Quadratic Equations Questions in English

(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,$x = 24$.

(C) Given equation: $16x^{2} - 24x - 1 = 0$.
Divide the entire equation by $16$ to make the coefficient of $x^{2}$ equal to $1$:
$x^{2} - \frac{24}{16}x - \frac{1}{16} = 0 \implies x^{2} - \frac{3}{2}x - \frac{1}{16} = 0$.
Move the constant term to the right side:
$x^{2} - \frac{3}{2}x = \frac{1}{16}$.
Add the square of half the coefficient of $x$ (which is $\frac{1}{2} \times \frac{3}{2} = \frac{3}{4}$,and its square is $\frac{9}{16}$) to both sides:
$x^{2} - \frac{3}{2}x + \left(\frac{3}{4}\right)^{2} = \frac{1}{16} + \frac{9}{16}$.
Rewrite the left side as a perfect square:
$\left(x - \frac{3}{4}\right)^{2} = \frac{10}{16}$.
Take the square root of both sides:
$x - \frac{3}{4} = \pm \sqrt{\frac{10}{16}} = \pm \frac{\sqrt{10}}{4}$.
Solve for $x$:
$x = \frac{3}{4} \pm \frac{\sqrt{10}}{4} = \frac{3 \pm \sqrt{10}}{4}$.
Thus,the roots are $\frac{3+\sqrt{10}}{4}$ and $\frac{3-\sqrt{10}}{4}$.

(D) Given equation: $3y^{2} + 7y - 20 = 0$.
Divide by $3$: $y^{2} + \frac{7}{3}y - \frac{20}{3} = 0$.
Add and subtract $(\frac{1}{2} \times \text{coefficient of } y)^{2} = (\frac{7}{6})^{2} = \frac{49}{36}$:
$y^{2} + \frac{7}{3}y + \frac{49}{36} - \frac{49}{36} - \frac{20}{3} = 0$.
$(y + \frac{7}{6})^{2} - (\frac{49 + 240}{36}) = 0$.
$(y + \frac{7}{6})^{2} = \frac{289}{36}$.
Taking square root: $y + \frac{7}{6} = \pm \frac{17}{6}$.
Case $1$: $y = \frac{17}{6} - \frac{7}{6} = \frac{10}{6} = \frac{5}{3}$.
Case $2$: $y = -\frac{17}{6} - \frac{7}{6} = -\frac{24}{6} = -4$.
Thus,the roots are $-4, \frac{5}{3}$.

(A) Given equation: $5x^{2} - 4x - 10 = 0$.
Divide the entire equation by $5$ to make the coefficient of $x^{2}$ equal to $1$:
$x^{2} - \frac{4}{5}x - 2 = 0$.
Shift the constant term to the right side:
$x^{2} - \frac{4}{5}x = 2$.
Add the square of half the coefficient of $x$ (which is $(\frac{1}{2} \times \frac{4}{5})^{2} = (\frac{2}{5})^{2} = \frac{4}{25}$) to both sides:
$x^{2} - \frac{4}{5}x + \frac{4}{25} = 2 + \frac{4}{25}$.
Rewrite the left side as a perfect square:
$(x - \frac{2}{5})^{2} = \frac{50 + 4}{25} = \frac{54}{25}$.
Take the square root of both sides:
$x - \frac{2}{5} = \pm \sqrt{\frac{54}{25}} = \pm \frac{3\sqrt{6}}{5}$.
Solve for $x$:
$x = \frac{2}{5} \pm \frac{3\sqrt{6}}{5} = \frac{2 \pm 3\sqrt{6}}{5}$.
Thus,the roots are $\frac{2 - 3\sqrt{6}}{5}$ and $\frac{2 + 3\sqrt{6}}{5}$.

(B) To solve the equation $m^{2} - 18m + 81 = 0$ by the method of completing the square:
$1$. The equation is already in the form $m^{2} - 18m = -81$.
$2$. To complete the square,we add the square of half the coefficient of $m$ to both sides. The coefficient of $m$ is $-18$,so half of it is $-9$,and $(-9)^{2} = 81$.
$3$. Adding $81$ to both sides: $m^{2} - 18m + 81 = -81 + 81$.
$4$. This simplifies to $(m - 9)^{2} = 0$.
$5$. Taking the square root of both sides,we get $m - 9 = 0$,which implies $m = 9$.
$6$. Since it is a quadratic equation,the roots are $9, 9$.

