Mix Examples - Quadratic Equations Questions in English

(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 $6$.

(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,$n = 24$.

(A) Let the numerator be $x$. Then the denominator is $2x - 1$. The ratio is $\frac{x}{2x - 1}$.
According to the problem,the sum of the ratio and its reciprocal is $2 \frac{4}{15} = \frac{34}{15}$.
So,$\frac{x}{2x - 1} + \frac{2x - 1}{x} = \frac{34}{15}$.
Let $y = \frac{x}{2x - 1}$. Then $y + \frac{1}{y} = \frac{34}{15}$.
$15y^2 - 34y + 15 = 0$.
$15y^2 - 25y - 9y + 15 = 0$.
$5y(3y - 5) - 3(3y - 5) = 0$.
$(5y - 3)(3y - 5) = 0$.
So,$y = 3/5$ or $y = 5/3$.
If $y = 3/5$,then $\frac{x}{2x - 1} = \frac{3}{5} \implies 5x = 6x - 3 \implies x = 3$. The ratio is $\frac{3}{2(3) - 1} = \frac{3}{5}$.
If $y = 5/3$,then $\frac{x}{2x - 1} = \frac{5}{3} \implies 3x = 10x - 5 \implies 7x = 5 \implies x = 5/7$. The ratio is $\frac{5/7}{2(5/7) - 1} = \frac{5/7}{3/7} = \frac{5}{3}$.

(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,$x = 50 \,km/hr$.

(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 $xy = 10$.
When $27$ is added to the number,the digits are interchanged,so the new number is $10y + x$.
According to the problem: $(10x + y) + 27 = 10y + x$.
Rearranging the terms: $9x - 9y = -27$,which simplifies to $x - y = -3$ or $y - x = 3$.
Substitute $y = x + 3$ into the product equation: $x(x + 3) = 10$.
$x^2 + 3x - 10 = 0$.
$(x + 5)(x - 2) = 0$.
Since $x$ must be a positive digit,$x = 2$.
Then $y = 2 + 3 = 5$.
The number is $10(2) + 5 = 25$.

(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.

(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 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 $17\,cm$.

(A) Let the cost price of the calculator be $x$ rupees.
Given that the profit percentage is equal to the cost price,the profit percentage is $x\%$.
Profit = $\text{Cost Price} \times \frac{\text{Profit Percentage}}{100} = x \times \frac{x}{100} = \frac{x^2}{100}$.
Selling Price = $\text{Cost Price} + \text{Profit} = x + \frac{x^2}{100}$.
Given that the selling price is Rs. $56$,we have the equation: $x + \frac{x^2}{100} = 56$.
Multiplying by $100$,we get $100x + x^2 = 5600$,which simplifies to $x^2 + 100x - 5600 = 0$.
Factoring the quadratic equation: $x^2 + 140x - 40x - 5600 = 0$.
$x(x + 140) - 40(x + 140) = 0$.
$(x - 40)(x + 140) = 0$.
Since the cost price cannot be negative,$x = 40$.
Therefore,the cost price of the calculator is Rs. $40$.

(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,$x = 25\, km/hr$.

(C) Let the two numbers be $x$ and $y$.
Given: $x + y = 15$ (Equation $1$).
Given: $\frac{1}{x} + \frac{1}{y} = \frac{3}{10}$ (Equation $2$).
From Equation $2$,we get $\frac{x + y}{xy} = \frac{3}{10}$.
Substituting the value of $x + y$ from Equation $1$ into this expression: $\frac{15}{xy} = \frac{3}{10}$.
Solving for $xy$: $3xy = 150$,so $xy = 50$.
Now we have $x + y = 15$ and $xy = 50$.
These are the roots of the quadratic equation $t^2 - (x + y)t + xy = 0$.
Substituting the values: $t^2 - 15t + 50 = 0$.
Factoring the quadratic: $(t - 5)(t - 10) = 0$.
Thus,$t = 5$ or $t = 10$.
The two numbers are $5$ and $10$.

