Nature of Roots Questions in English

(A) 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 $x^{2}+2x-2=0$ with the standard form,we have $a=1$,$b=2$,and $c=-2$.
Substituting these values into the formula:
$D = (2)^{2} - 4(1)(-2)$
$D = 4 - (-8)$
$D = 4 + 8$
$D = 12$.
Therefore,the discriminant is $12$.

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

(C) The given quadratic equation is in the form $ax^{2}+bx+c=0$.
Comparing $x^{2}-5x-14=0$ with the standard form,we get $a=1$,$b=-5$,and $c=-14$.
The discriminant $D$ is given by the formula $D=b^{2}-4ac$.
Substituting the values,we get $D=(-5)^{2}-4(1)(-14)$.
$D=25+56$.
$D=81$.
Therefore,the discriminant is $81$.

(D) For a quadratic equation in the form $ax^{2}+bx+c=0$,the discriminant $D$ is given by the formula $D = b^{2}-4ac$.
Comparing the given equation $x^{2}-7x+10=0$ with the standard form,we have $a=1$,$b=-7$,and $c=10$.
Substituting these values into the formula:
$D = (-7)^{2} - 4(1)(10)$
$D = 49 - 40$
$D = 9$
Thus,the discriminant is $9$.

(A) For a quadratic equation of the form $ax^{2} + bx + c = 0$,the discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Given the equation $3x^{2} - 12x + 16 = 0$,we identify the coefficients as $a = 3$,$b = -12$,and $c = 16$.
Substituting these values into the formula:
$D = (-12)^{2} - 4(3)(16)$
$D = 144 - 192$
$D = -48$
Thus,the discriminant of the given quadratic equation is $-48$.

(B) The given quadratic equation is in the form $ax^{2} + bx + c = 0$.
Comparing $5x^{2} - 4\sqrt{2}x - 1 = 0$ with the standard form,we get:
$a = 5$,$b = -4\sqrt{2}$,and $c = -1$.
The discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Substituting the values:
$D = (-4\sqrt{2})^{2} - 4(5)(-1)$
$D = (16 \times 2) + 20$
$D = 32 + 20$
$D = 52$.
Therefore,the discriminant is $52$.

(C) The given quadratic equation is $5x^{2} + 2x - 1 = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = 5$,$b = 2$,and $c = -1$.
The discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Substituting the values,we get $D = (2)^{2} - 4(5)(-1)$.
$D = 4 + 20 = 24$.
Therefore,the discriminant is $24$.

(D) For a quadratic equation of the form $ax^{2} + bx + c = 0$,the discriminant $D$ is given by the formula $D = b^{2} - 4ac$.
Given the equation: $\sqrt{5} x^{2} - 3 \sqrt{3} x - 2 \sqrt{5} = 0$.
Here,$a = \sqrt{5}$,$b = -3 \sqrt{3}$,and $c = -2 \sqrt{5}$.
Substituting these values into the formula:
$D = (-3 \sqrt{3})^{2} - 4(\sqrt{5})(-2 \sqrt{5})$
$D = (9 \times 3) - 4(-2 \times 5)$
$D = 27 - 4(-10)$
$D = 27 + 40$
$D = 67$.
Thus,the discriminant is $67$.

(A) For the quadratic equation $ax^{2} + bx + c = 0$,the quadratic formula is $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.
Here,$a = 5$,$b = -3$,and $c = -2$.
First,calculate the discriminant $D = b^{2} - 4ac = (-3)^{2} - 4(5)(-2) = 9 + 40 = 49$.
Since $D > 0$,real solutions exist.
Applying the formula: $x = \frac{-(-3) \pm \sqrt{49}}{2(5)} = \frac{3 \pm 7}{10}$.
Case $1$: $x = \frac{3 + 7}{10} = \frac{10}{10} = 1$.
Case $2$: $x = \frac{3 - 7}{10} = \frac{-4}{10} = -\frac{2}{5}$.
Thus,the solutions are $-\frac{2}{5}$ and $1$.