(C) Given equation: $6x^2 + 11x + 3 = 0$
Divide the entire equation by $6$ to make the coefficient of $x^2$ equal to $1$:
$x^2 + \frac{11}{6}x + \frac{3}{6} = 0 \implies x^2 + \frac{11}{6}x + \frac{1}{2} = 0$
Shift the constant term to the right side:
$x^2 + \frac{11}{6}x = -\frac{1}{2}$
Add the square of half the coefficient of $x$ (i.e.,$(\frac{1}{2} \cdot \frac{11}{6})^2 = (\frac{11}{12})^2 = \frac{121}{144}$) to both sides:
$x^2 + \frac{11}{6}x + \frac{121}{144} = -\frac{1}{2} + \frac{121}{144}$
$(x + \frac{11}{12})^2 = -\frac{72}{144} + \frac{121}{144} = \frac{49}{144}$
Taking the square root on both sides:
$x + \frac{11}{12} = \pm \frac{7}{12}$
Case $1$: $x = \frac{7}{12} - \frac{11}{12} = -\frac{4}{12} = -\frac{1}{3}$
Case $2$: $x = -\frac{7}{12} - \frac{11}{12} = -\frac{18}{12} = -\frac{3}{2}$
Thus,the roots are $-\frac{3}{2}$ and $-\frac{1}{3}$.

(D) Let the two parts be $x$ and $16 - x$. Assume $x$ is the larger part.
According to the problem,$2x^2 - (16 - x)^2 = 164$.
Expanding the equation: $2x^2 - (256 - 32x + x^2) = 164$.
$2x^2 - 256 + 32x - x^2 = 164$.
$x^2 + 32x - 420 = 0$.
Factoring the quadratic equation: $x^2 + 42x - 10x - 420 = 0$.
$x(x + 42) - 10(x + 42) = 0$.
$(x - 10)(x + 42) = 0$.
Since the parts must be positive,$x = 10$.
The other part is $16 - 10 = 6$.
Thus,the two parts are $10$ and $6$.

(C) Let the usual speed of the train be $x \,km/hr$.
Time taken to cover $900 \,km$ at usual speed is $T_1 = \frac{900}{x} \,hours$.
When the speed is decreased by $10 \,km/hr$,the new speed is $(x - 10) \,km/hr$.
Time taken to cover $900 \,km$ at the new speed is $T_2 = \frac{900}{x - 10} \,hours$.
According to the problem,$T_2 - T_1 = 4.5 \,hours$ (which is $9/2 \,hours$).
So,$\frac{900}{x - 10} - \frac{900}{x} = \frac{9}{2}$.
Dividing by $9$,we get $\frac{100}{x - 10} - \frac{100}{x} = \frac{1}{2}$.
$100 \left( \frac{x - (x - 10)}{x(x - 10)} \right) = \frac{1}{2}$.
$100 \left( \frac{10}{x^2 - 10x} \right) = \frac{1}{2}$.
$1000 \times 2 = x^2 - 10x$.
$x^2 - 10x - 2000 = 0$.
Solving the quadratic equation: $x^2 - 50x + 40x - 2000 = 0$.
$x(x - 50) + 40(x - 50) = 0$.
$(x - 50)(x + 40) = 0$.
Since speed cannot be negative,$x = 50 \,km/hr$.

(A) Let the original price of pure butter be $x$ Rs. per $kg$.
The quantity of butter purchased for Rs. $960$ at the original price is $\frac{960}{x} \ kg$.
When the price increases by Rs. $40$ per $kg$,the new price becomes $(x + 40)$ Rs. per $kg$.
The quantity of butter purchased for Rs. $960$ at the new price is $\frac{960}{x + 40} \ kg$.
According to the problem,the difference in quantity is $2 \ kg$:
$\frac{960}{x} - \frac{960}{x + 40} = 2$
Divide by $2$:
$\frac{480}{x} - \frac{480}{x + 40} = 1$
$480(x + 40) - 480x = x(x + 40)$
$480x + 19200 - 480x = x^2 + 40x$
$x^2 + 40x - 19200 = 0$
Solving the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-40 \pm \sqrt{1600 - 4(1)(-19200)}}{2}$
$x = \frac{-40 \pm \sqrt{1600 + 76800}}{2} = \frac{-40 \pm \sqrt{78400}}{2} = \frac{-40 \pm 280}{2}$
Since price cannot be negative,$x = \frac{240}{2} = 120$.
Thus,the original price is Rs. $120$ per $kg$.