(D) Let the speed of the current be $x\,km/h$.
Speed downstream = $(9 + x)\,km/h$.
Speed upstream = $(9 - x)\,km/h$.
Time taken downstream = $\frac{12}{9 + x}$.
Time taken upstream = $\frac{12}{9 - x}$.
Total time = $3\,hours$,so $\frac{12}{9 + x} + \frac{12}{9 - x} = 3$.
Dividing by $3$: $\frac{4}{9 + x} + \frac{4}{9 - x} = 1$.
$4(9 - x) + 4(9 + x) = (9 + x)(9 - x)$.
$36 - 4x + 36 + 4x = 81 - x^2$.
$72 = 81 - x^2$.
$x^2 = 81 - 72 = 9$.
$x = 3\,km/h$ (since speed cannot be negative).

(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 $x - 20 = 100 - 20 = 80\, km/hr$.

(B) 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 $xy = 15$.
When the digits are interchanged,the new number is $10y + x$.
According to the problem,the new number is $18$ more than the original number:
$(10y + x) - (10x + y) = 18$
$9y - 9x = 18$
$y - x = 2$,which implies $y = x + 2$.
Substitute $y = x + 2$ into the product equation $xy = 15$:
$x(x + 2) = 15$
$x^2 + 2x - 15 = 0$
$(x + 5)(x - 3) = 0$
Since $x$ must be a positive digit,$x = 3$.
Then $y = 3 + 2 = 5$.
The original number is $10(3) + 5 = 35$.

(C) Let the side of the smaller square be $x\,m$ and the side of the bigger square be $y\,m$.
The perimeter of the smaller square is $4x$ and the perimeter of the bigger square is $4y$.
According to the problem,$4y = 4x + 12$,which simplifies to $y = x + 3$.
The area of the smaller square is $x^2$ and the area of the bigger square is $y^2$.
According to the problem,$3x^2 = y^2 + 11$.
Substitute $y = x + 3$ into the second equation:
$3x^2 = (x + 3)^2 + 11$
$3x^2 = x^2 + 6x + 9 + 11$
$2x^2 - 6x - 20 = 0$
$x^2 - 3x - 10 = 0$
$(x - 5)(x + 2) = 0$
Since the side length cannot be negative,$x = 5$.
Therefore,the side of the bigger square is $y = x + 3 = 5 + 3 = 8\,m$.

(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. $70$.

(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 $10(2) + 7 = 27$.

(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.

(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 $2$.

(B) The given quadratic equation is $2x^{2} + 5x - k = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = 2$,$b = 5$,and $c = -k$.
The discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Given that $D = 81$,we substitute the values:
$81 = (5)^{2} - 4(2)(-k)$
$81 = 25 + 8k$
$81 - 25 = 8k$
$56 = 8k$
$k = \frac{56}{8} = 7$.
Therefore,the value of $k$ is $7$.

(C) The given quadratic equation is $k x^{2}-4 x-4=0$.
Comparing this with the standard form $a x^{2}+b x+c=0$,we get $a=k$,$b=-4$,and $c=-4$.
The discriminant $D$ is given by the formula $D = b^{2}-4ac$.
Given that $D = 64$,we substitute the values into the formula:
$64 = (-4)^{2}-4(k)(-4)$
$64 = 16 + 16k$
Subtract $16$ from both sides:
$48 = 16k$
Divide by $16$:
$k = 3$.

(D) The given equation is $(3x - 14)^2 = 0$.
Expanding this,we get $9x^2 - 84x + 196 = 0$.
Comparing this with the standard quadratic equation $ax^2 + bx + c = 0$,we have $a = 9$,$b = -84$,and $c = 196$.
The discriminant $D$ is given by the formula $D = b^2 - 4ac$.
Substituting the values,$D = (-84)^2 - 4(9)(196)$.
$D = 7056 - 7056 = 0$.
Alternatively,for any quadratic equation of the form $(px + q)^2 = 0$,the roots are equal,which implies the discriminant $D$ is always $0$.

(A) Given the equation $(x+2)(x-5)=0$.
Expanding the brackets: $x^2 - 5x + 2x - 10 = 0$.
This simplifies to the standard quadratic form: $x^2 - 3x - 10 = 0$.
Comparing this with the standard form $ax^2 + bx + c = 0$,we get $a=1$,$b=-3$,and $c=-10$.
The discriminant $D$ is given by the formula $D = b^2 - 4ac$.
Substituting the values: $D = (-3)^2 - 4(1)(-10)$.
$D = 9 + 40 = 49$.

