Quadratic Equations Questions in English

(A) Let the length and breadth of the park be $l$ and $b$ respectively.
The perimeter is given by $2(l + b) = 80\, m$,which simplifies to $l + b = 40$,or $b = 40 - l$.
The area is given by $l \times b = 400\, m^2$.
Substituting $b = 40 - l$ into the area equation: $l(40 - l) = 400$.
This gives $40l - l^2 = 400$,or $l^2 - 40l + 400 = 0$.
Comparing this with the standard quadratic equation $al^2 + bl + c = 0$,we have $a = 1, b = -40, c = 400$.
The discriminant $D = b^2 - 4ac = (-40)^2 - 4(1)(400) = 1600 - 1600 = 0$.
Since $D = 0$,the equation has real and equal roots,meaning the situation is possible.
The roots are given by $l = -b / (2a) = -(-40) / (2 \times 1) = 40 / 2 = 20$.
Thus,the length $l = 20\, m$ and the breadth $b = 40 - 20 = 20\, m$.

(A) The given quadratic equation is $x^{2}-3 \sqrt{5} x+10=0$.
Comparing this with the standard form $ax^{2}+bx+c=0$,we get:
$a=1, b=-3 \sqrt{5}, c=10$.
The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$.
Substituting the values:
$x = \frac{-(-3 \sqrt{5}) \pm \sqrt{(-3 \sqrt{5})^{2}-4(1)(10)}}{2(1)}$
$x = \frac{3 \sqrt{5} \pm \sqrt{45-40}}{2}$
$x = \frac{3 \sqrt{5} \pm \sqrt{5}}{2}$
Case $1$: $x = \frac{3 \sqrt{5} + \sqrt{5}}{2} = \frac{4 \sqrt{5}}{2} = 2 \sqrt{5}$.
Case $2$: $x = \frac{3 \sqrt{5} - \sqrt{5}}{2} = \frac{2 \sqrt{5}}{2} = \sqrt{5}$.
Thus,the roots of the equation are $2 \sqrt{5}$ and $\sqrt{5}$.

(B) Given equation is $\frac{1}{2} x^{2}-\sqrt{11} x+1=0$.
On comparing with $ax^{2}+bx+c=0$,we get:
$a=\frac{1}{2}, b=-\sqrt{11}, c=1$.
The quadratic formula is $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$.
Substituting the values:
$x = \frac{-(-\sqrt{11}) \pm \sqrt{(-\sqrt{11})^{2}-4 \times \frac{1}{2} \times 1}}{2 \times \frac{1}{2}}$
$x = \frac{\sqrt{11} \pm \sqrt{11-2}}{1}$
$x = \sqrt{11} \pm \sqrt{9}$
$x = \sqrt{11} \pm 3$.
Therefore,the roots are $3+\sqrt{11}$ and $\sqrt{11}-3$.

(B) Let the original average speed of the train be $x \, km/h$.
According to the problem,the time taken for the first part is $\frac{63}{x} \, hours$ and for the second part is $\frac{72}{x+6} \, hours$.
The total time taken is $3 \, hours$,so:
$\frac{63}{x} + \frac{72}{x+6} = 3$
Dividing the entire equation by $9$:
$\frac{7}{x} + \frac{8}{x+6} = \frac{1}{3}$
Multiplying by $3x(x+6)$ to clear the denominators:
$21(x+6) + 24x = x(x+6)$
$21x + 126 + 24x = x^2 + 6x$
$45x + 126 = x^2 + 6x$
$x^2 - 39x - 126 = 0$
Factoring the quadratic equation:
$(x - 42)(x + 3) = 0$
This gives $x = 42$ or $x = -3$.
Since speed cannot be negative,the original average speed is $42 \, km/h$.