(B) Let the original price of kerosene be $x$ Rs./litre.
Total amount spent = Rs. $360$.
Original quantity of kerosene = $\frac{360}{x}$ litres.
New price of kerosene = $(x + 2)$ Rs./litre.
New quantity of kerosene = $\frac{360}{x + 2}$ litres.
According to the problem,the difference in quantity is $2$ litres:
$\frac{360}{x} - \frac{360}{x + 2} = 2$
Divide by $2$:
$\frac{180}{x} - \frac{180}{x + 2} = 1$
$180(x + 2) - 180x = x(x + 2)$
$180x + 360 - 180x = x^2 + 2x$
$x^2 + 2x - 360 = 0$
$(x + 20)(x - 18) = 0$
Since price cannot be negative,$x = 18$.
Therefore,the original price of kerosene is Rs. $18$ per litre.

(D) Let the length of the hypotenuse be $h\,cm$,the base be $b\,cm$,and the altitude be $a\,cm$.
Given: $h = b + 2 \implies b = h - 2$.
Also,$h = 2a + 1 \implies a = \frac{h - 1}{2}$.
By the Pythagorean theorem,$h^2 = b^2 + a^2$.
Substituting the expressions for $b$ and $a$: $h^2 = (h - 2)^2 + \left(\frac{h - 1}{2}\right)^2$.
$h^2 = h^2 - 4h + 4 + \frac{h^2 - 2h + 1}{4}$.
Multiply by $4$: $4h^2 = 4h^2 - 16h + 16 + h^2 - 2h + 1$.
$0 = h^2 - 18h + 17$.
Factoring the quadratic equation: $(h - 17)(h - 1) = 0$.
Since $h$ must be greater than $2$ (as $b = h - 2 > 0$),we have $h = 17$.
Thus,the length of the hypotenuse is $17\,cm$.

(B) Let the usual speed of the train be $x\, km/hr$.
Time taken to cover $400\, km$ at usual speed is $T_1 = \frac{400}{x}$ hours.
When the speed is decreased by $5\, km/hr$,the new speed is $(x - 5)\, km/hr$.
Time taken at the new speed is $T_2 = \frac{400}{x - 5}$ hours.
According to the problem,$T_2 - T_1 = 4$.
So,$\frac{400}{x - 5} - \frac{400}{x} = 4$.
Dividing by $4$,we get $\frac{100}{x - 5} - \frac{100}{x} = 1$.
$100x - 100(x - 5) = x(x - 5)$.
$100x - 100x + 500 = x^2 - 5x$.
$x^2 - 5x - 500 = 0$.
$(x - 25)(x + 20) = 0$.
Since speed cannot be negative,$x = 25\, km/hr$.

(A) Let the speed of the faster train be $x\, km/hr$. Then the speed of the slower train is $(x - 20)\, km/hr$.
The time taken by the faster train is $T_1 = \frac{400}{x}$ hours.
The time taken by the slower train is $T_2 = \frac{400}{x - 20}$ hours.
According to the problem,the faster train takes $1$ hour less than the slower train,so $T_2 - T_1 = 1$.
$\frac{400}{x - 20} - \frac{400}{x} = 1$
$400 \left( \frac{x - (x - 20)}{x(x - 20)} \right) = 1$
$400 \times 20 = x^2 - 20x$
$x^2 - 20x - 8000 = 0$
$(x - 100)(x + 80) = 0$
Since speed cannot be negative,$x = 100\, km/hr$.
The speed of the slower train is $x - 20 = 100 - 20 = 80\, km/hr$.

(D) Let the tens digit be $x$ and the units digit be $y$. The number is $10x + y$.
Given that the product of the digits is $14$,so $xy = 14$.
When the digits are interchanged,the new number is $10y + x$.
According to the problem,the new number is $45$ more than the original number:
$(10y + x) - (10x + y) = 45$
$9y - 9x = 45$
$y - x = 5$,which implies $y = x + 5$.
Substitute $y = x + 5$ into $xy = 14$:
$x(x + 5) = 14$
$x^2 + 5x - 14 = 0$
$(x + 7)(x - 2) = 0$
Since $x$ must be a positive digit,$x = 2$.
Then $y = 2 + 5 = 7$.
The original number is $10(2) + 7 = 27$.

