Solution of a Quadratic Equation by Factorisation Questions in English

(B) Given equation: $x + 2 - \frac{6}{x + 2} = 1$
Let $y = x + 2$. Then the equation becomes: $y - \frac{6}{y} = 1$
Multiply by $y$: $y^2 - 6 = y$
Rearrange to standard form: $y^2 - y - 6 = 0$
Factorize: $y^2 - 3y + 2y - 6 = 0$
$y(y - 3) + 2(y - 3) = 0$
$(y - 3)(y + 2) = 0$
So,$y = 3$ or $y = -2$.
Substitute $y = x + 2$ back:
Case $1$: $x + 2 = 3 \implies x = 1$
Case $2$: $x + 2 = -2 \implies x = -4$
Thus,the solution set is $\{1, -4\}$.

(C) Given equation: $\frac{x}{x-1} + \frac{x-1}{x} = \frac{25}{12}$
Let $y = \frac{x}{x-1}$,then the equation becomes $y + \frac{1}{y} = \frac{25}{12}$.
Multiplying by $12y$,we get $12y^2 + 12 = 25y$,which rearranges to $12y^2 - 25y + 12 = 0$.
Factoring the quadratic: $12y^2 - 16y - 9y + 12 = 0 \implies 4y(3y - 4) - 3(3y - 4) = 0$.
So,$(4y - 3)(3y - 4) = 0$,giving $y = \frac{3}{4}$ or $y = \frac{4}{3}$.
Case $1$: $\frac{x}{x-1} = \frac{3}{4} \implies 4x = 3x - 3 \implies x = -3$.
Case $2$: $\frac{x}{x-1} = \frac{4}{3} \implies 3x = 4x - 4 \implies x = 4$.
Thus,the solution set is $\{4, -3\}$.

(D) Given equation: $\frac{9}{x-1} - \frac{2}{x-3} = \frac{5}{x+1}$
Taking $LCM$ on the left side: $\frac{9(x-3) - 2(x-1)}{(x-1)(x-3)} = \frac{5}{x+1}$
$\frac{9x - 27 - 2x + 2}{x^2 - 4x + 3} = \frac{5}{x+1}$
$\frac{7x - 25}{x^2 - 4x + 3} = \frac{5}{x+1}$
Cross-multiplying: $(7x - 25)(x + 1) = 5(x^2 - 4x + 3)$
$7x^2 + 7x - 25x - 25 = 5x^2 - 20x + 15$
$7x^2 - 18x - 25 = 5x^2 - 20x + 15$
$2x^2 + 2x - 40 = 0$
Dividing by $2$: $x^2 + x - 20 = 0$
Factorizing: $x^2 + 5x - 4x - 20 = 0$
$x(x + 5) - 4(x + 5) = 0$
$(x - 4)(x + 5) = 0$
Thus,$x = 4$ or $x = -5$.
The solution set is $\{-5, 4\}$.

(A) Let $y = \frac{x+1}{x-3}$. 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 equation: $2y^2 - 4y - y + 2 = 0 \implies 2y(y - 2) - 1(y - 2) = 0$.
This gives $(2y - 1)(y - 2) = 0$,so $y = \frac{1}{2}$ or $y = 2$.
Case $1$: $\frac{x+1}{x-3} = \frac{1}{2} \implies 2x + 2 = x - 3 \implies x = -5$.
Case $2$: $\frac{x+1}{x-3} = 2 \implies x + 1 = 2x - 6 \implies x = 7$.
Thus,the solution set is $\{-5, 7\}$.

(A) Let $AB = x$ and $BC = y$.
In rectangle $ABCD$,$\triangle ABC$ is a right-angled triangle with $\angle B = 90^\circ$.
By the Pythagorean theorem,$AB^2 + BC^2 = AC^2$.
Given $x^2 + y^2 = 29^2 = 841$.
Also given $x + y = 41$.
We know that $(x + y)^2 = x^2 + y^2 + 2xy$.
Substituting the values: $41^2 = 841 + 2xy$.
$1681 = 841 + 2xy$.
$2xy = 1681 - 841 = 840$.
$xy = 420$.
Now,we have $x + y = 41$ and $xy = 420$.
These are the roots of the quadratic equation $t^2 - (x+y)t + xy = 0$.
$t^2 - 41t + 420 = 0$.
Factoring the equation: $(t - 21)(t - 20) = 0$.
So,$t = 21$ or $t = 20$.
Since $AB > BC$,we have $AB = 21$ and $BC = 20$.
Thus,$AB = 21$.

