QUADRATIC EQUATION Questions in English

(B) $Quantity \, 1$: Let the speed of the current be $x \, m/min$.
The time taken against the current is $T_1 = \frac{200}{48-x}$ and the time taken with the current is $T_2 = \frac{200}{48+x}$.
Given $T_1 - T_2 = 10$,so $\frac{200}{48-x} - \frac{200}{48+x} = 10$.
Dividing by $10$: $\frac{20}{48-x} - \frac{20}{48+x} = 1$.
$20(48+x) - 20(48-x) = (48-x)(48+x) \Rightarrow 960 + 20x - 960 + 20x = 2304 - x^2$.
$x^2 + 40x - 2304 = 0$.
Solving the quadratic equation: $(x+72)(x-32) = 0$. Since speed cannot be negative,$x = 32 \, m/min$.
$Quantity \, 2$: The distance covered in $3$ rounds of a circular path is $3 \times (2 \pi r) = 3 \times 2 \times \frac{22}{7} \times 49 = 3 \times 2 \times 22 \times 7 = 924 \, m$.
Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{924}{14} = 66 \, m/min$.
Comparing the two,$32 < 66$,therefore Quantity $I < $ Quantity $II$.

(C) Let $m$,$w$,and $b$ be the efficiency of a man,a woman,and a boy respectively.
Given: $(10m + 15w) \times 8 = (12m + 8w) \times 10$
$80m + 120w = 120m + 80w$
$40w = 40m \implies m = w$.
Since a boy is $50\%$ less efficient than a man,$b = 0.5m$,which means $m = 2b$.
Total work = $(10m + 15m) \times 8 = 25m \times 8 = 200m$ units.
Quantity $I$: $2m + 4w + 18b = 2m + 4m + 9m = 15m$. Time = $200m / 15m = 40/3$ days.
Quantity $II$: $9m + 3w + 6b = 9m + 3m + 3m = 15m$. Time = $200m / 15m = 40/3$ days.
Therefore,Quantity $I =$ Quantity $II$.

(C) Let Babu's usual speed be $v \, km/h$ and the time he takes to reach the office be $T$. The distance from his house to the office is $D = v \times T$.
Babu meets his girlfriend at a point $P$ on the way. Let the time taken by the girlfriend to walk from $3:00 \, PM$ to the meeting point be $t_g$ and the distance covered be $d_g = 40 \times t_g$.
Babu travels from his house to point $P$ in time $t_b$. Since they meet at point $P$,the distance from the house to $P$ is $v \times t_b$.
Normally,Babu reaches the office at $5:00 \, PM$. If he leaves at $t_0$,then $T = 5 - t_0$.
In the scenario where they meet,they reach home $40 \, min$ $(2/3 \, hr)$ earlier. This means the total time saved is $40 \, min$.
Babu saves the time he would have spent traveling from point $P$ to the office and back to point $P$. The girlfriend walked for $2$ hours ($3:00 \, PM$ to $5:00 \, PM$ equivalent). By solving the relative motion equations,we find $v = 120 \, km/h$.
Thus,Quantity $I = 120 \, km/h$ and Quantity $II = 120 \, km/h$.
Therefore,Quantity $I = $ Quantity $II$ is not explicitly listed,but based on the provided options,the relationship is $I \leq II$ or $I \geq II$. Given the standard interpretation of such problems,$v = 120 \, km/h$ makes the quantities equal.

(B) Let the marked price $(MP)$ be $700x$.
Given that the cost price $(CP)$ is $79 \frac{2}{7} \%$ of $MP$,we have $CP = \frac{555}{700} \times 700x = 555x$.
The discount is $Rs. \,68$,so the selling price $(SP)$ is $700x - 68$.
Since there is a profit of $20 \%$,$SP = CP \times (1 + \frac{20}{100}) = 555x \times 1.2 = 666x$.
Equating the two expressions for $SP$: $666x = 700x - 68$.
$34x = 68$,which gives $x = 2$.
Therefore,Quantity $I$ $(CP)$ $= 555 \times 2 = 1110$.
Quantity $II = 1200$.
Since $1110 < 1200$,Quantity $I < $ Quantity $II$.

(B) Let the time taken by Manoj and Shubham to complete the work together be $x \, \text{hrs}$.
According to the problem:
Time taken by Manoj alone = $(x + 4.8) \, \text{hrs}$.
Time taken by Shubham alone = $(x + 10.8) \, \text{hrs}$.
Using the work-time formula:
$\frac{1}{x} = \frac{1}{x + 4.8} + \frac{1}{x + 10.8}$
For such problems,the shortcut formula is $x = \sqrt{a \times b}$,where $a$ and $b$ are the extra times taken by individuals compared to the combined time.
$x = \sqrt{4.8 \times 10.8}$
$x = \sqrt{\frac{48}{10} \times \frac{108}{10}}$
$x = \sqrt{\frac{5184}{100}}$
$x = \frac{72}{10} = 7.2 \, \text{hrs}$.
Quantity $1 = 7.2 \, \text{hrs}$.
Quantity $2 = 7.4 \, \text{hrs}$.
Since $7.2 < 7.4$,Quantity $1 < $ Quantity $2$.

