Quadratic Equations Questions in English

(A) Let the number of marbles John had be $x$.
Then the number of marbles Jivanti had $= 45 - x$.
The number of marbles left with John after losing $5$ marbles $= x - 5$.
The number of marbles left with Jivanti after losing $5$ marbles $= (45 - x) - 5 = 40 - x$.
According to the problem,the product of the remaining marbles is $124$:
$(x - 5)(40 - x) = 124$.
Expanding the product:
$40x - x^2 - 200 + 5x = 124$
$-x^2 + 45x - 200 = 124$
Rearranging the terms to form a standard quadratic equation:
$-x^2 + 45x - 200 - 124 = 0$
$-x^2 + 45x - 324 = 0$
Multiplying by $-1$:
$x^2 - 45x + 324 = 0$.
Thus,the number of marbles John had satisfies the quadratic equation $x^2 - 45x + 324 = 0$.

(N/A) Let the number of toys produced on that day be $x$.
Therefore,the cost of production (in rupees) of each toy that day $= 55 - x$.
So,the total cost of production (in rupees) that day $= x(55 - x)$.
Given that the total cost of production is ₹ $750$,we have:
$x(55 - x) = 750$
Expanding the equation:
$55x - x^2 = 750$
Rearranging the terms to form a standard quadratic equation:
$-x^2 + 55x - 750 = 0$
Multiplying by $-1$ to simplify:
$x^2 - 55x + 750 = 0$
Thus,the number of toys produced that day satisfies the quadratic equation $x^2 - 55x + 750 = 0$.

(A) Given equation: $(x-2)^{2}+1=2x-3$
Expand the left-hand side $(LHS)$:
$(x-2)^{2}+1 = (x^{2}-4x+4)+1 = x^{2}-4x+5$
Now,equate it to the right-hand side $(RHS)$:
$x^{2}-4x+5 = 2x-3$
Rearrange the terms to bring them to one side:
$x^{2}-4x-2x+5+3 = 0$
$x^{2}-6x+8 = 0$
Since this equation is in the form $ax^{2}+bx+c=0$ where $a=1$,$b=-6$,and $c=8$ $(a \neq 0)$,it is a quadratic equation.

(B) Given equation: $x(x+1)+8=(x+2)(x-2)$
Expanding the left side: $x^2+x+8$
Expanding the right side using the identity $(a+b)(a-b)=a^2-b^2$: $x^2-4$
Equating both sides: $x^2+x+8=x^2-4$
Subtracting $x^2$ from both sides: $x+8=-4$
Rearranging the terms: $x+12=0$
Since the highest power of the variable $x$ is $1$,it is a linear equation and not of the form $ax^2+bx+c=0$ (where $a \neq 0$).
Therefore,the given equation is not a quadratic equation.

(YES) Given equation: $x(2x + 3) = x^2 + 1$
First,expand the left-hand side $(LHS)$:
$x(2x + 3) = 2x^2 + 3x$
Now,substitute this back into the equation:
$2x^2 + 3x = x^2 + 1$
Subtract $(x^2 + 1)$ from both sides to bring the equation to the standard form $ax^2 + bx + c = 0$:
$2x^2 - x^2 + 3x - 1 = 0$
$x^2 + 3x - 1 = 0$
Since this equation is in the form $ax^2 + bx + c = 0$ where $a = 1$,$b = 3$,and $c = -1$ (and $a \neq 0$),the given equation is a quadratic equation.

(A) Given equation: $(x+2)^{3} = x^{3}-4$
Expanding the left-hand side using the identity $(a+b)^{3} = a^{3}+b^{3}+3ab(a+b)$:
$(x+2)^{3} = x^{3} + 2^{3} + 3(x)(2)(x+2) = x^{3} + 8 + 6x(x+2) = x^{3} + 8 + 6x^{2} + 12x$
Substituting this back into the equation:
$x^{3} + 6x^{2} + 12x + 8 = x^{3} - 4$
Subtracting $x^{3}$ from both sides:
$6x^{2} + 12x + 8 = -4$
Adding $4$ to both sides:
$6x^{2} + 12x + 12 = 0$
Dividing the entire equation by $6$:
$x^{2} + 2x + 2 = 0$
Since this equation is in the form $ax^{2} + bx + c = 0$ where $a \neq 0$,it is a quadratic equation.

(A) Given equation: $(x+1)^{2}=2(x-3)$
Expanding the left side using the identity $(a+b)^{2} = a^{2} + 2ab + b^{2}$:
$x^{2} + 2x + 1 = 2(x-3)$
Distributing the $2$ on the right side:
$x^{2} + 2x + 1 = 2x - 6$
Subtracting $2x$ from both sides:
$x^{2} + 1 = -6$
Adding $6$ to both sides:
$x^{2} + 7 = 0$
Since this equation is of the form $ax^{2} + bx + c = 0$ (where $a=1, b=0, c=7$),it is a quadratic equation.