(B) Given equation: $\frac{8}{x+5} = \frac{3}{x-1} + \frac{1}{x-5}$
First,simplify the right side: $\frac{8}{x+5} = \frac{3(x-5) + 1(x-1)}{(x-1)(x-5)}$
$\frac{8}{x+5} = \frac{3x - 15 + x - 1}{x^2 - 6x + 5}$
$\frac{8}{x+5} = \frac{4x - 16}{x^2 - 6x + 5}$
Cross-multiply: $8(x^2 - 6x + 5) = (x+5)(4x - 16)$
$8x^2 - 48x + 40 = 4x^2 - 16x + 20x - 80$
$8x^2 - 48x + 40 = 4x^2 + 4x - 80$
$4x^2 - 52x + 120 = 0$
Divide by $4$: $x^2 - 13x + 30 = 0$
Factorize: $x^2 - 10x - 3x + 30 = 0$
$x(x - 10) - 3(x - 10) = 0$
$(x - 3)(x - 10) = 0$
Thus,$x = 3$ or $x = 10$.
The solution set is $\{3, 10\}$.

(C) Given equation: $\frac{x}{x+1} + \frac{x+1}{x} = \frac{41}{20}$.
Let $y = \frac{x}{x+1}$,then the equation becomes $y + \frac{1}{y} = \frac{41}{20}$.
Multiplying by $20y$,we get $20y^2 + 20 = 41y$,which rearranges to $20y^2 - 41y + 20 = 0$.
Factorizing the quadratic: $20y^2 - 16y - 25y + 20 = 0 \Rightarrow 4y(5y - 4) - 5(5y - 4) = 0$.
So,$(4y - 5)(5y - 4) = 0$,giving $y = 5/4$ or $y = 4/5$.
Case $1$: $\frac{x}{x+1} = \frac{5}{4} \Rightarrow 4x = 5x + 5 \Rightarrow x = -5$.
Case $2$: $\frac{x}{x+1} = \frac{4}{5} \Rightarrow 5x = 4x + 4 \Rightarrow x = 4$.
Thus,the solution set is $\{4, -5\}$.

(D) Let $y = \frac{x}{1-x}$. Then the equation becomes $y + \frac{1}{y} = \frac{13}{6}$.
Multiplying by $6y$,we get $6y^2 + 6 = 13y$,which simplifies to $6y^2 - 13y + 6 = 0$.
Factoring the quadratic equation: $6y^2 - 9y - 4y + 6 = 0 \implies 3y(2y - 3) - 2(2y - 3) = 0$.
This gives $(3y - 2)(2y - 3) = 0$,so $y = \frac{2}{3}$ or $y = \frac{3}{2}$.
Case $1$: $\frac{x}{1-x} = \frac{2}{3} \implies 3x = 2 - 2x \implies 5x = 2 \implies x = \frac{2}{5}$.
Case $2$: $\frac{x}{1-x} = \frac{3}{2} \implies 2x = 3 - 3x \implies 5x = 3 \implies x = \frac{3}{5}$.
Thus,the solution set is $\{\frac{2}{5}, \frac{3}{5}\}$.

(A) Given equation: $\frac{x+1}{x-1} + \frac{x-2}{x+2} = 3$
Taking the common denominator: $\frac{(x+1)(x+2) + (x-2)(x-1)}{(x-1)(x+2)} = 3$
Expanding the numerators and denominators: $\frac{(x^2 + 3x + 2) + (x^2 - 3x + 2)}{x^2 + x - 2} = 3$
Simplifying the numerator: $\frac{2x^2 + 4}{x^2 + x - 2} = 3$
Cross-multiplying: $2x^2 + 4 = 3(x^2 + x - 2)$
$2x^2 + 4 = 3x^2 + 3x - 6$
Rearranging terms to form a quadratic equation: $x^2 + 3x - 10 = 0$
Factoring the quadratic equation: $x^2 + 5x - 2x - 10 = 0$
$x(x + 5) - 2(x + 5) = 0$
$(x - 2)(x + 5) = 0$
Thus,$x = 2$ or $x = -5$.
The solution set is $\{-5, 2\}$.

(B) Given equation: $\frac{x+1}{x-1} - \frac{x-1}{x+1} = \frac{5}{x}$
Taking $LCM$ on the left side: $\frac{(x+1)^2 - (x-1)^2}{(x-1)(x+1)} = \frac{5}{x}$
Simplify the numerator: $(x^2 + 2x + 1) - (x^2 - 2x + 1) = 4x$
Simplify the denominator: $(x-1)(x+1) = x^2 - 1$
So,$\frac{4x}{x^2 - 1} = \frac{5}{x}$
Cross-multiply: $4x^2 = 5(x^2 - 1)$
$4x^2 = 5x^2 - 5$
$x^2 = 5$
$x = \pm\sqrt{5}$
Thus,the solution set is $\{-\sqrt{5}, \sqrt{5}\}$.