(B) Let the original speed of the train be $x\, km/h$.
Then,the increased speed of the train is $(x+5)\, km/h$.
Distance $= 360\, km$.
Time taken at original speed $= \frac{360}{x}\, h$.
Time taken at increased speed $= \frac{360}{x+5}\, h$.
Given that the difference in time is $48\, min = \frac{48}{60}\, h = \frac{4}{5}\, h$.
According to the condition: $\frac{360}{x} - \frac{360}{x+5} = \frac{4}{5}$.
$\Rightarrow 360 \left( \frac{1}{x} - \frac{1}{x+5} \right) = \frac{4}{5}$.
$\Rightarrow 360 \left( \frac{x+5-x}{x(x+5)} \right) = \frac{4}{5}$.
$\Rightarrow \frac{360 \times 5}{x^2+5x} = \frac{4}{5}$.
$\Rightarrow 1800 \times 5 = 4(x^2+5x)$.
$\Rightarrow 9000 = 4x^2 + 20x$.
$\Rightarrow 4x^2 + 20x - 9000 = 0$.
Dividing by $4$: $x^2 + 5x - 2250 = 0$.
Solving by factorization: $x^2 + 50x - 45x - 2250 = 0$.
$x(x+50) - 45(x+50) = 0$.
$(x+50)(x-45) = 0$.
Since speed cannot be negative,$x = 45$.
Therefore,the original speed of the train is $45\, km/h$.

(D) Let Nisha's present age be $x$ years.
Then,Asha's present age $= x^2 + 2$ years.
The time taken for Nisha to reach her mother's present age is $(x^2 + 2) - x$ years.
After this time,Asha's age will be $(x^2 + 2) + (x^2 + 2 - x) = 2x^2 - x + 4$ years.
According to the problem,this age is $10x - 1$.
So,$2x^2 - x + 4 = 10x - 1$.
$2x^2 - 11x + 5 = 0$.
$2x^2 - 10x - x + 5 = 0$.
$2x(x - 5) - 1(x - 5) = 0$.
$(x - 5)(2x - 1) = 0$.
Since $x = 1/2$ is not practical for the given conditions,$x = 5$.
Nisha's present age $= 5$ years.
Asha's present age $= 5^2 + 2 = 27$ years.

(B) We know that the time between $2\, pm$ and $3\, pm$ is $60\, minutes$.
Given that at $t$ minutes past $2\, pm$, the time remaining to reach $3\, pm$ is $\left(\frac{t^{2}}{4} - 3\right)$ minutes.
The sum of the time passed and the time remaining must equal the total duration of one hour $(60\, minutes)$:
$t + \left(\frac{t^{2}}{4} - 3\right) = 60$
Multiply the entire equation by $4$ to clear the fraction:
$4t + t^{2} - 12 = 240$
Rearrange into the standard quadratic form $at^{2} + bt + c = 0$:
$t^{2} + 4t - 252 = 0$
Factor the quadratic equation:
$t^{2} + 18t - 14t - 252 = 0$
$t(t + 18) - 14(t + 18) = 0$
$(t + 18)(t - 14) = 0$
This gives $t = -18$ or $t = 14$.
Since time cannot be negative, we discard $t = -18$.
Therefore, $t = 14\, minutes$.

(NO) Let the polynomial be $p(x) = x^{4}-5x^{2}+3x-1$.
The degree of a polynomial is the highest power of the variable present in the expression.
In the given polynomial $p(x)$,the highest power of $x$ is $4$.
$A$ quadratic equation is defined as an equation of the form $ax^{2}+bx+c=0$,where $a \neq 0$ and the degree of the equation is $2$.
Since the degree of the given polynomial is $4$,it is a biquadratic equation,not a quadratic equation.
Therefore,$x^{4}-5x^{2}+3x-1=0$ is not a quadratic equation.

(B) Given equation: $9x = 3x^3$.
Rearranging the terms to one side,we get: $3x^3 - 9x = 0$.
$A$ quadratic equation is defined as an equation of the form $ax^2 + bx + c = 0$,where $a \neq 0$ and the highest power (degree) of the variable is $2$.
In the equation $3x^3 - 9x = 0$,the highest power of the variable $x$ is $3$.
Since the degree of the polynomial is $3$,it is a cubic equation,not a quadratic equation.
Therefore,$9x = 3x^3$ is not a quadratic equation.

(B) Given equation: $x^{2}+\frac{1}{x^{2}}=-2$ $(x \neq 0)$
Multiply the entire equation by $x^{2}$:
$x^{2}(x^{2}) + x^{2}(\frac{1}{x^{2}}) = -2(x^{2})$
$x^{4} + 1 = -2x^{2}$
Rearrange the terms to form a polynomial equation:
$x^{4} + 2x^{2} + 1 = 0$
Let $p(x) = x^{4} + 2x^{2} + 1$.
The degree of this polynomial is $4$.
$A$ quadratic equation must have a degree of $2$.
Since the degree is $4$,the given equation is not a quadratic equation.