(B) Given the quadratic equation $k x^{2} + 3x - 4 = 0$.
Since $x = 2$ is a solution,we substitute $x = 2$ into the equation:
$k(2)^{2} + 3(2) - 4 = 0$
$4k + 6 - 4 = 0$
$4k + 2 = 0$
$4k = -2$
$k = -2/4$
$k = -1/2$
Therefore,the value of $k$ is $-1/2$.

(C) To determine which quadratic equation has a root $x=3$,we substitute $x=3$ into each equation:
For option $C$: $x^{2}-8x+15=0$
Substituting $x=3$: $(3)^{2}-8(3)+15 = 9-24+15 = 24-24 = 0$.
Since the result is $0$,$x=3$ is a root of the equation $x^{2}-8x+15=0$.

(C) Given the quadratic equation $x^{2} + 6x + k = 0$.
Since $-4$ is a root of the equation,it must satisfy the equation.
Substitute $x = -4$ into the equation:
$(-4)^{2} + 6(-4) + k = 0$
$16 - 24 + k = 0$
$-8 + k = 0$
$k = 8$
Therefore,the value of $k$ is $8$.

(A) Given the quadratic equation $k x^{2} - 7x + 6 = 0$.
Since $\frac{3}{2}$ is a root of the equation,it must satisfy the equation.
Substitute $x = \frac{3}{2}$ into the equation:
$k(\frac{3}{2})^{2} - 7(\frac{3}{2}) + 6 = 0$
$k(\frac{9}{4}) - \frac{21}{2} + 6 = 0$
Multiply the entire equation by $4$ to clear the denominators:
$9k - 42 + 24 = 0$
$9k - 18 = 0$
$9k = 18$
$k = 2$
Therefore,the value of $k$ is $2$.

(A) Let the roots of the quadratic equation $x^{2}-2x-c=0$ be $\alpha$ and $\beta$.
Given that one root $\alpha = 5$.
The sum of the roots of a quadratic equation $ax^{2}+bx+c=0$ is given by $-\frac{b}{a}$.
Here,$a = 1$,$b = -2$,and $c = -c$.
Sum of roots $\alpha + \beta = -\frac{-2}{1} = 2$.
Substituting $\alpha = 5$,we get $5 + \beta = 2$.
Therefore,$\beta = 2 - 5 = -3$.
Thus,the other root is $-3$.

(B) For a quadratic equation of the form $ax^{2} + bx + c = 0$,the product of the roots is given by $\frac{c}{a}$.
Here,the equation is $x^{2} + bx - 12 = 0$,where $a = 1$ and $c = -12$.
Let the roots be $\alpha$ and $\beta$. We are given $\alpha = 2$.
The product of the roots $\alpha \cdot \beta = \frac{c}{a} = \frac{-12}{1} = -12$.
Substituting the value of $\alpha$:
$2 \cdot \beta = -12$.
$\beta = \frac{-12}{2} = -6$.
Thus,the other root is $-6$.

(C) Given the quadratic equation $2x^{2}-5x+k=0$.
Since $x = \frac{7}{2}$ is a root of the equation,it must satisfy the equation.
Substitute $x = \frac{7}{2}$ into the equation:
$2(\frac{7}{2})^{2} - 5(\frac{7}{2}) + k = 0$
$2(\frac{49}{4}) - \frac{35}{2} + k = 0$
$\frac{49}{2} - \frac{35}{2} + k = 0$
$\frac{14}{2} + k = 0$
$7 + k = 0$
$k = -7$

(C) quadratic equation in one variable $x$ is an equation of the form $ax^2 + bx + c = 0$,where $a, b,$ and $c$ are real numbers and $a \neq 0$.
If $a = 0$,the equation becomes a linear equation $(bx + c = 0)$,which is not quadratic.
Therefore,the standard form is $ax^2 + bx + c = 0$ with the condition $a \neq 0$.