(B) $Quantity \, 1:$
Let the actual cost price per unit be $1$. When buying,he gets $110$ units for the price of $100$ units. So,effective cost price per unit $= \frac{100}{110}$.
When selling,he gives $90$ units for every $100$ units promised. He also offers a $10 \%$ discount on the selling price. Let the marked price be $1$ per unit. Selling price $= 0.9$ per unit.
Effective profit calculation: He buys $110$ units for $100$ and sells $90$ units for $0.9 \times 90 = 81$.
Profit $\% = \frac{81 - 100/110}{100/110} \times 100 = \frac{81 \times 110 - 100}{100} = \frac{8910 - 100}{100} = 88.1 \%$.
Correction: Using standard ratio method,Profit $\% = (\frac{110}{100} \times \frac{100}{90} \times \frac{90}{100} - 1) \times 100 = (1.1 - 1) \times 100 = 10 \%$.
$Quantity \, 2:$
Cost Price $= CP$. Marked Price $= 1.25 CP$. Selling Price $= 1.25 CP \times 0.9 = 1.125 CP$.
Profit on Selling Price $= \frac{SP - CP}{SP} \times 100 = \frac{1.125 CP - CP}{1.125 CP} \times 100 = \frac{0.125}{1.125} \times 100 = \frac{1}{9} \times 100 = 11.11 \%$.
Since $10 \% < 11.11 \%$,Quantity $I < $ Quantity $II$.

(A) $Quantity \, 1:$
Let the original duration be $t$ hours and original speed be $s$ km/hr.
Given,$s \times t = 3000 \implies s = \frac{3000}{t}$.
According to the problem,$(s - 100)(t + 1) = 3000$.
Substituting $s = \frac{3000}{t}$ into the equation:
$(\frac{3000}{t} - 100)(t + 1) = 3000$
$3000 + \frac{3000}{t} - 100t - 100 = 3000$
$\frac{3000}{t} - 100t - 100 = 0$
Dividing by $100$: $\frac{30}{t} - t - 1 = 0$
$30 - t^2 - t = 0 \implies t^2 + t - 30 = 0$
$(t + 6)(t - 5) = 0$. Since time cannot be negative,$t = 5 \, hours$.
$Quantity \, 2:$
Let the usual speed be $v$ and usual time be $T$.
New speed $= \frac{3}{4}v$. Since distance is constant,time is inversely proportional to speed.
New time $= \frac{4}{3}T$.
Given,$\frac{4}{3}T - T = 20 \, minutes$.
$\frac{1}{3}T = 20 \, minutes \implies T = 60 \, minutes = 1 \, hour$.
Comparing $Quantity \, 1$ $(5 \, hours)$ and $Quantity \, 2$ $(1 \, hour)$,we get $Quantity \, I > Quantity \, II$.

(D) The circumference of the smaller wheel is $C_1 = \pi d_1 = 7\pi \,cm$.
The circumference of the larger wheel is $C_2 = \pi d_2 = 14\pi \,cm$.
Let $n$ be the number of revolutions per second made by both wheels.
The speed of the smaller wheel is $v_1 = 7\pi n \,cm/s$ and the speed of the larger wheel is $v_2 = 14\pi n \,cm/s$.
Since they move towards each other,their relative speed is $v_1 + v_2 = (7\pi n + 14\pi n) = 21\pi n \,cm/s$.
They meet after $10 \,s$ over a distance of $1990.50 \,cm$.
Therefore,$(21\pi n) \times 10 = 1990.50$.
$210\pi n = 1990.50$.
$n = \frac{1990.50}{210\pi} = \frac{1990.5}{210 \times (22/7)} = \frac{1990.5 \times 7}{210 \times 22} = \frac{13933.5}{4620} = 3.0159 \dots$ (approx $3$ revolutions per second).
Speed of smaller wheel $= 7\pi \times 3 = 21\pi \,cm/s$.
Thus,Quantity $1 = 21\pi \,cm/s$,which is equal to Quantity $2$.

(B) For the quadratic equation $ax^{2}+bx+c=0$,the roots are equal if the discriminant $D = b^{2}-4ac = 0$.
Given equation: $2 k x^{2}+5 k x+2=0$.
Here,$a=2k$,$b=5k$,and $c=2$.
Substituting these values into the discriminant formula:
$(5k)^{2} - 4(2k)(2) = 0$
$25k^{2} - 16k = 0$
Factor out $k$:
$k(25k - 16) = 0$
This gives two possible values for $k$: $k=0$ or $25k-16=0$.
If $k=0$,the equation becomes $2=0$,which is not a quadratic equation. Therefore,$k \neq 0$.
Thus,$25k = 16$,which implies $k = \frac{16}{25}$.