(A) Given equation: $x^{2}-2x=(-2)(3-x)$
Expand the right side: $x^{2}-2x = -6 + 2x$
Rearrange the terms to one side: $x^{2}-2x - 2x + 6 = 0$
Simplify: $x^{2}-4x + 6 = 0$
Since this equation is in the form $ax^{2}+bx+c=0$ where $a \neq 0$,it is a quadratic equation.

(B) Expand both sides of the equation:
$(x-2)(x+1) = x^2 + x - 2x - 2 = x^2 - x - 2$
$(x-1)(x+3) = x^2 + 3x - x - 3 = x^2 + 2x - 3$
Equating both sides:
$x^2 - x - 2 = x^2 + 2x - 3$
Subtract $x^2$ from both sides:
$-x - 2 = 2x - 3$
Rearranging the terms:
$3x - 1 = 0$
Since the highest power of the variable $x$ is $1$,it is not of the form $ax^2 + bx + c = 0$ (where $a \neq 0$).
Therefore,the given equation is not a quadratic equation.

(A) Expand the left side: $(x-3)(2x+1) = 2x^2 + x - 6x - 3 = 2x^2 - 5x - 3$.
Expand the right side: $x(x+5) = x^2 + 5x$.
Equating both sides: $2x^2 - 5x - 3 = x^2 + 5x$.
Rearranging the terms to one side: $2x^2 - x^2 - 5x - 5x - 3 = 0$.
Simplifying the expression: $x^2 - 10x - 3 = 0$.
Since this equation is in the form $ax^2 + bx + c = 0$ where $a \neq 0$,it is a quadratic equation.

(A) Expand both sides of the equation:
$(2x - 1)(x - 3) = 2x^2 - 6x - x + 3 = 2x^2 - 7x + 3$
$(x + 5)(x - 1) = x^2 - x + 5x - 5 = x^2 + 4x - 5$
Equating both sides:
$2x^2 - 7x + 3 = x^2 + 4x - 5$
Rearranging the terms to one side:
$2x^2 - x^2 - 7x - 4x + 3 + 5 = 0$
$x^2 - 11x + 8 = 0$
Since this equation is in the form $ax^2 + bx + c = 0$ where $a \neq 0$,it is a quadratic equation.

(B) Given equation: $x^{2}+3x+1=(x-2)^{2}$
Expand the right side using the identity $(a-b)^{2} = a^{2}-2ab+b^{2}$:
$x^{2}+3x+1 = x^{2}-4x+4$
Subtract $x^{2}$ from both sides:
$3x+1 = -4x+4$
Rearrange the terms to one side:
$3x+4x+1-4 = 0$
$7x-3 = 0$
The resulting equation is of the form $ax+b=0$,which is a linear equation,not a quadratic equation of the form $ax^{2}+bx+c=0$ (where $a \neq 0$).
Therefore,the given equation is not a quadratic equation.

(A) Given equation: $x^{3}-4x^{2}-x+1=(x-2)^{3}$
Expand the right side using the identity $(a-b)^{3} = a^{3}-b^{3}-3a^{2}b+3ab^{2}$:
$(x-2)^{3} = x^{3}-2^{3}-3(x^{2})(2)+3(x)(2^{2}) = x^{3}-8-6x^{2}+12x$
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 the 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.

(D) Let the breadth of the plot be $x \ m$.
According to the problem,the length of the plot is $(2x + 1) \ m$.
We know that the area of a rectangle is given by the formula: $\text{Area} = \text{Length} \times \text{Breadth}$.
Substituting the given values,we get: $528 = x(2x + 1)$.
Expanding the equation: $528 = 2x^{2} + x$.
Rearranging the terms to form a standard quadratic equation $ax^{2} + bx + c = 0$,we get: $2x^{2} + x - 528 = 0$.

(N/A) Let the two consecutive positive integers be $x$ and $x+1$.
According to the problem,the product of these two integers is $306$.
Therefore,$x(x+1) = 306$.
Expanding the equation,we get $x^2 + x = 306$.
Subtracting $306$ from both sides,we obtain the quadratic equation: $x^2 + x - 306 = 0$.

(N/A) Let Rohan's present age be $x$ years.
Then,his mother's present age $= x + 26$ years.
After $3$ years:
Rohan's age $= x + 3$ years.
Mother's age $= (x + 26) + 3 = x + 29$ years.
According to the problem,the product of their ages after $3$ years is $360$.
Therefore,$(x + 3)(x + 29) = 360$.
Expanding the equation: $x^2 + 29x + 3x + 87 = 360$.
$x^2 + 32x + 87 - 360 = 0$.
$x^2 + 32x - 273 = 0$.