(B) quadratic equation is a polynomial equation of degree $2$.
The standard form of a quadratic equation is $ax^2 + bx + c = 0$.
For the equation to be quadratic,the coefficient of the $x^2$ term must not be zero.
If $a = 0$,the equation becomes $bx + c = 0$,which is a linear equation,not a quadratic one.
Therefore,the necessary condition is $a \neq 0$.

(D) quadratic equation is defined as a polynomial equation of degree $2$.
For an equation to be quadratic,the coefficient of the $x^2$ term must not be zero,because if $a = 0$,the equation reduces to a linear equation $(bx + c = 0)$.
Therefore,the condition for $ax^2 + bx + c = 0$ to be a quadratic equation is $a \neq 0$,where $a, b,$ and $c$ are real numbers $(a, b, c \in R)$.

(A) quadratic equation in one variable is an equation of the form $ax^{2} + bx + c = 0$,where $a, b, c$ are real numbers and $a \neq 0$.
Option $A$: $3x^{2} + \sqrt{2}x - 7 = 0$ is a quadratic equation in one variable $(x)$.
Option $B$: $3x^{2} + 2y - 6 = 0$ has two variables ($x$ and $y$).
Option $C$: $4x - y = 5$ is a linear equation in two variables.
Option $D$: $5x^{3} - 1 = 2$ is a cubic equation in one variable.

(C) quadratic equation is an equation of the form $ax^2 + bx + c = 0$,where $a \neq 0$.
Let us simplify each option:
$A$: $x(3x + 7) = (x + 1)(x - 1) \implies 3x^2 + 7x = x^2 - 1 \implies 2x^2 + 7x + 1 = 0$. This is a quadratic equation.
$B$: $x^2 - 2x + 1 = 0$. This is already in the form $ax^2 + bx + c = 0$ with $a = 1$. This is a quadratic equation.
$C$: $2x(3x - 5) + 1 = 3x(2x + 5) + 3 \implies 6x^2 - 10x + 1 = 6x^2 + 15x + 3 \implies -10x + 1 = 15x + 3 \implies 25x + 2 = 0$. This is a linear equation,not a quadratic equation.
$D$: $4 - 3x - 2x^2 = 0$. This is a quadratic equation with $a = -2$.

(B) To determine the nature of the given expression,we expand the product:
$(x+6)(x+5) = x^2 + 5x + 6x + 30 = x^2 + 11x + 30$.
Since the equation is given as $(x+6)(x+5) = 0$,we can write it as $x^2 + 11x + 30 = 0$.
An equation of the form $ax^2 + bx + c = 0$,where $a \neq 0$,is defined as a quadratic equation.
Since the highest power of the variable $x$ is $2$,it is a quadratic equation.

(A) quadratic equation is an equation of the form $ax^2 + bx + c = 0$,where $a, b, c$ are real numbers and $a \neq 0$.
Let us analyze each option:
$A$: $x + \frac{1}{x} = 2 \implies x^2 + 1 = 2x \implies x^2 - 2x + 1 = 0$. This is a quadratic equation.
$B$: $x^2 + \frac{1}{x} = 2 \implies x^3 + 1 = 2x \implies x^3 - 2x + 1 = 0$. This is a cubic equation.
$C$: $x + \frac{1}{x^2} = 3 \implies x^3 + 1 = 3x^2 \implies x^3 - 3x^2 + 1 = 0$. This is a cubic equation.
$D$: $x(x + 1) = 2 \implies x^2 + x = 2 \implies x^2 + x - 2 = 0$. This is also a quadratic equation.
Note: In the original provided options,$A$ is the standard form representation. If multiple options are valid,$A$ is the most direct representation.

(A) To determine which equation has $x=2$ as a solution,we substitute $x=2$ into each given equation and check if the result is $0$.
For option $A$: $x^{2}-3x+2=0$
Substitute $x=2$: $(2)^{2}-3(2)+2 = 4-6+2 = 0$.
Since the result is $0$,$x=2$ is a solution for this equation.
For option $B$: $x^{2}+x+2=0$
Substitute $x=2$: $(2)^{2}+2+2 = 4+2+2 = 8 \neq 0$.
For option $C$: $x^{2}-2x+2=0$
Substitute $x=2$: $(2)^{2}-2(2)+2 = 4-4+2 = 2 \neq 0$.
For option $D$: $2x^{2}+2x+1=0$
Substitute $x=2$: $2(2)^{2}+2(2)+1 = 2(4)+4+1 = 8+4+1 = 13 \neq 0$.
Therefore,the correct option is $A$.