(B) Let the breadth of the rectangle be $b \text{ cm}$.
Then,the length of the rectangle is $l = b + 3 \text{ cm}$.
The area of the rectangle is $A = l \times b = (b + 3)b = b^2 + 3b$.
The area of the circle with radius $r = \frac{\sqrt{35}}{\sqrt{11}} \text{ cm}$ is $A = \pi r^2 = \frac{22}{7} \times \left(\frac{\sqrt{35}}{\sqrt{11}}\right)^2$.
$A = \frac{22}{7} \times \frac{35}{11} = 2 \times 5 = 10 \text{ cm}^2$.
Equating the areas: $b^2 + 3b = 10$.
$b^2 + 3b - 10 = 0$.
$(b + 5)(b - 2) = 0$.
Since breadth cannot be negative,$b = 2 \text{ cm}$.
Therefore,the length is $l = 2 + 3 = 5 \text{ cm}$.
The dimensions of the rectangle are $5 \text{ cm} \times 2 \text{ cm}$.

(C) Let $r$ be the radius and $l$ be the length of the pipe.
Given that the surface area of the pipe (curved surface area) is $2 \pi r l = 628 \, m^2$.
Taking $\pi \approx 3.14$,we have $2 \times 3.14 \times r \times l = 628$,which simplifies to $r \times l = 100$.
Given that the difference between the length and radius is $15 \, m$,we have $l - r = 15$,so $l = r + 15$.
Substituting $l$ in the equation $r \times l = 100$,we get $r(r + 15) = 100$,which is $r^2 + 15r - 100 = 0$.
Factoring the quadratic equation: $(r + 20)(r - 5) = 0$.
Since the radius cannot be negative,$r = 5 \, m$.
Then,$l = 5 + 15 = 20 \, m$.
If the pipe is closed at one end,it acts as a cylinder with volume $V = \pi r^2 l$.
$V = 3.14 \times (5)^2 \times 20 = 3.14 \times 25 \times 20 = 3.14 \times 500 = 1570 \, m^3$.

(C) For a quadratic equation $ax^{2} + bx + c = 0$,the roots are equal if the discriminant $D = b^{2} - 4ac = 0$.
Given the equation $2 k x^{2} + 5 k x + 2 = 0$,we have $a = 2k$,$b = 5k$,and $c = 2$.
Substituting these values into the discriminant formula:
$(5k)^{2} - 4(2k)(2) = 0$
$25k^{2} - 16k = 0$
Factor out $k$:
$k(25k - 16) = 0$
This gives two possibilities: $k = 0$ or $25k - 16 = 0$.
If $k = 0$,the equation becomes $2 = 0$,which is impossible. Thus,$k \neq 0$.
Therefore,$25k - 16 = 0$,which implies $k = \frac{16}{25}$.

(C) Given the quadratic equation $ax^{2} + bx + c = 0$ with roots $\alpha$ and $2\alpha$.
From the properties of roots:
Sum of roots $= \alpha + 2\alpha = 3\alpha = -\frac{b}{a} \implies \alpha = -\frac{b}{3a} \quad (1)$
Product of roots $= \alpha \cdot 2\alpha = 2\alpha^{2} = \frac{c}{a} \implies \alpha^{2} = \frac{c}{2a} \quad (2)$
Substituting $(1)$ into $(2)$:
$\left(-\frac{b}{3a}\right)^{2} = \frac{c}{2a}$
$\frac{b^{2}}{9a^{2}} = \frac{c}{2a}$
$2b^{2} = 9a^{2} \cdot \frac{c}{a}$
$2b^{2} = 9ac$

(D) Given equation: $\frac{1}{x+a} + \frac{1}{x+b} = \frac{1}{c}$
Taking the common denominator on the left side:
$\frac{(x+b) + (x+a)}{(x+a)(x+b)} = \frac{1}{c}$
$\frac{2x + a + b}{x^2 + x(a+b) + ab} = \frac{1}{c}$
Cross-multiplying:
$c(2x + a + b) = x^2 + x(a+b) + ab$
$2cx + c(a+b) = x^2 + x(a+b) + ab$
Rearranging into standard quadratic form $Ax^2 + Bx + C = 0$:
$x^2 + x(a+b-2c) + (ab - c(a+b)) = 0$
For a quadratic equation $Ax^2 + Bx + C = 0$,the sum of roots is $-B/A$. Given the sum of roots is $0$:
$-(a+b-2c) = 0 \Rightarrow a+b = 2c$
The product of the roots is $C/A = ab - c(a+b)$.
Substituting $c = \frac{a+b}{2}$:
Product $= ab - \left(\frac{a+b}{2}\right)(a+b) = ab - \frac{(a+b)^2}{2}$
$= \frac{2ab - (a^2 + 2ab + b^2)}{2} = \frac{2ab - a^2 - 2ab - b^2}{2} = \frac{-(a^2 + b^2)}{2} = -\frac{1}{2}(a^2 + b^2)$

(D) Given equation: $\sqrt{3y+1} = \sqrt{y-1}$.
Squaring both sides,we get:
$3y + 1 = y - 1$
Rearranging the terms to solve for $y$:
$3y - y = -1 - 1$
$2y = -2$
$y = -1$
Now,check the validity of the root by substituting $y = -1$ back into the original equation:
$LHS$: $\sqrt{3(-1) + 1} = \sqrt{-3 + 1} = \sqrt{-2}$
Since $\sqrt{-2}$ is not a real number,$y = -1$ is an extraneous root.
Therefore,there is no real root for the given equation.
Thus,the correct option is $D$.