(A) Given equation: $(x+6)(x+5)=0$
Expanding the left side:
$x(x+5) + 6(x+5) = 0$
$x^2 + 5x + 6x + 30 = 0$
$x^2 + 11x + 30 = 0$
$A$ quadratic equation is an equation of the form $ax^2 + bx + c = 0$,where $a \neq 0$.
In the equation $x^2 + 11x + 30 = 0$,the highest power of the variable $x$ is $2$.
Since the equation is in the form $ax^2 + bx + c = 0$ with $a=1$,it is a quadratic equation.

(B) Given equation: $(3x + 1)(3x + 2) = (9x - 1)(x + 1)$
Expanding both sides:
$9x^2 + 6x + 3x + 2 = 9x^2 + 9x - x - 1$
$9x^2 + 9x + 2 = 9x^2 + 8x - 1$
Subtracting $(9x^2 + 8x - 1)$ from both sides:
$9x^2 - 9x^2 + 9x - 8x + 2 + 1 = 0$
$x + 3 = 0$
The degree of the resulting polynomial $p(x) = x + 3$ is $1$.
Since the degree of the equation is not $2$,it is not a quadratic equation. It is a linear equation.

(B) Given equation: $x+\frac{1}{x}=x^{2}$
Multiply both sides by $x$ (since $x \neq 0$):
$x^{2}+1=x^{3}$
Rearranging the terms to one side:
$x^{3}-x^{2}-1=0$
Let $p(x) = x^{3}-x^{2}-1$.
The degree of the polynomial $p(x)$ is $3$.
$A$ quadratic equation must have a degree of $2$.
Since the degree of this equation is $3$,it is a cubic equation,not a quadratic equation.
Therefore,$x+\frac{1}{x}=x^{2}$ is not a quadratic equation.

(B) quadratic equation in the variable $x$ is an equation of the form $ax^{2} + bx + c = 0$,where $a, b, c$ are real numbers and $a \neq 0$. The exponent of the variable $x$ must be a non-negative integer,specifically $2$.
Given equation: $3x^{2} + 5\sqrt{x} + 3 = 0$.
We can rewrite the term $\sqrt{x}$ as $x^{\frac{1}{2}}$.
So,the equation becomes $3x^{2} + 5x^{\frac{1}{2}} + 3 = 0$.
In this equation,the exponent of $x$ in the second term is $\frac{1}{2}$,which is not a whole number (integer).
Since the definition of a quadratic equation requires the exponents of the variable to be non-negative integers,this equation does not satisfy the condition.
Therefore,$3x^{2} + 5\sqrt{x} + 3 = 0$ is not a quadratic equation.

(A) Given equation: $x - \frac{1}{x} = 5$
Multiply the entire equation by $x$ to eliminate the denominator:
$x(x) - x(\frac{1}{x}) = 5(x)$
$x^{2} - 1 = 5x$
Rearrange the terms to the standard form $ax^{2} + bx + c = 0$:
$x^{2} - 5x - 1 = 0$
Here,the highest power (degree) of the variable $x$ is $2$.
Since the equation is in the form $ax^{2} + bx + c = 0$ where $a \neq 0$,it is a quadratic equation.

(B) Given equation: $x^{2}-3=\frac{2}{x}$
Multiply both sides by $x$ (where $x \neq 0$):
$x(x^{2}-3) = 2$
$x^{3}-3x = 2$
Rearranging the terms to standard form:
$x^{3}-3x-2 = 0$
Here,the highest power (degree) of the variable $x$ is $3$.
$A$ quadratic equation must have a degree of $2$.
Since the degree is $3$,the given equation is a cubic equation,not a quadratic equation.

(A) The given equation is $p(x) = \sqrt{2} x^{2} - 5 \sqrt{3} x + 6 = 0$.
Here,the highest power (degree) of the variable $x$ in the polynomial $p(x) = \sqrt{2} x^{2} - 5 \sqrt{3} x + 6$ is $2$.
Since the degree of the polynomial is $2$,it is a quadratic polynomial.
Therefore,the given equation is a quadratic equation.