(B) Given that $3$ is a root of the quadratic equation $x^{2}-kx+6=0$.
Substituting $x=3$ into the equation:
$(3)^{2}-k(3)+6=0$
$9-3k+6=0$
$15-3k=0$
$3k=15$
$k=5$
Therefore,the value of $k$ is $5$.

(B) Since $-3$ is a root of the quadratic equation $x^{2}+3(k+2)x-9=0$,it must satisfy the equation.
Substituting $x=-3$ into the equation:
$(-3)^{2}+3(k+2)(-3)-9=0$
$9-9(k+2)-9=0$
$9-9k-18-9=0$
$-9k-18=0$
$-9k=18$
$k=-2$

(A) Given that $1$ is the solution (root) of the quadratic equation $3x^{2}-kx+2=0$.
Substituting $x=1$ into the equation:
$3(1)^{2}-k(1)+2=0$
$3(1)-k+2=0$
$3-k+2=0$
$5-k=0$
Therefore,$k=5$.

(B) Given that $x=-2$ is a solution of the quadratic equation $k x^{2}+5 x+2=0$.
Substituting $x=-2$ into the equation:
$k(-2)^{2}+5(-2)+2=0$
$k(4)-10+2=0$
$4k-8=0$
$4k=8$
$k=2$
Therefore,the value of $k$ is $2$.

(B) Given that $-3$ is a solution of the quadratic equation $2x^{2} + 5x + k = 0$.
Substituting $x = -3$ into the equation:
$2(-3)^{2} + 5(-3) + k = 0$
$2(9) - 15 + k = 0$
$18 - 15 + k = 0$
$3 + k = 0$
$k = -3$

(D) quadratic expression $ax^2 + bx + c$ is a perfect square if its discriminant $D = b^2 - 4ac$ is equal to $0$.
Here,$a = 16$,$b = 40$,and $c = k$.
Setting the discriminant to zero: $D = (40)^2 - 4(16)(k) = 0$.
$1600 - 64k = 0$.
$64k = 1600$.
$k = \frac{1600}{64} = 25$.
Alternatively,for a perfect square of the form $(mx + n)^2 = m^2x^2 + 2mnx + n^2$,we compare $16x^2 + 40x + k$ with $(4x + n)^2 = 16x^2 + 8nx + n^2$.
Equating the middle terms: $8n = 40$,so $n = 5$.
Then $k = n^2 = 5^2 = 25$.

(B) The golden ratio,denoted by $\phi$,is defined as the ratio of two quantities such that the ratio of the sum of the quantities to the larger quantity is equal to the ratio of the larger quantity to the smaller one.
Mathematically,for two quantities $a$ and $b$ where $a > b > 0$,the golden ratio is defined as $\frac{a+b}{a} = \frac{a}{b} = \phi$.
This leads to the quadratic equation $\phi^2 - \phi - 1 = 0$.
Using the quadratic formula $\phi = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $\phi = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{5}}{2}$.
Since the ratio must be positive,we take the positive root: $\phi = \frac{1+\sqrt{5}}{2} \approx 1.618$.

(C) Let $x = \frac{1+\sqrt{5}}{2}$.
Multiply both sides by $2$: $2x = 1+\sqrt{5}$.
Subtract $1$ from both sides: $2x-1 = \sqrt{5}$.
Square both sides: $(2x-1)^{2} = (\sqrt{5})^{2}$.
Expand the left side: $4x^{2}-4x+1 = 5$.
Rearrange the terms: $4x^{2}-4x-4 = 0$.
Divide the entire equation by $4$: $x^{2}-x-1 = 0$.
Thus,the golden number is a root of the quadratic equation $x^{2}-x-1=0$.

(C) Given the quadratic equation $x^{2} = 49$.
Subtracting $49$ from both sides,we get $x^{2} - 49 = 0$.
This is in the form of $a^{2} - b^{2} = (a - b)(a + b)$,where $a = x$ and $b = 7$.
So,$(x - 7)(x + 7) = 0$.
Setting each factor to zero,we get $x - 7 = 0$ or $x + 7 = 0$.
Therefore,$x = 7$ or $x = -7$.