(B) Given the quadratic equation $x^{2}+3ax+c=0$.
Let $\alpha$ and $\beta$ be the roots of the equation.
From the properties of roots,we have the sum of roots $\alpha+\beta = -3a$ and the product of roots $\alpha\beta = c$.
We are given $\alpha^{2}+\beta^{2}=5$.
We know the identity $\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2} - 2\alpha\beta$.
Substituting the known values into the identity: $5 = (-3a)^{2} - 2(c)$.
$5 = 9a^{2} - 2c$.
$9a^{2} = 5 + 2c$.
$a^{2} = \frac{5+2c}{9}$.
Therefore,$a = \pm \sqrt{\frac{5+2c}{9}}$.
Comparing with the given options,the correct value is $\sqrt{\frac{5+2c}{9}}$.

(D) For the quadratic equation $x^{2}+3ax+2a^{2}=0$,the sum of the roots $\alpha+\beta = -\frac{b}{a_{coeff}} = -3a$ and the product of the roots $\alpha\beta = \frac{c}{a_{coeff}} = 2a^{2}$.
We know that $\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2} - 2\alpha\beta$.
Given $\alpha^{2}+\beta^{2} = 5$,we substitute the values:
$5 = (-3a)^{2} - 2(2a^{2})$
$5 = 9a^{2} - 4a^{2}$
$5 = 5a^{2}$
$a^{2} = 1$
$a = \pm 1$.

(B) Given that $x+\frac{1}{x}=2.$
To find the value of $x^{2}+\frac{1}{x^{2}},$ we square both sides of the given equation:
$\left(x+\frac{1}{x}\right)^{2} = 2^{2}$
Using the algebraic identity $(a+b)^{2} = a^{2}+b^{2}+2ab,$
$x^{2} + \frac{1}{x^{2}} + 2(x)\left(\frac{1}{x}\right) = 4$
$x^{2} + \frac{1}{x^{2}} + 2 = 4$
$x^{2} + \frac{1}{x^{2}} = 4 - 2$
$x^{2} + \frac{1}{x^{2}} = 2$

(D) Given $x+\frac{1}{x}=3.$
Squaring both sides,we get $\left(x+\frac{1}{x}\right)^{2}=3^{2} \Rightarrow x^{2}+\frac{1}{x^{2}}+2=9 \Rightarrow x^{2}+\frac{1}{x^{2}}=7.$
Now,squaring again,$\left(x^{2}+\frac{1}{x^{2}}\right)^{2}=7^{2} \Rightarrow x^{4}+\frac{1}{x^{4}}+2=49 \Rightarrow x^{4}+\frac{1}{x^{4}}=47.$
Finally,squaring once more,$\left(x^{4}+\frac{1}{x^{4}}\right)^{2}=47^{2} \Rightarrow x^{8}+\frac{1}{x^{8}}+2=2209 \Rightarrow x^{8}+\frac{1}{x^{8}}=2207.$

(D) Given: $x+y=3$ and $xy=2$.
We know that $(x-y)^2 = (x+y)^2 - 4xy$.
Substituting the values: $(x-y)^2 = 3^2 - 4(2) = 9 - 8 = 1$.
Thus,$x-y = 1$ (assuming $x > y$).
Now,we use the identity $x^3-y^3 = (x-y)(x^2+y^2+xy)$.
We can rewrite this as $x^3-y^3 = (x-y)((x+y)^2 - xy)$.
Substituting the known values: $x^3-y^3 = 1 \times (3^2 - 2) = 1 \times (9 - 2) = 7$.

(C) Given: $x-\frac{1}{x}=\sqrt{21}$.
We know that $(x+\frac{1}{x})^2 = (x-\frac{1}{x})^2 + 4$.
Substituting the value: $(x+\frac{1}{x})^2 = (\sqrt{21})^2 + 4 = 21 + 4 = 25$.
Therefore,$x+\frac{1}{x} = \sqrt{25} = 5$.
Now,$x^2+\frac{1}{x^2} = (x-\frac{1}{x})^2 + 2 = (\sqrt{21})^2 + 2 = 21 + 2 = 23$.
Finally,$(x^2+\frac{1}{x^2})(x+\frac{1}{x}) = 23 \times 5 = 115$.

(B) Given that $a+b+c=0$.
We know the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.
Since $a+b+c=0$,the entire right side of the equation becomes zero:
$a^{3}+b^{3}+c^{3}-3abc = 0 \times (a^{2}+b^{2}+c^{2}-ab-bc-ca) = 0$.
Therefore,$a^{3}+b^{3}+c^{3} = 3abc$.