(B) quadratic equation in the variable $x$ is an equation of the form $ax^{2} + bx + c = 0$,where $a, b, c$ are real numbers and $a \neq 0$. The exponent of the variable $x$ must be a non-negative integer.
Given equation: $x^{2} + 2\sqrt{x} + 3 = 0$.
We can rewrite the term $\sqrt{x}$ as $x^{\frac{1}{2}}$.
So,the equation becomes $x^{2} + 2x^{\frac{1}{2}} + 3 = 0$.
In this equation,the exponent of $x$ in the second term is $\frac{1}{2}$,which is not a whole number (non-negative integer).
Since the variable $x$ does not have a power of $2$ as its highest degree with integer exponents,the given equation is not a polynomial,and therefore,it is not a quadratic equation.

(A) To verify if $x = -\sqrt{2}$ is a solution,substitute $x = -\sqrt{2}$ into the quadratic equation $\sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0$.
$LHS$ $= \sqrt{2}(-\sqrt{2})^{2} + 7(-\sqrt{2}) + 5 \sqrt{2}$
$= \sqrt{2}(2) - 7 \sqrt{2} + 5 \sqrt{2}$
$= 2 \sqrt{2} - 7 \sqrt{2} + 5 \sqrt{2}$
$= (2 - 7 + 5) \sqrt{2}$
$= 0 \times \sqrt{2} = 0$
Since $LHS$ = $RHS$,$x = -\sqrt{2}$ is a solution of the given quadratic equation.

(B) To verify if $x=3$ is a solution of the quadratic equation $6x^{2}-x-2=0$,we substitute $x=3$ into the expression $6x^{2}-x-2$.
Substituting $x=3$:
$6(3)^{2} - (3) - 2$
$= 6(9) - 3 - 2$
$= 54 - 3 - 2$
$= 49$
Since the result $49 \neq 0$,the value $x=3$ does not satisfy the equation.
Therefore,$x=3$ is not a solution of the quadratic equation $6x^{2}-x-2=0$.

(YES) To verify if $x = 5$ is a solution,we substitute $x = 5$ into the left-hand side $(LHS)$ of the equation.
$LHS$ = $\frac{x+1}{x-1} - \frac{x-1}{x+1}$
Substituting $x = 5$:
$LHS$ = $\frac{5+1}{5-1} - \frac{5-1}{5+1}$
$LHS$ = $\frac{6}{4} - \frac{4}{6}$
Simplifying the fractions:
$LHS$ = $\frac{3}{2} - \frac{2}{3}$
Finding a common denominator (which is $6$):
$LHS$ = $\frac{3 \times 3}{6} - \frac{2 \times 2}{6} = \frac{9}{6} - \frac{4}{6} = \frac{5}{6}$
Since $LHS$ = $RHS$ (which is $\frac{5}{6}$),the given value $x = 5$ is a solution of the quadratic equation.

(N/A) Given that $x=\sqrt{2}$ is one of the roots of the quadratic equation $ax^{2}+\sqrt{2}bx+2c=0$.
Since $x=\sqrt{2}$ is a root,it must satisfy the equation.
Substituting $x=\sqrt{2}$ into the equation:
$a(\sqrt{2})^{2} + \sqrt{2}b(\sqrt{2}) + 2c = 0$
Simplifying the terms:
$a(2) + 2b + 2c = 0$
$2a + 2b + 2c = 0$
Dividing the entire equation by $2$:
$a + b + c = 0$
Hence,it is proved that $a+b+c=0$.

(B) Since $\sqrt{2}$ is a root of the equation $x^{2}-(1+k)x+\sqrt{2}=0$,it must satisfy the equation.
Substituting $x = \sqrt{2}$ into the equation:
$(\sqrt{2})^{2} - (1+k)(\sqrt{2}) + \sqrt{2} = 0$
$2 - \sqrt{2} - \sqrt{2}k + \sqrt{2} = 0$
$2 - \sqrt{2}k = 0$
$2 = \sqrt{2}k$
$k = \frac{2}{\sqrt{2}}$
$k = \sqrt{2}$

(B) To determine if the equation is quadratic,we simplify both sides:
Left Hand Side $(LHS)$: $(2x + 1)(3x + 2) = 6x^2 + 4x + 3x + 2 = 6x^2 + 7x + 2$
Right Hand Side $(RHS)$: $6(x - 1)(x - 2) = 6(x^2 - 2x - x + 2) = 6(x^2 - 3x + 2) = 6x^2 - 18x + 12$
Equating $LHS$ and $RHS$: $6x^2 + 7x + 2 = 6x^2 - 18x + 12$
Subtracting $6x^2$ from both sides: $7x + 2 = -18x + 12$
Rearranging the terms: $7x + 18x + 2 - 12 = 0$
$25x - 10 = 0$
Since the highest power of the variable $x$ is $1$ and not $2$,the equation is a linear equation,not a quadratic equation.