(C) Given quadratic equation is $m^{2}-18m+81=0$.
Comparing this with the standard form $a^{2}-2ab+b^{2}=(a-b)^{2}$,we can write:
$m^{2}-2(m)(9)+(9)^{2}=0$
This simplifies to:
$(m-9)^{2}=0$
Therefore:
$(m-9)(m-9)=0$
Setting each factor to zero:
$m-9=0$ or $m-9=0$
Solving for $m$:
$m=9$ or $m=9$
Thus,the roots of the given quadratic equation are $9$ and $9$.

(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 $\frac{1}{3}$.

(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 $x = \frac{a}{2}$.

(A) Given the formula for the sum of the first $n$ natural numbers is $S = \frac{n(n+1)}{2}$.
We are given $S = 300$,so we have the equation $\frac{n(n+1)}{2} = 300$.
Multiplying both sides by $2$,we get $n(n+1) = 600$.
Expanding this,we get $n^2 + n - 600 = 0$.
To solve this quadratic equation,we look for two numbers that multiply to $-600$ and add to $1$. These numbers are $25$ and $-24$.
So,$(n + 25)(n - 24) = 0$.
This gives $n = -25$ or $n = 24$.
Since $n$ represents the count of natural numbers,it must be positive.
Therefore,$n = 24$.

(C) The area of a right-angled triangle is given by the formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
Given,$\text{base} = x \, m$ and $\text{height} = (x+2) \, m$.
$\text{Area} = 84 \, m^2$.
So,$84 = \frac{1}{2} \times x \times (x+2)$.
$168 = x^2 + 2x$.
$x^2 + 2x - 168 = 0$.
Factoring the quadratic equation: $x^2 + 14x - 12x - 168 = 0$.
$x(x+14) - 12(x+14) = 0$.
$(x-12)(x+14) = 0$.
Since the length cannot be negative,$x = 12$.
The sides are $x = 12 \, m$ and $x+2 = 14 \, m$.

(D) Let the two natural numbers be $x$ and $y$.
Given that the sum of the numbers is $x + y = 5$,which implies $y = 5 - x$.
The sum of their reciprocals is $\frac{1}{x} + \frac{1}{y} = \frac{5}{6}$.
Substituting $y = 5 - x$ into the equation: $\frac{1}{x} + \frac{1}{5 - x} = \frac{5}{6}$.
Taking the common denominator: $\frac{(5 - x) + x}{x(5 - x)} = \frac{5}{6}$.
This simplifies to $\frac{5}{5x - x^2} = \frac{5}{6}$.
Dividing both sides by $5$,we get $\frac{1}{5x - x^2} = \frac{1}{6}$.
Therefore,$5x - x^2 = 6$,which rearranges to the quadratic equation $x^2 - 5x + 6 = 0$.
Factoring the quadratic equation: $(x - 2)(x - 3) = 0$.
Thus,$x = 2$ or $x = 3$.
If $x = 2$,then $y = 5 - 2 = 3$. If $x = 3$,then $y = 5 - 3 = 2$.
The two numbers are $2$ and $3$.

(A) Let the two parts be $x$ and $(29 - x)$.
According to the problem,the sum of their squares is $425$:
$x^2 + (29 - x)^2 = 425$
$x^2 + (841 - 58x + x^2) = 425$
$2x^2 - 58x + 841 - 425 = 0$
$2x^2 - 58x + 416 = 0$
Dividing by $2$:
$x^2 - 29x + 208 = 0$
Factoring the quadratic equation:
$x^2 - 13x - 16x + 208 = 0$
$x(x - 13) - 16(x - 13) = 0$
$(x - 13)(x - 16) = 0$
So,$x = 13$ or $x = 16$.
If $x = 13$,the other part is $29 - 13 = 16$.
If $x = 16$,the other part is $29 - 16 = 13$.
Thus,the two parts are $13$ and $16$.

(B) Let the two consecutive odd numbers be $x$ and $x + 2$.
According to the problem,the sum of their squares is $514$:
$x^2 + (x + 2)^2 = 514$
$x^2 + x^2 + 4x + 4 = 514$
$2x^2 + 4x + 4 = 514$
$2x^2 + 4x - 510 = 0$
Dividing by $2$:
$x^2 + 2x - 255 = 0$
Factoring the quadratic equation:
$x^2 + 17x - 15x - 255 = 0$
$x(x + 17) - 15(x + 17) = 0$
$(x - 15)(x + 17) = 0$
Since the numbers are positive,$x = 15$.
The two numbers are $15$ and $15 + 2 = 17$.