(C) Given: $x = \frac{\sqrt{3}+1}{\sqrt{3}-1}$ and $y = \frac{\sqrt{3}-1}{\sqrt{3}+1}$.
Rationalizing $x$:
$x = \frac{(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{3+1+2\sqrt{3}}{3-1} = \frac{4+2\sqrt{3}}{2} = 2+\sqrt{3}$.
Rationalizing $y$:
$y = \frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{3+1-2\sqrt{3}}{3-1} = \frac{4-2\sqrt{3}}{2} = 2-\sqrt{3}$.
Now,calculate $x^{2}+y^{2}$:
$x^{2}+y^{2} = (2+\sqrt{3})^{2} + (2-\sqrt{3})^{2}$.
Using the identity $(a+b)^{2} + (a-b)^{2} = 2(a^{2}+b^{2})$:
$x^{2}+y^{2} = 2(2^{2} + (\sqrt{3})^{2}) = 2(4+3) = 2(7) = 14$.

(D) Given expression: $(a^{4}+b^{4})(a^{2}+b^{2})(a+b)(a-b)$
Step $1$: Use the identity $(a+b)(a-b) = a^{2}-b^{2}$.
So,$(a+b)(a-b) = a^{2}-b^{2}$.
Step $2$: Substitute this into the expression:
$(a^{4}+b^{4})(a^{2}+b^{2})(a^{2}-b^{2})$
Step $3$: Use the identity $(x+y)(x-y) = x^{2}-y^{2}$ where $x=a^{2}$ and $y=b^{2}$:
$(a^{2}+b^{2})(a^{2}-b^{2}) = (a^{2})^{2}-(b^{2})^{2} = a^{4}-b^{4}$.
Step $4$: Now multiply the remaining terms:
$(a^{4}+b^{4})(a^{4}-b^{4})$
Step $5$: Apply the identity $(x+y)(x-y) = x^{2}-y^{2}$ again where $x=a^{4}$ and $y=b^{4}$:
$(a^{4})^{2}-(b^{4})^{2} = a^{8}-b^{8}$.

(C) Given that $x-\frac{1}{x}=a$.
We know the algebraic identity $(x-y)^{3}=x^{3}-y^{3}-3xy(x-y)$.
Substituting $y = \frac{1}{x}$ into the identity,we get:
$\left(x-\frac{1}{x}\right)^{3}=x^{3}-\left(\frac{1}{x}\right)^{3}-3(x)\left(\frac{1}{x}\right)\left(x-\frac{1}{x}\right)$.
Simplifying the expression:
$\left(x-\frac{1}{x}\right)^{3}=x^{3}-\frac{1}{x^{3}}-3\left(x-\frac{1}{x}\right)$.
Now,substitute the given value $x-\frac{1}{x}=a$ into the equation:
$a^{3}=x^{3}-\frac{1}{x^{3}}-3(a)$.
Rearranging the terms to solve for $x^{3}-\frac{1}{x^{3}}$:
$x^{3}-\frac{1}{x^{3}}=a^{3}+3a$.

(C) Given $x = 2^{1/3} - 2^{-1/3}$.
We need to find the value of $2x^3 + 6x$.
First,cube both sides of the equation $x = 2^{1/3} - 2^{-1/3}$:
$x^3 = (2^{1/3} - 2^{-1/3})^3$
Using the identity $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$:
$x^3 = (2^{1/3})^3 - (2^{-1/3})^3 - 3(2^{1/3})(2^{-1/3})(2^{1/3} - 2^{-1/3})$
$x^3 = 2 - 2^{-1} - 3(1)(x)$
$x^3 = 2 - 1/2 - 3x$
$x^3 = 3/2 - 3x$
Multiply by $2$:
$2x^3 = 3 - 6x$
Rearranging the terms:
$2x^3 + 6x = 3$.

(D) Let the given expression be $E = \frac{(a-b)^{2}}{(b-c)(c-a)} + \frac{(b-c)^{2}}{(a-b)(c-a)} + \frac{(a-c)^{2}}{(a-b)(b-c)}$.
To add these fractions,we take the common denominator as $(a-b)(b-c)(c-a)$.
Note that $(c-a) = -(a-c)$,so $(a-c)^2 = (c-a)^2$.
Rewriting the expression with the common denominator:
$E = \frac{(a-b)^3 + (b-c)^3 + (a-c)^2(a-b)(b-c)/(a-b)(b-c)}{(a-b)(b-c)(c-a)}$
Actually,a simpler way is to observe the denominators:
$E = \frac{-(a-b)^3 - (b-c)^3 - (c-a)^3}{(a-b)(b-c)(c-a)}$
Using the algebraic identity: If $x+y+z = 0$,then $x^3+y^3+z^3 = 3xyz$.
Here,let $x = (a-b)$,$y = (b-c)$,and $z = (c-a)$.
Then $x+y+z = (a-b) + (b-c) + (c-a) = 0$.
Therefore,$(a-b)^3 + (b-c)^3 + (c-a)^3 = 3(a-b)(b-c)(c-a)$.
Substituting this into the expression:
$E = \frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)} = 3$.