(A) To determine if the equation $16x^2 - 3 = (2x + 5)(5x - 3)$ is quadratic,we first expand the right side:
$(2x + 5)(5x - 3) = 2x(5x) + 2x(-3) + 5(5x) + 5(-3)$
$= 10x^2 - 6x + 25x - 15$
$= 10x^2 + 19x - 15$
Now,substitute this back into the original equation:
$16x^2 - 3 = 10x^2 + 19x - 15$
Rearrange all terms to one side to set the equation to $0$:
$16x^2 - 10x^2 - 19x - 3 + 15 = 0$
$6x^2 - 19x + 12 = 0$
Since this equation is in the form $ax^2 + bx + c = 0$ where $a \neq 0$,it is a quadratic equation.

(A) Given equation: $(x-2)^{2}+1=2x-3$
Expand the left side using the identity $(a-b)^{2} = a^{2}-2ab+b^{2}$:
$(x^{2}-4x+4)+1=2x-3$
$x^{2}-4x+5=2x-3$
Bring all terms to one side to set the equation to zero:
$x^{2}-4x-2x+5+3=0$
$x^{2}-6x+8=0$
Since this equation is in the form $ax^{2}+bx+c=0$ where $a \neq 0$,it is a quadratic equation.

(B) Given equation: $x(x+1)+8=(x+2)(x-2)$
Expanding both sides:
$x^2 + x + 8 = x^2 - 4$
Subtracting $x^2$ from both sides:
$x + 8 = -4$
$x + 12 = 0$
Since the highest power of the variable $x$ is $1$,this is a linear equation,not a quadratic equation. $A$ quadratic equation must have the form $ax^2 + bx + c = 0$ where $a \neq 0$.

(B) Given equation: $x + \frac{1}{x} = x^2$ $(x \neq 0)$.
Multiply the entire equation by $x$ to eliminate the fraction:
$x(x) + x(\frac{1}{x}) = x(x^2)$
$x^2 + 1 = x^3$
Rearranging the terms to one side:
$x^3 - x^2 - 1 = 0$.
$A$ quadratic equation is defined as an equation of the form $ax^2 + bx + c = 0$,where $a \neq 0$. The highest power (degree) of the variable $x$ in the given equation is $3$. Since the degree of the equation is $3$,it is a cubic equation,not a quadratic equation.

(A) To determine if the equation is quadratic,we simplify it:
$(x-3)(2x+1) = x(x+5)$
$2x^2 + x - 6x - 3 = x^2 + 5x$
$2x^2 - 5x - 3 = x^2 + 5x$
Subtracting $(x^2 + 5x)$ from both sides:
$2x^2 - x^2 - 5x - 5x - 3 = 0$
$x^2 - 10x - 3 = 0$
Since this equation is in the form $ax^2 + bx + c = 0$ where $a \neq 0$,it is a quadratic equation.

(A) To determine if the equation $(x+2)^{3} = x(x^{2}-1)$ is quadratic,we expand both sides.
Using the identity $(a+b)^{3} = a^{3} + b^{3} + 3ab(a+b)$,the left side becomes:
$(x+2)^{3} = x^{3} + 2^{3} + 3(x)(2)(x+2) = x^{3} + 8 + 6x(x+2) = x^{3} + 8 + 6x^{2} + 12x$.
The right side is:
$x(x^{2}-1) = x^{3} - x$.
Equating both sides:
$x^{3} + 6x^{2} + 12x + 8 = x^{3} - x$.
Subtracting $x^{3}$ from both sides:
$6x^{2} + 12x + 8 = -x$.
Rearranging the terms to the form $ax^{2} + bx + c = 0$:
$6x^{2} + 13x + 8 = 0$.
Since this equation is of the form $ax^{2} + bx + c = 0$ where $a \neq 0$,it is a quadratic equation.