(N/A) Let the speed of the train be $x \, km/h$.
Time taken to travel $480 \, km = \frac{480}{x} \, hours$.
In the second condition,the speed of the train is $(x - 8) \, km/h$.
It is given that the train takes $3 \, hours$ more to cover the same distance.
Therefore,the time taken to travel $480 \, km = \left( \frac{480}{x} + 3 \right) \, hours$.
Using the relation: $\text{Speed} \times \text{Time} = \text{Distance}$.
$(x - 8) \left( \frac{480}{x} + 3 \right) = 480$
$480 + 3x - \frac{3840}{x} - 24 = 480$
$3x - \frac{3840}{x} - 24 = 0$
Multiplying the entire equation by $x$:
$3x^2 - 24x - 3840 = 0$
Dividing by $3$:
$x^2 - 8x - 1280 = 0$

(N/A) Suppose the breadth of the hall is $x \ m$. Then,its length should be $(2x + 1) \ m$.
Now,the area of the hall $= \text{length} \times \text{breadth} = (2x + 1) \times x = (2x^2 + x) \ m^2$.
Given that the area is $300 \ m^2$,we have:
$2x^2 + x = 300$
$2x^2 + x - 300 = 0$
Applying the factorization method,we split the middle term:
$2x^2 - 24x + 25x - 300 = 0$
$2x(x - 12) + 25(x - 12) = 0$
$(x - 12)(2x + 25) = 0$
So,the roots of the equation are $x = 12$ or $x = -12.5$.
Since $x$ represents the breadth,it cannot be negative. Therefore,$x = 12$.
The breadth of the hall is $12 \ m$ and its length is $2(12) + 1 = 25 \ m$.

(N/A) The given equation is $2x^{2}-5x+3=0$.
Divide the entire equation by $2$ to make the coefficient of $x^{2}$ equal to $1$:
$x^{2}-\frac{5}{2}x+\frac{3}{2}=0$
Now,complete the square by adding and subtracting the square of half the coefficient of $x$,which is $(\frac{1}{2} \times \frac{5}{2})^{2} = (\frac{5}{4})^{2} = \frac{25}{16}$:
$x^{2}-\frac{5}{2}x + \frac{25}{16} - \frac{25}{16} + \frac{3}{2} = 0$
$(x-\frac{5}{4})^{2} - \frac{25}{16} + \frac{24}{16} = 0$
$(x-\frac{5}{4})^{2} - \frac{1}{16} = 0$
$(x-\frac{5}{4})^{2} = \frac{1}{16}$
Taking the square root on both sides:
$x-\frac{5}{4} = \pm \frac{1}{4}$
Case $1$: $x-\frac{5}{4} = \frac{1}{4} \implies x = \frac{5}{4} + \frac{1}{4} = \frac{6}{4} = \frac{3}{2}$
Case $2$: $x-\frac{5}{4} = -\frac{1}{4} \implies x = \frac{5}{4} - \frac{1}{4} = \frac{4}{4} = 1$
Thus,the solutions are $x = \frac{3}{2}$ and $x = 1$.

(N/A) Multiplying the equation throughout by $5$,we get:
$25x^{2}-30x-10=0$
This can be rewritten as:
$(5x)^{2}-2 \times (5x) \times 3 + 3^{2} - 3^{2} - 10 = 0$
$(5x-3)^{2} - 9 - 10 = 0$
$(5x-3)^{2} - 19 = 0$
$(5x-3)^{2} = 19$
Taking the square root on both sides:
$5x-3 = \pm \sqrt{19}$
$5x = 3 \pm \sqrt{19}$
$x = \frac{3 \pm \sqrt{19}}{5}$
Therefore,the roots are $\frac{3+\sqrt{19}}{5}$ and $\frac{3-\sqrt{19}}{5}$.

(16 M, 33 M) Let the breadth of the plot be $x \ m$. Then the length is $(2x + 1) \ m$.
Given that the area is $528 \ m^2$,we have $x(2x + 1) = 528$,which simplifies to $2x^2 + x - 528 = 0$.
This is a quadratic equation of the form $ax^2 + bx + c = 0$,where $a = 2, b = 1, c = -528$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-528)}}{2(2)} = \frac{-1 \pm \sqrt{1 + 4224}}{4} = \frac{-1 \pm \sqrt{4225}}{4} = \frac{-1 \pm 65}{4}$.
This gives two possible values for $x$: $x = \frac{64}{4} = 16$ or $x = \frac{-66}{4} = -16.5$.
Since the breadth cannot be negative,we take $x = 16 \ m$.
Therefore,the breadth is $16 \ m$ and the length is $2(16) + 1 = 33 \ m$.