(A) Given $a+b+c=0.$
We need to find the value of $\frac{1}{b^{2}+c^{2}-a^{2}}+\frac{1}{c^{2}+a^{2}-b^{2}}+\frac{1}{a^{2}+b^{2}-c^{2}}.$
Since $a+b+c=0,$ we have $a+b=-c.$
Squaring both sides,$(a+b)^{2}=(-c)^{2},$ which gives $a^{2}+b^{2}+2ab=c^{2}.$
Rearranging,$b^{2}+c^{2}-a^{2} = b^{2}+(a^{2}+2ab+b^{2})-a^{2} = 2b^{2}+2ab = 2b(a+b).$
Since $a+b=-c,$ we get $b^{2}+c^{2}-a^{2} = 2b(-c) = -2bc.$
Similarly,$c^{2}+a^{2}-b^{2} = -2ac$ and $a^{2}+b^{2}-c^{2} = -2ab.$
Substituting these into the expression:
$\frac{1}{-2bc} + \frac{1}{-2ac} + \frac{1}{-2ab} = \frac{-a-b-c}{2abc}.$
Since $a+b+c=0,$ the numerator is $0.$
Thus,the value of the expression is $0.$

(C) Given that $x+y+z=0.$
From this,we can derive the following relations:
$x+y = -z$
$y+z = -x$
$z+x = -y$
Now,substitute these values into the expression $\frac{(x+y)(y+z)(z+x)}{x y z}$:
$= \frac{(-z)(-x)(-y)}{x y z}$
$= \frac{-(x y z)}{x y z}$
$= -1$

(C) Given $a+b+c=0.$
Squaring both sides,we get $(a+b+c)^2 = 0^2.$
$a^2+b^2+c^2+2(ab+bc+ca) = 0 \Rightarrow a^2+b^2+c^2 = -2(ab+bc+ca).$
Squaring again,$(a^2+b^2+c^2)^2 = [-2(ab+bc+ca)]^2.$
$a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2) = 4(a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)).$
Since $a+b+c=0,$ the term $2abc(a+b+c) = 0.$
So,$a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2) = 4(a^2b^2+b^2c^2+c^2a^2).$
$a^4+b^4+c^4 = 4(a^2b^2+b^2c^2+c^2a^2) - 2(a^2b^2+b^2c^2+c^2a^2) = 2(a^2b^2+b^2c^2+c^2a^2).$
Therefore,$\frac{a^4+b^4+c^4}{a^2b^2+b^2c^2+c^2a^2} = 2.$

(B) Given the equation $x+y=2z$.
We can rewrite this as $y-z = z-x$.
Now,substitute this into the expression $\frac{x}{x-z} + \frac{z}{y-z}$.
$\frac{x}{x-z} + \frac{z}{z-x} = \frac{x}{x-z} - \frac{z}{x-z}$.
$= \frac{x-z}{x-z} = 1$.

(B) Given equations are $x+\frac{1}{y}=1$ and $y+\frac{1}{z}=1$.
From the first equation,$x=1-\frac{1}{y} = \frac{y-1}{y}$.
Therefore,$\frac{1}{x} = \frac{y}{y-1}$.
From the second equation,$\frac{1}{z} = 1-y$.
Therefore,$z = \frac{1}{1-y}$.
Now,substitute these into the expression $z+\frac{1}{x}$:
$z+\frac{1}{x} = \frac{1}{1-y} + \frac{y}{y-1}$.
Since $y-1 = -(1-y)$,we can write:
$z+\frac{1}{x} = \frac{1}{1-y} - \frac{y}{1-y} = \frac{1-y}{1-y} = 1$.

(C) Given that $a=b=c.$
Substitute $b=a$ and $c=a$ into the expression $\frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}.$
The numerator becomes $(a+a+a)^{2} = (3a)^{2} = 9a^{2}.$
The denominator becomes $a^{2}+a^{2}+a^{2} = 3a^{2}.$
Therefore,the expression is $\frac{9a^{2}}{3a^{2}} = 3.$

(D) The given expression is $a^{2}+b^{2}+c^{2}-2ab+2ac-2bc$.
This can be factored as $(a-b+c)^{2}$.
Given $a=x+y$,$b=x-y$,and $c=2x-1$.
Substitute the values into the expression:
$(a-b+c)^{2} = [(x+y) - (x-y) + (2x-1)]^{2}$
$= [x+y-x+y+2x-1]^{2}$
$= [2x+2y-1]^{2}$
Note that $[2x+2y-1]^{2} = [-(1-2x-2y)]^{2} = (1-2x-2y)^{2}$.

(B) We are given the identity: $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$.
First,we find the value of $x^{2}+y^{2}+z^{2}$ using the identity $(x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2(xy+yz+zx)$.
Substituting the given values: $16^{2} = x^{2}+y^{2}+z^{2} + 2(78)$.
$256 = x^{2}+y^{2}+z^{2} + 156$.
$x^{2}+y^{2}+z^{2} = 256 - 156 = 100$.
Now,substitute these values into the original identity:
$x^{3}+y^{3}+z^{3}-3xyz = (16)(100 - 78)$.
$x^{3}+y^{3}+z^{3}-3xyz = 16(22) = 352$.