(B) quadratic equation in the variable $x$ is an equation of the form $ax^{2} + bx + c = 0$,where $a, b, c$ are real numbers and $a \neq 0$.
In the given equation $x^{2} + 5\sqrt{x} - 7 = 0$,the term $5\sqrt{x}$ can be written as $5x^{1/2}$.
Since the exponent of the variable $x$ is $1/2$,which is not a non-negative integer,this equation is not a polynomial equation of degree $2$.
Therefore,it is not a quadratic equation.

(A) Given equation: $x^{3}-4x^{2}-x+1=(x-2)^{3}$.
Expanding the right side using the identity $(a-b)^{3} = a^{3}-3a^{2}b+3ab^{2}-b^{3}$:
$(x-2)^{3} = x^{3}-3(x^{2})(2)+3(x)(2^{2})-(2)^{3} = x^{3}-6x^{2}+12x-8$.
Now,substitute this back into the equation:
$x^{3}-4x^{2}-x+1 = x^{3}-6x^{2}+12x-8$.
Subtract $x^{3}$ from both sides:
$-4x^{2}-x+1 = -6x^{2}+12x-8$.
Rearrange all terms to one side:
$(-4x^{2}+6x^{2}) + (-x-12x) + (1+8) = 0$.
$2x^{2}-13x+9 = 0$.
Since the equation is in the form $ax^{2}+bx+c=0$ where $a \neq 0$,it is a quadratic equation.

(A) To determine if the equation $7x = 2x^2$ is quadratic,we rewrite it in the standard form $ax^2 + bx + c = 0$.
Rearranging the terms,we get $2x^2 - 7x = 0$.
Comparing this with the standard form $ax^2 + bx + c = 0$,we have $a = 2$,$b = -7$,and $c = 0$.
Since the highest power of the variable $x$ is $2$ and $a \neq 0$,the given equation is a quadratic equation.

(A) Given equation: $\frac{x-2}{x+2} - \frac{x+2}{x-2} = \frac{3}{7}$.
Taking $LCM$ on the left side: $\frac{(x-2)^2 - (x+2)^2}{(x+2)(x-2)} = \frac{3}{7}$.
Expanding the terms: $\frac{(x^2 - 4x + 4) - (x^2 + 4x + 4)}{x^2 - 4} = \frac{3}{7}$.
Simplifying the numerator: $\frac{-8x}{x^2 - 4} = \frac{3}{7}$.
Cross-multiplying: $3(x^2 - 4) = -56x$.
$3x^2 - 12 = -56x$.
Rearranging into standard form $ax^2 + bx + c = 0$: $3x^2 + 56x - 12 = 0$.
Since the equation is in the form $ax^2 + bx + c = 0$ where $a \neq 0$,it is a quadratic equation.

(A) To determine if the equation $(3x - 4)^2 - (2x - 3)^2 = 7$ is quadratic,we expand both squares:
$(3x - 4)^2 = 9x^2 - 24x + 16$
$(2x - 3)^2 = 4x^2 - 12x + 9$
Now,substitute these into the original equation:
$(9x^2 - 24x + 16) - (4x^2 - 12x + 9) = 7$
$9x^2 - 24x + 16 - 4x^2 + 12x - 9 = 7$
Combine like terms:
$(9x^2 - 4x^2) + (-24x + 12x) + (16 - 9) = 7$
$5x^2 - 12x + 7 = 7$
Subtract $7$ from both sides:
$5x^2 - 12x = 0$
Since the equation is in the form $ax^2 + bx + c = 0$ where $a = 5 \neq 0$,it is a quadratic equation.

(B) To verify if $x = \frac{1}{2}$ is a solution,substitute the value into the quadratic equation $2x^{2} - 6x + 3 = 0$.
Substitute $x = \frac{1}{2}$:
$2(\frac{1}{2})^{2} - 6(\frac{1}{2}) + 3$
$= 2(\frac{1}{4}) - 3 + 3$
$= \frac{1}{2} - 3 + 3$
$= \frac{1}{2}$
Since the result is $\frac{1}{2}$ and not $0$,$x = \frac{1}{2}$ is not a solution to the given quadratic equation.

(A) To verify if $x = \frac{2}{3}$ is a solution,substitute the value of $x$ into the quadratic equation $6x^{2} - x - 2 = 0$.
$LHS$ = $6(\frac{2}{3})^{2} - (\frac{2}{3}) - 2$
$LHS$ = $6(\frac{4}{9}) - \frac{2}{3} - 2$
$LHS$ = $\frac{24}{9} - \frac{2}{3} - 2$
$LHS$ = $\frac{8}{3} - \frac{2}{3} - 2$
$LHS$ = $\frac{6}{3} - 2$
$LHS$ = $2 - 2 = 0$
Since $LHS$ = $RHS$,$x = \frac{2}{3}$ is a solution of the given quadratic equation.