(D) Given equation: $(x+2)^{2} = 16$
Taking the square root on both sides: $x+2 = \pm 4$
Case $1$: $x+2 = 4 \implies x = 4-2 = 2$
Case $2$: $x+2 = -4 \implies x = -4-2 = -6$
Therefore,the roots are $2$ and $-6$.

(D) quadratic equation with roots $\alpha$ and $\beta$ is given by the formula: $(x - \alpha)(x - \beta) = 0$.
Given the roots are $\alpha = -1$ and $\beta = 2$.
Substituting these values into the formula,we get: $(x - (-1))(x - 2) = 0$.
This simplifies to: $(x + 1)(x - 2) = 0$.
Expanding the expression: $x(x) + x(-2) + 1(x) + 1(-2) = 0$.
$x^{2} - 2x + x - 2 = 0$.
$x^{2} - x - 2 = 0$.
Therefore,the correct quadratic equation is $x^{2} - x - 2 = 0$.

(D) quadratic equation with roots $\alpha$ and $\beta$ is given by the formula: $x^{2} - (\alpha + \beta)x + (\alpha \cdot \beta) = 0$.
Here,the roots are $\alpha = -8$ and $\beta = 8$.
Sum of the roots: $\alpha + \beta = -8 + 8 = 0$.
Product of the roots: $\alpha \cdot \beta = (-8) \times (8) = -64$.
Substituting these values into the formula:
$x^{2} - (0)x + (-64) = 0$
$x^{2} - 64 = 0$
$x^{2} = 64$.
Thus,the correct option is $D$.

(C) The standard form of a quadratic equation is given by $ax^{2}+bx+c=0$.
Given the equation $5x^{2}-4x-1=0$.
Comparing this with the standard form $ax^{2}+bx+c=0$,we identify the coefficients:
$a = 5$
$b = -4$
$c = -1$
Therefore,the value of $b$ is $-4$.

(C) The given quadratic equation is $x(2x - 1) - 5 = 0$.
Expanding the equation,we get $2x^2 - x - 5 = 0$.
The standard form of a quadratic equation is $ax^2 + bx + c = 0$.
Comparing $2x^2 - x - 5 = 0$ with $ax^2 + bx + c = 0$,we identify the coefficients as $a = 2$,$b = -1$,and $c = -5$.
Therefore,the value of $a$ is $2$.

(D) Since $-4$ is a root of the quadratic equation $x^{2}-ax-8=0$,it must satisfy the equation.
Substituting $x = -4$ into the equation:
$(-4)^{2} - a(-4) - 8 = 0$
$16 + 4a - 8 = 0$
$8 + 4a = 0$
$4a = -8$
$a = -2$

(A) For a quadratic equation of the form $ax^{2}+bx+c=0$,the sum of the roots is given by $\alpha+\beta = -\frac{b}{a}$.
Here,$a=1$,$b=4$,and $c=m$.
Given that one root $\alpha = 2$,let the other root be $\beta$.
Substituting the values into the sum formula: $2 + \beta = -\frac{4}{1}$.
$2 + \beta = -4$.
$\beta = -4 - 2$.
$\beta = -6$.
Therefore,the other root is $-6$.

(B) Let the quadratic equation be $x^{2}+bx-15=0$.
Given that one root is $\alpha = 3$.
Let the other root be $\beta$.
For a quadratic equation $ax^{2}+bx+c=0$,the product of the roots is given by $\alpha \cdot \beta = \frac{c}{a}$.
Here,$a = 1$,$b = b$,and $c = -15$.
Substituting the values,we get $3 \cdot \beta = \frac{-15}{1}$.
$3 \cdot \beta = -15$.
$\beta = \frac{-15}{3} = -5$.
Therefore,the other root is $-5$.

(D) quadratic equation with roots $\alpha$ and $\beta$ is given by the formula: $x^{2} - (\alpha + \beta)x + (\alpha \cdot \beta) = 0$.
Given roots are $\alpha = -4$ and $\beta = 5$.
Sum of roots: $\alpha + \beta = -4 + 5 = 1$.
Product of roots: $\alpha \cdot \beta = (-4) \times 5 = -20$.
Substituting these values into the formula: $x^{2} - (1)x + (-20) = 0$.
This simplifies to: $x^{2} - x - 20 = 0$.
Therefore,the correct option is $D$.