(B) We know the standard algebraic identity:
$a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$
Multiplying and dividing the right side by $2$,we get:
$a^{3}+b^{3}+c^{3}-3abc = \frac{1}{2}(a+b+c)(2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca)$
This can be rewritten as:
$a^{3}+b^{3}+c^{3}-3abc = \frac{1}{2}(a+b+c)\{(a^{2}-2ab+b^{2})+(b^{2}-2bc+c^{2})+(c^{2}-2ca+a^{2})\}$
$a^{3}+b^{3}+c^{3}-3abc = \frac{1}{2}(a+b+c)\{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\}$
Therefore,the given expression is equal to $a^{3}+b^{3}+c^{3}-3abc$.

(C) The algebraic identity for the expression is given by:
$x^{3}+y^{3}+z^{3}-3xyz = \frac{1}{2}(x+y+z) \{(x-y)^{2} + (y-z)^{2} + (z-x)^{2}\}$
Given values are $x=89, y=87, z=84$.
Step $1$: Calculate the sum $(x+y+z)$:
$x+y+z = 89+87+84 = 260$
Step $2$: Calculate the differences:
$(x-y) = 89-87 = 2$
$(y-z) = 87-84 = 3$
$(z-x) = 84-89 = -5$
Step $3$: Substitute these values into the identity:
$= \frac{1}{2} \times 260 \times \{2^{2} + 3^{2} + (-5)^{2}\}$
$= 130 \times \{4 + 9 + 25\}$
$= 130 \times 38$
$= 4940$

(A) Given: $x=a(b-c), y=b(c-a), z=c(a-b)$.
Dividing by $a, b, c$ respectively,we get:
$\frac{x}{a} = b-c, \frac{y}{b} = c-a, \frac{z}{c} = a-b$.
Let $p = \frac{x}{a}, q = \frac{y}{b}, r = \frac{z}{c}$.
Then $p+q+r = (b-c) + (c-a) + (a-b) = 0$.
We know the algebraic identity: If $p+q+r=0$,then $p^3+q^3+r^3 = 3pqr$.
Substituting the values back:
$\left(\frac{x}{a}\right)^3 + \left(\frac{y}{b}\right)^3 + \left(\frac{z}{c}\right)^3 = 3 \left(\frac{x}{a}\right) \left(\frac{y}{b}\right) \left(\frac{z}{c}\right) = \frac{3xyz}{abc}$.

(D) Given that $x+\frac{1}{x}=3$.
To find the value of $x^{2}+\frac{1}{x^{2}}$,we square the given equation on both sides:
$\left(x+\frac{1}{x}\right)^{2} = 3^{2}$
Using the algebraic identity $(a+b)^{2} = a^{2}+b^{2}+2ab$,we get:
$x^{2} + \frac{1}{x^{2}} + 2(x)\left(\frac{1}{x}\right) = 9$
$x^{2} + \frac{1}{x^{2}} + 2 = 9$
Subtracting $2$ from both sides:
$x^{2} + \frac{1}{x^{2}} = 9 - 2 = 7$
Therefore,the value is $7$.

(D) Given values are $a=17, b=15, c=13$.
We need to evaluate the expression $E = a^{2}+b^{2}+c^{2}-2ab-2bc-2ca$.
Substitute the values directly into the expression:
$E = (17)^{2} + (15)^{2} + (13)^{2} - 2(17)(15) - 2(15)(13) - 2(13)(17)$
$E = 289 + 225 + 169 - 510 - 390 - 442$
$E = 683 - 1342$
$E = -659$.

(B) We know the algebraic identity: $a^{3} + b^{3} + c^{3} - 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)$.
The denominator is given as $(ab + bc + ca - a^{2} - b^{2} - c^{2})$,which can be written as $-(a^{2} + b^{2} + c^{2} - ab - bc - ca)$.
Substituting these into the expression:
$\frac{(a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)}{-(a^{2} + b^{2} + c^{2} - ab - bc - ca)} = -(a + b + c)$.
Given $a = -5, b = -6, c = 10$:
$-(a + b + c) = -(-5 - 6 + 10) = -(-1) = 1$.

(C) We know the algebraic identity: $(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a)$.
Rearranging this identity,we get: $(a+b+c)^3 - 3(a+b)(b+c)(c+a) = a^3 + b^3 + c^3$.
Given $a=5, b=3, c=2$.
Substituting these values into the simplified expression:
$a^3 + b^3 + c^3 = 5^3 + 3^3 + 2^3$.
$= 125 + 27 + 8$.
$= 160$.

(C) The given quadratic equation is $Px^2 + 4x + 1 = 0$.
Comparing this with the standard form $Ax^2 + Bx + C = 0$,we get $A = P$,$B = 4$,and $C = 1$.
For a quadratic equation to have real roots,the discriminant $D$ must be greater than or equal to zero $(D \geq 0)$.
$D = B^2 - 4AC \geq 0$.
Substituting the values,we get $4^2 - 4(P)(1) \geq 0$.
$16 - 4P \geq 0$.
$16 \geq 4P$.
Dividing by $4$,we get $P \leq 4$.
Additionally,for the equation to be quadratic,the coefficient of $x^2$ must not be zero,so $P \neq 0$. Thus,the set of values is $P \leq 4$ and $P \neq 0$.