(A) Let the smaller of the two consecutive odd positive integers be $x$. Then,the second integer will be $x+2$.
According to the question,
$x^{2} + (x+2)^{2} = 290$
$x^{2} + x^{2} + 4x + 4 = 290$
$2x^{2} + 4x - 286 = 0$
Dividing by $2$,we get:
$x^{2} + 2x - 143 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$:
$x = \frac{-2 \pm \sqrt{2^{2} - 4(1)(-143)}}{2(1)}$
$x = \frac{-2 \pm \sqrt{4 + 572}}{2} = \frac{-2 \pm \sqrt{576}}{2} = \frac{-2 \pm 24}{2}$
$x = \frac{22}{2} = 11$ or $x = \frac{-26}{2} = -13$
Since $x$ must be a positive integer,we take $x = 11$.
The two consecutive odd integers are $11$ and $13$.
Verification: $11^{2} + 13^{2} = 121 + 169 = 290$.

(N/A) Let the breadth of the rectangular park be $x\, m$.
Then,its length is $(x+3)\, m$.
Therefore,the area of the rectangular park is $x(x+3) = (x^2 + 3x)\, m^2$.
The area of the isosceles triangle is $\frac{1}{2} \times \text{base} \times \text{altitude} = \frac{1}{2} \times x \times 12 = 6x\, m^2$.
According to the problem,the area of the rectangular park is $4\, m^2$ more than the area of the triangular park:
$x^2 + 3x = 6x + 4$
$x^2 - 3x - 4 = 0$
$(x - 4)(x + 1) = 0$
So,$x = 4$ or $x = -1$.
Since breadth cannot be negative,$x = 4$.
Thus,the breadth of the park is $4\, m$ and its length is $4 + 3 = 7\, m$.

(A) The given quadratic equation is $3x^{2} - 5x + 2 = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = 3$,$b = -5$,and $c = 2$.
First,calculate the discriminant $D = b^{2} - 4ac$:
$D = (-5)^{2} - 4(3)(2) = 25 - 24 = 1$.
Since $D > 0$,real and distinct roots exist.
Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:
$x = \frac{-(-5) \pm \sqrt{1}}{2(3)} = \frac{5 \pm 1}{6}$.
Case $1$: $x = \frac{5 + 1}{6} = \frac{6}{6} = 1$.
Case $2$: $x = \frac{5 - 1}{6} = \frac{4}{6} = \frac{2}{3}$.
Thus,the roots are $1$ and $\frac{2}{3}$.

(A) To determine if an equation is quadratic,we simplify it to the standard form $ax^{2}+bx+c=0$,where $a \neq 0$.
Check option $(A)$: $(x+2)(x-1) = x^{2}-x+2x-2 = x^{2}+x-2$. Given equation: $x^{2}+x-2 = x^{2}-2x-3$. Simplifying,$3x+1=0$. This is a linear equation,not a quadratic equation.
Check option $(B)$: $(x+2)^{2} = 2(x+3) \implies x^{2}+4x+4 = 2x+6 \implies x^{2}+2x-2=0$. This is a quadratic equation.
Check option $(C)$: $x^{2}+3x = (-1)(1-3x)^{2} \implies x^{2}+3x = -(1-6x+9x^{2}) \implies x^{2}+3x = -1+6x-9x^{2} \implies 10x^{2}-3x+1=0$. This is a quadratic equation.
Check option $(D)$: $x^{3}-x^{2}+2x+1 = (x+1)^{3} \implies x^{3}-x^{2}+2x+1 = x^{3}+3x^{2}+3x+1 \implies -x^{2}+2x = 3x^{2}+3x \implies 4x^{2}+x=0$. This is a quadratic equation.
Therefore,the equation in option $(A)$ is not a quadratic equation.

(B) To solve the quadratic equation $4x^{2} - \sqrt{3}x - 5 = 0$ by the method of completing the square,we first ensure the coefficient of $x^{2}$ is $1$ or a perfect square. Here,$4x^{2} = (2x)^{2}$.
We rewrite the equation as $(2x)^{2} - 2(2x)(\frac{\sqrt{3}}{4}) - 5 = 0$.
To complete the square in the form $(a - b)^{2} = a^{2} - 2ab + b^{2}$,we identify $a = 2x$ and $b = \frac{\sqrt{3}}{4}$.
The term to be added and subtracted is $b^{2} = (\frac{\sqrt{3}}{4})^{2} = \frac{3}{16}$.
Thus,the constant is $\frac{3}{16}$.