(A) The given quadratic equation is $9x^2 + 6x + 4 = 0$.
Comparing this with the standard form $ax^2 + bx + c = 0$,we get $a = 9$,$b = 6$,and $c = 4$.
The discriminant $D$ is given by $D = b^2 - 4ac$.
Substituting the values: $D = (6)^2 - 4(9)(4) = 36 - 144 = -108$.
Since the discriminant $D < 0$,the square root of $D$ is not a real number.
Therefore,the equation has no real solution in $R$.

(D) For the quadratic equation $am^{2} + bm + c = 0$,the quadratic formula is $m = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.
Here,$a = 48$,$b = -13$,and $c = -1$.
First,calculate the discriminant $D = b^{2} - 4ac = (-13)^{2} - 4(48)(-1) = 169 + 192 = 361$.
Since $D > 0$,real solutions exist.
$m = \frac{-(-13) \pm \sqrt{361}}{2(48)} = \frac{13 \pm 19}{96}$.
Case $1$: $m = \frac{13 + 19}{96} = \frac{32}{96} = \frac{1}{3}$.
Case $2$: $m = \frac{13 - 19}{96} = \frac{-6}{96} = -\frac{1}{16}$.
Thus,the solutions are $\frac{1}{3}$ and $-\frac{1}{16}$.

(A) For the quadratic equation $ax^{2} + bx + c = 0$,the discriminant $D$ is given by $D = b^{2} - 4ac$.
Here,$a = 5$,$b = 12$,and $c = 10$.
$D = (12)^{2} - 4(5)(10)$
$D = 144 - 200$
$D = -56$.
Since the discriminant $D < 0$,the equation has no real roots in the set of real numbers $R$.

(D) For a quadratic equation of the form $ax^2 + bx + c = 0$,the roots are given by the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Here,$a = 25$,$b = 20$,and $c = 7$.
First,calculate the discriminant $D = b^2 - 4ac$:
$D = (20)^2 - 4(25)(7)$
$D = 400 - 700$
$D = -300$.
Since the discriminant $D < 0$,the quadratic equation has no real roots.

(D) For the quadratic equation $ax^{2} + bx + c = 0$, the roots are given by the quadratic formula: $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.
Here, $a = 3$, $b = -2$, and $c = 2$.
First, calculate the discriminant $D = b^{2} - 4ac$.
$D = (-2)^{2} - 4(3)(2) = 4 - 24 = -20$.
Since the discriminant $D < 0$, the equation has no real roots.
The complex roots are $x = \frac{-(-2) \pm \sqrt{-20}}{2(3)} = \frac{2 \pm 2i\sqrt{5}}{6} = \frac{1 \pm i\sqrt{5}}{3}$.

(B) The given quadratic equation is $2x^{2} - 2\sqrt{2}x + 1 = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$, we get $a = 2$, $b = -2\sqrt{2}$, and $c = 1$.
The quadratic formula is $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.
First, calculate the discriminant $D = b^{2} - 4ac = (-2\sqrt{2})^{2} - 4(2)(1) = 8 - 8 = 0$.
Since $D = 0$, the equation has two equal real roots.
The roots are $x = \frac{-(-2\sqrt{2}) \pm \sqrt{0}}{2(2)} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Thus, the roots are $\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$.

(A) The given quadratic equation is $\sqrt{3}x^{2} - 2x + \sqrt{3} = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = \sqrt{3}$,$b = -2$,and $c = \sqrt{3}$.
First,we calculate the discriminant $D = b^{2} - 4ac$.
$D = (-2)^{2} - 4(\sqrt{3})(\sqrt{3})$
$D = 4 - 4(3) = 4 - 12 = -8$.
Since the discriminant $D < 0$,the equation has no real roots in the set of real numbers $R$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(B) 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,$m = 6$.