + 10x + 25) = 125$
Combine like terms:
$2x^{2} + 10x + 25 = 125$
Subtract $125$ from both sides to set the equation to zero:
$2x^{2} + 10x - 100 = 0$
Divide the entire equation by $2$ to simplify:
$x^{2} + 5x - 50 = 0$
Factorize the quadratic equation by finding two numbers that multiply to $-50$ and add to $5$:
These numbers are $10$ and $-5$:
$x^{2} + 10x - 5x - 50 = 0$
$x(x + 10) - 5(x + 10) = 0$
$(x - 5)(x + 10) = 0$
Setting each factor to zero:
$x - 5 = 0 \implies x = 5$
$x + 10 = 0 \implies x = -10$
Thus,the solutions are $5, -10$.

(D) Given equation: $\frac{x}{x+1} + \frac{x+1}{x} = \frac{5}{2}$.
Let $y = \frac{x}{x+1}$. Then the equation becomes $y + \frac{1}{y} = \frac{5}{2}$.
Multiplying by $2y$,we get $2y^2 + 2 = 5y$,which simplifies to $2y^2 - 5y + 2 = 0$.
Factoring the quadratic: $2y^2 - 4y - y + 2 = 0 \implies 2y(y - 2) - 1(y - 2) = 0$.
So,$(2y - 1)(y - 2) = 0$,giving $y = 1/2$ or $y = 2$.
Case $1$: $\frac{x}{x+1} = \frac{1}{2} \implies 2x = x + 1 \implies x = 1$.
Case $2$: $\frac{x}{x+1} = 2 \implies x = 2x + 2 \implies x = -2$.
The solution set is $\{1, -2\}$.

(D) Given equation: $\frac{x+2}{x} + \frac{x}{x+2} = \frac{10}{3}$
Taking $LCM$: $\frac{(x+2)^2 + x^2}{x(x+2)} = \frac{10}{3}$
Expand the numerator: $\frac{x^2 + 4x + 4 + x^2}{x^2 + 2x} = \frac{10}{3}$
$\frac{2x^2 + 4x + 4}{x^2 + 2x} = \frac{10}{3}$
Cross-multiply: $3(2x^2 + 4x + 4) = 10(x^2 + 2x)$
$6x^2 + 12x + 12 = 10x^2 + 20x$
Rearrange terms: $4x^2 + 8x - 12 = 0$
Divide by $4$: $x^2 + 2x - 3 = 0$
Factorize: $x^2 + 3x - x - 3 = 0$
$x(x+3) - 1(x+3) = 0$
$(x-1)(x+3) = 0$
Thus,$x = 1$ or $x = -3$.
The solution set is $\{1, -3\}$.

(A) Given equation: $\frac{x+1}{x+2} + \frac{1}{x} = \frac{5}{4}$
Taking $LCM$ on the left side: $\frac{x(x+1) + 1(x+2)}{x(x+2)} = \frac{5}{4}$
$\frac{x^2 + x + x + 2}{x^2 + 2x} = \frac{5}{4}$
$\frac{x^2 + 2x + 2}{x^2 + 2x} = \frac{5}{4}$
Cross-multiplying: $4(x^2 + 2x + 2) = 5(x^2 + 2x)$
$4x^2 + 8x + 8 = 5x^2 + 10x$
Rearranging terms to form a quadratic equation: $x^2 + 2x - 8 = 0$
Factorizing the quadratic equation: $x^2 + 4x - 2x - 8 = 0$
$x(x+4) - 2(x+4) = 0$
$(x-2)(x+4) = 0$
Thus,$x = 2$ or $x = -4$.
The solution set is $\{2, -4\}$.