(A) To verify if $x = -2\sqrt{2}$ is a solution,substitute the value of $x$ into the quadratic equation $x^{2} + \sqrt{2}x - 4 = 0$.
$LHS$ $= (-2\sqrt{2})^{2} + \sqrt{2}(-2\sqrt{2}) - 4$
$= (4 \times 2) + (-2 \times 2) - 4$
$= 8 - 4 - 4$
$= 0$
Since $LHS$ $=$ $RHS$,the given value $x = -2\sqrt{2}$ is a solution of the quadratic equation.

(B) To verify if $x = \frac{1}{2}$ is a solution,substitute $x = \frac{1}{2}$ into the quadratic equation $2x^2 - 5x + 3 = 0$.
$LHS$ $= 2(\frac{1}{2})^2 - 5(\frac{1}{2}) + 3$
$= 2(\frac{1}{4}) - \frac{5}{2} + 3$
$= \frac{1}{2} - \frac{5}{2} + 3$
$= -\frac{4}{2} + 3$
$= -2 + 3 = 1$
Since $LHS$ $\neq$ $RHS$ (where $RHS$ $= 0$),$x = \frac{1}{2}$ is not a solution of the given quadratic equation.

To verify whether $x = \frac{-2}{m+n}$ is a solution, we substitute it into the quadratic equation.
Given equation:
$(m+n)^2 x^2 + (m+n)x - 2 = 0$
Substitute $x = \frac{-2}{m+n}$:
$(m+n)^2 \left(\frac{-2}{m+n}\right)^2 + (m+n)\left(\frac{-2}{m+n}\right) - 2 = 0$
Simplifying:
$(m+n)^2 \cdot \frac{4}{(m+n)^2} + (m+n)\cdot \frac{-2}{m+n} - 2 = 0$
$4 - 2 - 2 = 0$
$0 = 0$
Since both sides are equal, $x = \frac{-2}{m+n}$ is a solution of the given quadratic equation.

(B) To verify if $x=2$ is a solution,substitute $x=2$ into the given quadratic equation: $(x-2)(x+3)+1=0$.
Substituting $x=2$:
$(2-2)(2+3)+1 = (0)(5)+1 = 0+1 = 1$.
Since the result is $1$ and not $0$,the left-hand side is not equal to the right-hand side.
Therefore,$x=2$ is not a solution of the given quadratic equation.

(A) To verify if $x = -\frac{1}{2}$ is a solution,substitute the value into the left-hand side $(LHS)$ of the equation.
$LHS$ = $\frac{1}{3 - 2(-\frac{1}{2})} + \frac{1}{5 + 2(-\frac{1}{2})}$
$LHS$ = $\frac{1}{3 - (-1)} + \frac{1}{5 + (-1)}$
$LHS$ = $\frac{1}{3 + 1} + \frac{1}{5 - 1}$
$LHS$ = $\frac{1}{4} + \frac{1}{4}$
$LHS$ = $\frac{2}{4} = \frac{1}{2}$
Since $LHS$ = $RHS$ (which is $\frac{1}{2}$),the given value $x = -\frac{1}{2}$ is a solution of the quadratic equation.

(A) To verify if $x = \frac{\sqrt{3}}{4}$ is a solution,substitute the value of $x$ into the quadratic equation $4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3} = 0$.
$LHS$ = $4 \sqrt{3} \left( \frac{\sqrt{3}}{4} \right)^{2} + 5 \left( \frac{\sqrt{3}}{4} \right) - 2 \sqrt{3}$
$LHS$ = $4 \sqrt{3} \left( \frac{3}{16} \right) + \frac{5 \sqrt{3}}{4} - 2 \sqrt{3}$
$LHS$ = $\frac{3 \sqrt{3}}{4} + \frac{5 \sqrt{3}}{4} - 2 \sqrt{3}$
$LHS$ = $\frac{8 \sqrt{3}}{4} - 2 \sqrt{3}$
$LHS$ = $2 \sqrt{3} - 2 \sqrt{3} = 0$
Since $LHS$ = $RHS$,the given value $x = \frac{\sqrt{3}}{4}$ is a solution of the quadratic equation.