(B) Let the roots of the quadratic equation $ax^2 + bx + c = 0$ be $\alpha$ and $\beta$.
Given that the roots are reciprocal of each other,we have $\beta = 1/\alpha$,which implies $\alpha \cdot \beta = 1$.
For a quadratic equation $ax^2 + bx + c = 0$,the product of the roots is given by $\alpha \cdot \beta = c/a$.
Comparing $6x^2 - 13x + m = 0$ with $ax^2 + bx + c = 0$,we get $a = 6$,$b = -13$,and $c = m$.
Substituting these values into the product of roots formula: $1 = m/6$.
Therefore,$m = 6$.

(C) quadratic polynomial is a polynomial of degree $2$. According to the Fundamental Theorem of Algebra,a polynomial of degree $n$ has at most $n$ roots or zeros. Therefore,a quadratic polynomial $(n=2)$ has at most $2$ zeros.

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

(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) For a quadratic equation of the form $ax^2 + bx + c = 0$,the discriminant $D$ is given by the formula $D = b^2 - 4ac$.
Comparing the given equation $3x^2 - 6x + 4 = 0$ with the standard form,we have $a = 3$,$b = -6$,and $c = 4$.
Substituting these values into the formula:
$D = (-6)^2 - 4(3)(4)$
$D = 36 - 48$
$D = -12$
Therefore,the discriminant is $-12$.

(D) The given quadratic equation is $x^{2} + 12x + 36 = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = 1$,$b = 12$,and $c = 36$.
The discriminant $D$ is calculated as $D = b^{2} - 4ac$.
$D = (12)^{2} - 4(1)(36) = 144 - 144 = 0$.
Since the discriminant $D = 0$,the roots of the quadratic equation are real and equal.
Alternatively,the equation can be written as $(x + 6)^{2} = 0$,which gives $x = -6, -6$.

(A) To determine the nature of the roots of the quadratic equation $ax^2 + bx + c = 0$,we calculate the discriminant $D = b^2 - 4ac$.
Here,$a = 9$,$b = 30$,and $c = 25$.
Substituting these values into the discriminant formula:
$D = (30)^2 - 4(9)(25)$
$D = 900 - 900$
$D = 0$
Since the discriminant $D = 0$,the quadratic equation has real and equal roots.

(B) The given quadratic equation is $x^{2}-10x+(2k-1)=0$.
Comparing this with the standard form $ax^{2}+bx+c=0$,we get $a=1$,$b=-10$,and $c=2k-1$.
The discriminant $D$ is given by the formula $D=b^{2}-4ac$.
Given $D=40$,we substitute the values:
$40 = (-10)^{2} - 4(1)(2k-1)$
$40 = 100 - 4(2k-1)$
$40 = 100 - 8k + 4$
$40 = 104 - 8k$
$8k = 104 - 40$
$8k = 64$
$k = 8$.

(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:
$2(\frac{7}{2})^{2} - 5(\frac{7}{2}) + k = 0$
$2(\frac{49}{4}) - \frac{35}{2} + k = 0$
$\frac{49}{2} - \frac{35}{2} + k = 0$
$\frac{14}{2} + k = 0$
$7 + k = 0$
$k = -7$

(D) For a quadratic equation $ax^{2} + bx + c = 0$,the product of the roots is given by $\alpha \cdot \beta = c/a$.
Given the equation $2x^{2} + 5x + 3 = 0$,we have $a = 2$,$b = 5$,and $c = 3$.
Let the roots be $\alpha = -1$ and $\beta$ be the other root.
Using the product of roots formula: $(-1) \cdot \beta = 3/2$.
Therefore,$\beta = -3/2$.
Alternatively,using the sum of roots: $\alpha + \beta = -b/a$.
$-1 + \beta = -5/2$.
$\beta = -5/2 + 1 = -3/2$.

(D) quadratic polynomial in one variable $x$ is an algebraic expression of the form $ax^2 + bx + c$,where $a, b,$ and $c$ are real numbers $(a, b, c \in R)$ and the coefficient of the squared term must be non-zero $(a \neq 0)$.
If we set the polynomial equal to zero,it becomes a quadratic equation $(ax^2 + bx + c = 0)$.
Since the question asks for the standard form of a quadratic polynomial,the correct expression is $ax^2 + bx + c$ with the condition $a \neq 0$.

(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 $(a, b, c \in R)$.