(C) Let the cost price of the book 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.
$119 = x + \frac{x^2}{100}$.
Multiplying by $100$,we get $11900 = 100x + x^2$.
Rearranging into a quadratic equation: $x^2 + 100x - 11900 = 0$.
Factoring the equation: $x^2 + 170x - 70x - 11900 = 0$.
$x(x + 170) - 70(x + 170) = 0$.
$(x - 70)(x + 170) = 0$.
Since the cost price cannot be negative,$x = 70$.
Therefore,the cost price of the book is Rs. $70$.

(A) To find the solution of the quadratic equation $x^{2} + 7x + 12 = 0$,we factorize the quadratic expression:
$x^{2} + 4x + 3x + 12 = 0$
$x(x + 4) + 3(x + 4) = 0$
$(x + 4)(x + 3) = 0$
Setting each factor to zero:
$x + 4 = 0 \implies x = -4$
$x + 3 = 0 \implies x = -3$
Comparing this with the given options,$-3$ is a solution.

(D) To find the roots of the quadratic equation $3x^2 - 4x + 1 = 0$,we can use the splitting the middle term method.
We need two numbers that multiply to $(3 \times 1) = 3$ and add to $-4$.
These numbers are $-3$ and $-1$.
Now,rewrite the equation: $3x^2 - 3x - x + 1 = 0$.
Factor by grouping: $3x(x - 1) - 1(x - 1) = 0$.
This gives $(3x - 1)(x - 1) = 0$.
Setting each factor to zero:
$3x - 1 = 0 \implies x = \frac{1}{3}$
$x - 1 = 0 \implies x = 1$
Therefore,the roots are $1, \frac{1}{3}$.

(B) Given the quadratic equation: $x^{2} + 5x + 6 = 0$.
To solve this,we factorize the quadratic expression by splitting the middle term:
$x^{2} + 2x + 3x + 6 = 0$
$x(x + 2) + 3(x + 2) = 0$
$(x + 2)(x + 3) = 0$
Setting each factor to zero:
$x + 2 = 0 \implies x = -2$
$x + 3 = 0 \implies x = -3$
Therefore,the solution set is $\{-2, -3\}$.

(A) To find the roots of the quadratic equation $x^{2}-2x-15=0$,we can use the factorization method.
We need to find two numbers whose product is $-15$ and whose sum is $-2$.
These two numbers are $-5$ and $3$,since $(-5) \times 3 = -15$ and $(-5) + 3 = -2$.
Now,rewrite the equation as: $x^{2}-5x+3x-15=0$.
Factor by grouping: $x(x-5)+3(x-5)=0$.
This gives: $(x-5)(x+3)=0$.
Setting each factor to zero,we get $x-5=0$ or $x+3=0$.
Therefore,$x=5$ or $x=-3$.
The roots of the equation are $5$ and $-3$.

(D) To find the solution set of the quadratic equation $x^{2}+5x-14=0$,we factorize the quadratic expression.
We look for two numbers whose product is $-14$ and whose sum is $5$.
These numbers are $7$ and $-2$,since $7 \times (-2) = -14$ and $7 + (-2) = 5$.
Thus,we can write the equation as: $x^{2} + 7x - 2x - 14 = 0$.
Grouping the terms,we get: $x(x + 7) - 2(x + 7) = 0$.
Factoring out $(x + 7)$,we have: $(x + 7)(x - 2) = 0$.
Setting each factor to zero,we get $x + 7 = 0$ or $x - 2 = 0$.
Therefore,$x = -7$ or $x = 2$.
The solution set is $\{-7, 2\}$.

(B) Given equation: $(x-7)^{2}-16=0$
Method $1$: Using the identity $a^2 - b^2 = (a-b)(a+b)$
$(x-7)^2 - 4^2 = 0$
$(x-7-4)(x-7+4) = 0$
$(x-11)(x-3) = 0$
Therefore,$x-11=0$ or $x-3=0$,which gives $x=11$ or $x=3$.
Method $2$: Expanding the square
$x^2 - 14x + 49 - 16 = 0$
$x^2 - 14x + 33 = 0$
$x^2 - 11x - 3x + 33 = 0$
$x(x-11) - 3(x-11) = 0$
$(x-11)(x-3) = 0$
Thus,the roots are $3$ and $11$.