(A) To verify if $x = 2$ is a solution,substitute $x = 2$ into the left-hand side $(LHS)$ of the equation:
$LHS$ = $\frac{1}{2+4} - \frac{1}{2-7}$
$LHS$ = $\frac{1}{6} - \frac{1}{-5}$
$LHS$ = $\frac{1}{6} + \frac{1}{5}$
$LHS$ = $\frac{5+6}{30} = \frac{11}{30}$
Since $LHS$ = $RHS$ $(\frac{11}{30})$,the value $x = 2$ is a solution of the given quadratic equation.

(A) Given equation: $x + \frac{1}{x} = 3 \frac{1}{3}$.
Convert the mixed fraction to an improper fraction: $3 \frac{1}{3} = \frac{3 \times 3 + 1}{3} = \frac{10}{3}$.
So,the equation is $x + \frac{1}{x} = \frac{10}{3}$.
Substitute $x = \frac{1}{3}$ into the left-hand side $(LHS)$ of the equation:
$LHS$ $= \frac{1}{3} + \frac{1}{1/3} = \frac{1}{3} + 3$.
$LHS$ $= \frac{1 + 9}{3} = \frac{10}{3}$.
Since $LHS$ $= \frac{10}{3}$ and $RHS$ $= \frac{10}{3}$,$LHS$ $=$ $RHS$.
Therefore,$x = \frac{1}{3}$ is a solution of the given quadratic equation.

(A) Given the quadratic equation is $kx^{2} + 2x - 3 = 0$.
Since $x = 2$ is a root of the equation,it must satisfy the equation.
Substitute $x = 2$ into the equation:
$k(2)^{2} + 2(2) - 3 = 0$
$k(4) + 4 - 3 = 0$
$4k + 1 = 0$
$4k = -1$
$k = -\frac{1}{4}$

(B) Given the quadratic equation $3x^{2} + 2kx - 3 = 0$.
Since $x = -\frac{1}{2}$ is a root of the equation,it must satisfy the equation.
Substitute $x = -\frac{1}{2}$ into the equation:
$3(-\frac{1}{2})^{2} + 2k(-\frac{1}{2}) - 3 = 0$
$3(\frac{1}{4}) - k - 3 = 0$
$\frac{3}{4} - k - 3 = 0$
$-k = 3 - \frac{3}{4}$
$-k = \frac{12 - 3}{4}$
$-k = \frac{9}{4}$
$k = -\frac{9}{4}$

(N/A) Given the quadratic equation is $x^{2} - \sqrt{p}x + q = 0$.
Since $x = -\sqrt{p}$ is a root of the equation,it must satisfy the equation.
Substitute $x = -\sqrt{p}$ into the equation:
$(-\sqrt{p})^{2} - \sqrt{p}(-\sqrt{p}) + q = 0$
$p - (-\sqrt{p} \cdot \sqrt{p}) + q = 0$
$p - (-p) + q = 0$
$p + p + q = 0$
$2p + q = 0$
Hence,it is proved.

(N/A) Given the quadratic equation is $px^{2} + qx + r = 0$.
Since $x = -2$ is a root of this equation,it must satisfy the equation.
Substituting $x = -2$ into the equation,we get:
$p(-2)^{2} + q(-2) + r = 0$
$p(4) - 2q + r = 0$
$4p - 2q + r = 0$
Hence,it is proved that $4p - 2q + r = 0$.

(A) Let $AB = x$ and $BC = y$. Given $x + y = 47$ and $x^2 + y^2 = AC^2 = 37^2 = 1369$.
We know that $(x + y)^2 = x^2 + y^2 + 2xy$.
Substituting the values: $47^2 = 1369 + 2xy$.
$2209 = 1369 + 2xy$,which gives $2xy = 840$,so $xy = 420$.
Now,we have $x + y = 47$ and $xy = 420$. These are the roots of the quadratic equation $t^2 - 47t + 420 = 0$.
Solving the quadratic equation: $t^2 - 35t - 12t + 420 = 0$.
$t(t - 35) - 12(t - 35) = 0$,so $(t - 35)(t - 12) = 0$.
The roots are $t = 35$ and $t = 12$.
Since $AB > BC$,we have $AB = 35$ and $BC = 12$.