(A) The given quadratic equation is $2x^{2} + Px + 4 = 0$.
Since $2$ is a root of the equation,it must satisfy the equation.
Substituting $x = 2$ in the equation:
$2(2)^{2} + P(2) + 4 = 0$
$2(4) + 2P + 4 = 0$
$8 + 2P + 4 = 0$
$2P + 12 = 0$
$2P = -12$
$P = -6$.
Now,substitute $P = -6$ back into the original equation:
$2x^{2} - 6x + 4 = 0$.
Divide the entire equation by $2$:
$x^{2} - 3x + 2 = 0$.
Factorizing the quadratic equation:
$x^{2} - 2x - x + 2 = 0$
$x(x - 2) - 1(x - 2) = 0$
$(x - 1)(x - 2) = 0$.
So,the roots are $x = 1$ and $x = 2$.
Since one root is given as $2$,the second root is $1$.
Thus,the second root is $1$ and the value of $P$ is $-6$.

(A) The given quadratic equation is $x^{2}-5x+6=0$.
To find the roots,we factorize the quadratic expression:
$x^{2}-2x-3x+6=0$
$x(x-2)-3(x-2)=0$
$(x-2)(x-3)=0$
Setting each factor to zero,we get:
$x-2=0$ or $x-3=0$
$x=2$ or $x=3$.
Since one root is given as $3$,the other root is $2$.

(C) Given equation: $\sqrt{7} x^{2}-6 x-13 \sqrt{7}=0$
To solve by splitting the middle term,we need two numbers whose product is $(\sqrt{7}) \times (-13 \sqrt{7}) = -91$ and whose sum is $-6$.
These numbers are $-13$ and $7$.
$\sqrt{7} x^{2}-13 x+7 x-13 \sqrt{7}=0$
Taking common terms:
$x(\sqrt{7} x-13)+\sqrt{7}(\sqrt{7} x-13)=0$
$(x+\sqrt{7})(\sqrt{7} x-13)=0$
Setting each factor to zero:
$x+\sqrt{7}=0 \Rightarrow x=-\sqrt{7}$
$\sqrt{7} x-13=0 \Rightarrow x=\frac{13}{\sqrt{7}} = \frac{13 \sqrt{7}}{7}$
Thus,the roots are $-\sqrt{7}$ and $\frac{13 \sqrt{7}}{7}$.

(A) The given quadratic equation is $3 a^{2} x^{2}-a b x-2 b^{2}=0$.
We can factorize the middle term $-abx$ as $-3abx + 2abx$:
$3 a^{2} x^{2}-3 a b x+2 a b x-2 b^{2}=0$
Taking common terms from the groups:
$3 a x(a x-b)+2 b(a x-b)=0$
$(a x-b)(3 a x+2 b)=0$
Setting each factor to zero:
$a x-b=0 \Rightarrow x=b/a$
$3 a x+2 b=0 \Rightarrow x=-2b/3a$
Thus,the roots are $b/a$ and $-2b/3a$.

(B) The given quadratic equation is $a^{2} x^{2}-3 a b x+2 b^{2}=0$.
We can factorize the quadratic expression by splitting the middle term:
$a^{2} x^{2}-2 a b x-a b x+2 b^{2}=0$
Grouping the terms,we get:
$a x(a x-2 b)-b(a x-2 b)=0$
Taking $(a x-2 b)$ as a common factor:
$(a x-2 b)(a x-b)=0$
Setting each factor to zero:
$a x-2 b=0 \Rightarrow x=\frac{2 b}{a}$
$a x-b=0 \Rightarrow x=\frac{b}{a}$
Thus,the roots of the equation are $\frac{2 b}{a}$ and $\frac{b}{a}$.

(A) The general form of a quadratic equation is $x^{2} - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Given roots are $\alpha = \sqrt{2}$ and $\beta = 2\sqrt{2}$.
Sum of roots $= \alpha + \beta = \sqrt{2} + 2\sqrt{2} = 3\sqrt{2}$.
Product of roots $= \alpha \cdot \beta = (\sqrt{2}) \cdot (2\sqrt{2}) = 2 \cdot 2 = 4$.
Substituting these values into the general form,we get:
$x^{2} - (3\sqrt{2})x + 4 = 0$.

(B) The given quadratic equation is $ax^{2} + (4a^{2} - 3b)x - 12ab = 0$.
Expanding the middle term,we get $ax^{2} + 4a^{2}x - 3bx - 12ab = 0$.
Grouping the terms,we have $ax(x + 4a) - 3b(x + 4a) = 0$.
Factoring out $(x + 4a)$,we get $(ax - 3b)(x + 4a) = 0$.
Setting each factor to zero,we get $ax - 3b = 0$ or $x + 4a = 0$.
Solving for $x$,we find $x = \frac{3b}{a}$ or $x = -4a$.
Thus,the roots of the equation are $-4a$ and $\frac{3b}{a}$.