(C) For an equation to be a quadratic equation,it must be in the form $ax^{2}+bx+c=0$,where $a \neq 0$.
$(A)$ $x^{2}+2x+1 = (4-x)^{2}+3$
$\Rightarrow x^{2}+2x+1 = 16-8x+x^{2}+3$
$\Rightarrow 10x-18=0$. This is a linear equation.
$(B)$ $-2x^{2} = (5-x)(2x-\frac{2}{5})$
$\Rightarrow -2x^{2} = 10x-2-2x^{2}+\frac{2x}{5}$
$\Rightarrow 0 = \frac{52x}{5}-2$. This is a linear equation.
$(C)$ $x^{3}-x^{2} = (x-1)^{3}$
$\Rightarrow x^{3}-x^{2} = x^{3}-3x^{2}+3x-1$
$\Rightarrow 2x^{2}-3x+1=0$. This is a quadratic equation as it is in the form $ax^{2}+bx+c=0$ with $a=2 \neq 0$.
$(D)$ $(k+1)x^{2}+\frac{3}{2}x=7$,where $k=-1$
$\Rightarrow (-1+1)x^{2}+\frac{3}{2}x=7$
$\Rightarrow \frac{3}{2}x-7=0$. This is a linear equation.

(D) quadratic equation is of the form $ax^{2}+bx+c=0$,where $a \neq 0$.
$(A)$ $2(x-1)^{2}=4 x^{2}-2 x+1 \Rightarrow 2(x^{2}-2x+1)=4x^{2}-2x+1 \Rightarrow 2x^{2}-4x+2=4x^{2}-2x+1 \Rightarrow 2x^{2}+2x-1=0$. This is a quadratic equation.
$(B)$ $2 x-x^{2}=x^{2}+5 \Rightarrow 2x^{2}-2x+5=0$. This is a quadratic equation.
$(C)$ $(-\sqrt{2} x+\sqrt{3})^{2}=3 x^{2}-5 x \Rightarrow 2x^{2}-2\sqrt{6}x+3=3x^{2}-5x \Rightarrow x^{2}-(5-2\sqrt{6})x-3=0$. This is a quadratic equation.
$(D)$ $(x^{2}+2 x)^{2}=x^{4}+3+4 x^{2} \Rightarrow x^{4}+4x^{3}+4x^{2}=x^{4}+3+4x^{2} \Rightarrow 4x^{3}-3=0$. Since the highest power of $x$ is $3$,this is a cubic equation,not a quadratic equation.

(A) To determine if $x = 2$ is a root of an equation,we substitute $x = 2$ into each equation and check if the result is $0$.
$(A)$ For $2x^2 - 7x + 6 = 0$:
$2(2)^2 - 7(2) + 6 = 2(4) - 14 + 6 = 8 - 14 + 6 = 0$.
Since the result is $0$,$x = 2$ is a root of this equation.
$(B)$ For $x^2 + 3x - 12 = 0$:
$(2)^2 + 3(2) - 12 = 4 + 6 - 12 = -2 \neq 0$.
So,$x = 2$ is not a root.
$(C)$ For $x^2 - 4x + 5 = 0$:
$(2)^2 - 4(2) + 5 = 4 - 8 + 5 = 1 \neq 0$.
So,$x = 2$ is not a root.
$(D)$ For $3x^2 - 6x - 2 = 0$:
$3(2)^2 - 6(2) - 2 = 12 - 12 - 2 = -2 \neq 0$.
So,$x = 2$ is not a root.
Therefore,the correct equation is $2x^2 - 7x + 6 = 0$.

(B) Since $\frac{1}{2}$ is a root of the quadratic equation $x^{2}+kx-\frac{5}{4}=0$,it must satisfy the equation.
Substituting $x = \frac{1}{2}$ into the equation:
$(\frac{1}{2})^{2} + k(\frac{1}{2}) - \frac{5}{4} = 0$
$\frac{1}{4} + \frac{k}{2} - \frac{5}{4} = 0$
$\frac{1 + 2k - 5}{4} = 0$
$2k - 4 = 0$
$2k = 4$
$k = 2$

(B) To determine if $0.2$ is a root of the equation $x^{2}-0.4=0$,we substitute $x = 0.2$ into the equation.
Substituting $x = 0.2$ into the left-hand side $(LHS)$ gives:
$(0.2)^{2} - 0.4 = 0.04 - 0.4 = -0.36$.
Since the result is $-0.36$ and not $0$,the value $0.2$ does not satisfy the equation.
Therefore,$0.2$ is not a root of the equation $x^{2}-0.4=0$.