(B) Given the quadratic equation: $x(x+1)-6=0$
Expanding the equation: $x^{2}+x-6=0$
To find the roots,we factorize the quadratic expression:
$x^{2}+3x-2x-6=0$
$x(x+3)-2(x+3)=0$
$(x+3)(x-2)=0$
Setting each factor to zero:
$x+3=0 \implies x=-3$
$x-2=0 \implies x=2$
Therefore,the roots are $-3$ and $2$.

(C) Given quadratic equation: $6 x^{2}-13 x+6=0$
Splitting the middle term: $6 x^{2}-9 x-4 x+6=0$
Factoring by grouping: $3 x(2 x-3)-2(2 x-3)=0$
$(2 x-3)(3 x-2)=0$
Setting each factor to zero: $2 x-3=0$ or $3 x-2=0$
Therefore,the roots are $x=\frac{3}{2}$ or $x=\frac{2}{3}$.

(A) Given equation: $x + \frac{2}{x} = 3$
Multiply the entire equation by $x$ to eliminate the denominator:
$x^2 + 2 = 3x$
Rearrange the equation into the standard form $ax^2 + bx + c = 0$:
$x^2 - 3x + 2 = 0$
Factor the quadratic equation:
$x^2 - 2x - x + 2 = 0$
$x(x - 2) - 1(x - 2) = 0$
$(x - 2)(x - 1) = 0$
Setting each factor to zero:
$x - 2 = 0 \implies x = 2$
$x - 1 = 0 \implies x = 1$
Thus,the roots are $2$ and $1$.

(B) To find the solution of the quadratic equation $x^{2} + 7x + 12 = 0$,we factorize the quadratic expression.
We look for two numbers whose product is $12$ and whose sum is $7$.
These numbers are $3$ and $4$.
So,the equation can be written as $x^{2} + 3x + 4x + 12 = 0$.
Grouping the terms,we get $x(x + 3) + 4(x + 3) = 0$.
Factoring out $(x + 3)$,we get $(x + 3)(x + 4) = 0$.
Setting each factor to zero,we get $x + 3 = 0$ or $x + 4 = 0$.
Thus,$x = -3$ or $x = -4$.
Comparing this with the given options,$-3$ is a solution.

(C) Given quadratic equation is $4x^{2} + 2x + \frac{1}{4} = 0$.
To simplify,multiply the entire equation by $4$:
$16x^{2} + 8x + 1 = 0$.
This is in the form of $(ax + b)^{2} = a^{2}x^{2} + 2abx + b^{2}$.
Here,$(4x)^{2} + 2(4x)(1) + (1)^{2} = 0$.
This simplifies to $(4x + 1)^{2} = 0$.
Therefore,$4x + 1 = 0$,which gives $4x = -1$.
So,$x = -\frac{1}{4}$.
The roots are $-\frac{1}{4}, -\frac{1}{4}$.

(C) Given the quadratic equation: $(x-3)^{2}-4=0$
We can rewrite this as: $(x-3)^{2}=4$
Taking the square root on both sides,we get: $x-3 = \pm \sqrt{4}$
$x-3 = \pm 2$
This gives two possible cases:
Case $1$: $x-3 = 2 \implies x = 2+3 = 5$
Case $2$: $x-3 = -2 \implies x = -2+3 = 1$
The solutions are $x=5$ and $x=1$.
Comparing this with the given options,$1$ is one of the solutions.

(B) Let the natural number be $x$.
According to the problem,the sum of the number and its reciprocal is $\frac{5}{2}$.
So,$x + \frac{1}{x} = \frac{5}{2}$.
Multiplying the entire equation by $2x$ to clear the denominators,we get:
$2x^2 + 2 = 5x$.
Rearranging the terms to form a quadratic equation:
$2x^2 - 5x + 2 = 0$.
Factoring the quadratic equation:
$2x^2 - 4x - x + 2 = 0$.
$2x(x - 2) - 1(x - 2) = 0$.
$(2x - 1)(x - 2) = 0$.
This gives two possible values for $x$:
$2x - 1 = 0 \implies x = \frac{1}{2}$ (not a natural number).
$x - 2 = 0 \implies x = 2$ (a natural number).
Therefore,the required natural number is $2$.

(A) Let $AB = x$ and $BC = y$.
Given that $x + y = 23$ and $xy = 120$.
We can form a quadratic equation $t^2 - (x+y)t + xy = 0$,which becomes $t^2 - 23t + 120 = 0$.
Factoring the quadratic equation: $t^2 - 15t - 8t + 120 = 0$.
$t(t - 15) - 8(t - 15) = 0$.
$(t - 15)(t - 8) = 0$.
So,the roots are $t = 15$ and $t = 8$.
Since $AB > BC$,we assign $AB = 15$ and $BC = 8$.
Therefore,$AB = 15$.