(B) 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,$a=2$,$b=-\sqrt{5}$,and $c=-2$.
First,calculate the discriminant $D = b^{2}-4ac$:
$D = (-\sqrt{5})^{2} - 4 \times 2 \times (-2) = 5 + 16 = 21$.
Now,substitute the values into the quadratic formula:
$x = \frac{-(-\sqrt{5}) \pm \sqrt{21}}{2 \times 2} = \frac{\sqrt{5} \pm \sqrt{21}}{4}$.
Thus,the roots are $\frac{\sqrt{5}+\sqrt{21}}{4}$ and $\frac{\sqrt{5}-\sqrt{21}}{4}$.

(D) The given equation is $2x^{2} - 3x - 5 = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get:
$a = 2, b = -3, c = -5$.
The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.
Substituting the values:
$x = \frac{-(-3) \pm \sqrt{(-3)^{2} - 4(2)(-5)}}{2(2)}$
$x = \frac{3 \pm \sqrt{9 + 40}}{4}$
$x = \frac{3 \pm \sqrt{49}}{4}$
$x = \frac{3 \pm 7}{4}$.
Calculating the two roots:
$x_{1} = \frac{3 + 7}{4} = \frac{10}{4} = \frac{5}{2}$
$x_{2} = \frac{3 - 7}{4} = \frac{-4}{4} = -1$.
Thus,the roots of the equation are $\frac{5}{2}$ and $-1$.

(A) Given equation is $5x^{2} + 13x + 8 = 0$.
On comparing with the standard form $ax^{2} + bx + c = 0$,we get:
$a = 5, b = 13, c = 8$.
The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.
Substituting the values:
$x = \frac{-(13) \pm \sqrt{(13)^{2} - 4(5)(8)}}{2(5)}$
$x = \frac{-13 \pm \sqrt{169 - 160}}{10}$
$x = \frac{-13 \pm \sqrt{9}}{10}$
$x = \frac{-13 \pm 3}{10}$.
Case $1$: $x = \frac{-13 + 3}{10} = \frac{-10}{10} = -1$.
Case $2$: $x = \frac{-13 - 3}{10} = \frac{-16}{10} = -\frac{8}{5}$.
Thus,the roots are $-1$ and $-\frac{8}{5}$.

(B) The given equation is $-3x^{2} + 5x + 12 = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = -3$,$b = 5$,and $c = 12$.
The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.
Substituting the values,we get $x = \frac{-(5) \pm \sqrt{(5)^{2} - 4(-3)(12)}}{2(-3)}$.
Simplifying the expression,$x = \frac{-5 \pm \sqrt{25 + 144}}{-6} = \frac{-5 \pm \sqrt{169}}{-6}$.
Since $\sqrt{169} = 13$,we have $x = \frac{-5 \pm 13}{-6}$.
Taking the positive sign: $x = \frac{-5 + 13}{-6} = \frac{8}{-6} = -\frac{4}{3}$.
Taking the negative sign: $x = \frac{-5 - 13}{-6} = \frac{-18}{-6} = 3$.
Thus,the roots of the equation are $-\frac{4}{3}$ and $3$.

(C) Given equation is $-x^{2}+7x-10=0$.
On comparing with the standard form $ax^{2}+bx+c=0$,we get:
$a = -1, b = 7, c = -10$.
The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$.
Substituting the values:
$x = \frac{-(7) \pm \sqrt{(7)^{2}-4(-1)(-10)}}{2(-1)}$
$x = \frac{-7 \pm \sqrt{49-40}}{-2}$
$x = \frac{-7 \pm \sqrt{9}}{-2}$
$x = \frac{-7 \pm 3}{-2}$.
Calculating the two roots:
$x_1 = \frac{-7 + 3}{-2} = \frac{-4}{-2} = 2$
$x_2 = \frac{-7 - 3}{-2} = \frac{-10}{-2} = 5$.
Thus,the roots of the equation are $2$ and $5$.

(D) The given equation is $x^{2}+2 \sqrt{2} x-6=0$.
Comparing this with the standard form $ax^{2}+bx+c=0$,we get $a=1, b=2 \sqrt{2}$,and $c=-6$.
The quadratic formula is $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$.
Substituting the values,we get $x = \frac{-(2 \sqrt{2}) \pm \sqrt{(2 \sqrt{2})^{2}-4(1)(-6)}}{2(1)}$.
$x = \frac{-2 \sqrt{2} \pm \sqrt{8+24}}{2} = \frac{-2 \sqrt{2} \pm \sqrt{32}}{2}$.
Since $\sqrt{32} = 4 \sqrt{2}$,we have $x = \frac{-2 \sqrt{2} \pm 4 \sqrt{2}}{2}$.
Calculating the two roots:
$x_1 = \frac{-2 \sqrt{2} + 4 \sqrt{2}}{2} = \frac{2 \sqrt{2}}{2} = \sqrt{2}$.
$x_2 = \frac{-2 \sqrt{2} - 4 \sqrt{2}}{2} = \frac{-6 \sqrt{2}}{2} = -3 \sqrt{2}$.
Thus,the roots are $\sqrt{2}$ and $-3 \sqrt{2}$.

(N/A) Given the equation $x + \frac{1}{x} = 3$.
Multiplying throughout by $x$,we get:
$x^2 + 1 = 3x$
Rearranging the terms,we get the quadratic equation:
$x^2 - 3x + 1 = 0$
Comparing this with the standard form $ax^2 + bx + c = 0$,we have $a = 1, b = -3, c = 1$.
The discriminant $D = b^2 - 4ac = (-3)^2 - 4(1)(1) = 9 - 4 = 5$.
Since $D > 0$,the roots are real and distinct.
Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$,we get:
$x = \frac{-(-3) \pm \sqrt{5}}{2(1)} = \frac{3 \pm \sqrt{5}}{2}$.
Thus,the roots are $\frac{3 + \sqrt{5}}{2}$ and $\frac{3 - \sqrt{5}}{2}$.

(N/A) Given the equation: $\frac{1}{x} - \frac{1}{x-2} = 3$ for $x \neq 0, 2$.
Multiplying both sides by $x(x-2)$,we get:
$(x-2) - x = 3x(x-2)$
Simplifying the left side:
$-2 = 3x^2 - 6x$
Rearranging into the standard quadratic form $ax^2 + bx + c = 0$:
$3x^2 - 6x + 2 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a = 3, b = -6, c = 2$:
Discriminant $D = b^2 - 4ac = (-6)^2 - 4(3)(2) = 36 - 24 = 12$.
Thus,$x = \frac{6 \pm \sqrt{12}}{2(3)} = \frac{6 \pm 2\sqrt{3}}{6}$.
Dividing by $2$:
$x = \frac{3 \pm \sqrt{3}}{3}$.
Therefore,the roots are $\frac{3 + \sqrt{3}}{3}$ and $\frac{3 - \sqrt{3}}{3}$.

(3, 1/2) Given equation: $2x^2 - 7x + 3 = 0$.
Step $1$: Move the constant term to the right side:
$2x^2 - 7x = -3$.
Step $2$: Divide the entire equation by the coefficient of $x^2$ (which is $2$):
$x^2 - \frac{7}{2}x = -\frac{3}{2}$.
Step $3$: Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $-\frac{7}{2}$,so half of it is $-\frac{7}{4}$. Adding $(-\frac{7}{4})^2 = \frac{49}{16}$ to both sides:
$x^2 - 2(x)(\frac{7}{4}) + (\frac{7}{4})^2 = -\frac{3}{2} + \frac{49}{16}$.
Step $4$: Write the left side as a perfect square:
$(x - \frac{7}{4})^2 = \frac{-24 + 49}{16} = \frac{25}{16}$.
Step $5$: Take the square root of both sides:
$x - \frac{7}{4} = \pm \frac{5}{4}$.
Step $6$: Solve for $x$:
Case $1$: $x = \frac{7}{4} + \frac{5}{4} = \frac{12}{4} = 3$.
Case $2$: $x = \frac{7}{4} - \frac{5}{4} = \frac{2}{4} = \frac{1}{2}$.
Thus,the roots are $3$ and $\frac{1}{2}$.

(N/A) Given equation: $2x^{2} + x - 4 = 0$.
Step $1$: Divide the entire equation by $2$ to make the coefficient of $x^{2}$ equal to $1$:
$x^{2} + \frac{1}{2}x - 2 = 0 \Rightarrow x^{2} + \frac{1}{2}x = 2$.
Step $2$: Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $\frac{1}{2}$,so half of it is $\frac{1}{4}$. Adding $(\frac{1}{4})^{2}$ to both sides:
$x^{2} + 2(x)(\frac{1}{4}) + (\frac{1}{4})^{2} = 2 + (\frac{1}{4})^{2}$.
Step $3$: Write the left side as a perfect square:
$(x + \frac{1}{4})^{2} = 2 + \frac{1}{16} = \frac{32 + 1}{16} = \frac{33}{16}$.
Step $4$: Take the square root of both sides:
$x + \frac{1}{4} = \pm \sqrt{\frac{33}{16}} = \pm \frac{\sqrt{33}}{4}$.
Step $5$: Solve for $x$:
$x = -\frac{1}{4} \pm \frac{\sqrt{33}}{4} = \frac{-1 \pm \sqrt{33}}{4}$.
Thus,the roots are $x = \frac{-1 + \sqrt{33}}{4}$ and $x = \frac{-1 - \sqrt{33}}{4}$.

(N/A) Given equation: $4x^{2} + 4\sqrt{3}x + 3 = 0$.
We can rewrite the equation as:
$(2x)^{2} + 2(2x)(\sqrt{3}) + (\sqrt{3})^{2} = 0$.
This is in the form of the algebraic identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$,where $a = 2x$ and $b = \sqrt{3}$.
Therefore,$(2x + \sqrt{3})^{2} = 0$.
Taking the square root on both sides,we get:
$2x + \sqrt{3} = 0$.
Solving for $x$:
$2x = -\sqrt{3} \Rightarrow x = -\frac{\sqrt{3}}{2}$.
Since the quadratic equation has a repeated root,the roots are $x = -\frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}$.

(N/A) Given equation: $2 x^{2}-7 x+3=0$
On comparing this equation with the standard form $a x^{2}+b x+c=0$,we get:
$a=2, b=-7, c=3$
The quadratic formula is given by:
$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$
Substituting the values of $a, b,$ and $c$:
$x=\frac{-(-7) \pm \sqrt{(-7)^{2}-4(2)(3)}}{2(2)}$
$x=\frac{7 \pm \sqrt{49-24}}{4}$
$x=\frac{7 \pm \sqrt{25}}{4}$
$x=\frac{7 \pm 5}{4}$
Now,solving for both cases:
Case $1$: $x = \frac{7+5}{4} = \frac{12}{4} = 3$
Case $2$: $x = \frac{7-5}{4} = \frac{2}{4} = \frac{1}{2}$
Therefore,the roots of the equation are $3$ and $\frac{1}{2}$.

(N/A) Given equation: $2 x^{2} + x - 4 = 0$
On comparing this equation with the standard form $a x^{2} + b x + c = 0$,we obtain:
$a = 2, b = 1, c = -4$
By using the quadratic formula,$x = \frac{-b \pm \sqrt{b^{2} - 4 a c}}{2 a}$:
Substituting the values:
$x = \frac{-1 \pm \sqrt{(1)^{2} - 4(2)(-4)}}{2(2)}$
$x = \frac{-1 \pm \sqrt{1 + 32}}{4}$
$x = \frac{-1 \pm \sqrt{33}}{4}$
Thus,the roots are $x = \frac{-1 + \sqrt{33}}{4}$ and $x = \frac{-1 - \sqrt{33}}{4}$.

(N/A) Given equation: $x - \frac{1}{x} = 3$
Multiply throughout by $x$ to get: $x^{2} - 1 = 3x$
Rearrange into standard quadratic form: $x^{2} - 3x - 1 = 0$
Comparing this with $ax^{2} + bx + c = 0$,we get $a = 1, b = -3, c = -1$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$:
$x = \frac{-(-3) \pm \sqrt{(-3)^{2} - 4(1)(-1)}}{2(1)}$
$x = \frac{3 \pm \sqrt{9 + 4}}{2}$
$x = \frac{3 \pm \sqrt{13}}{2}$
Thus,the roots are $x = \frac{3 + \sqrt{13}}{2}$ and $x = \frac{3 - \sqrt{13}}{2}$.

(C) Let the present age of Rehman be $x$ years.
Three years ago,his age was $(x-3)$ years.
Five years hence,his age will be $(x+5)$ years.
It is given that the sum of the reciprocals of Rehman's ages $3$ years ago and $5$ years from now is $\frac{1}{3}$.
$\therefore \frac{1}{x-3} + \frac{1}{x+5} = \frac{1}{3}$
$\frac{x+5+x-3}{(x-3)(x+5)} = \frac{1}{3}$
$\frac{2x+2}{x^2+2x-15} = \frac{1}{3}$
$3(2x+2) = x^2+2x-15$
$6x+6 = x^2+2x-15$
$x^2-4x-21 = 0$
$x^2-7x+3x-21 = 0$
$x(x-7)+3(x-7) = 0$
$(x-7)(x+3) = 0$
$x = 7$ or $x = -3$.
Since age cannot be negative,the present age of Rehman is $7$ years.

(C) Let the marks in Mathematics be $x$.
Then,the marks in English will be $30-x$.
According to the given condition:
$(x+2)(30-x-3) = 210$
$(x+2)(27-x) = 210$
$27x - x^2 + 54 - 2x = 210$
$-x^2 + 25x + 54 = 210$
$x^2 - 25x + 156 = 0$
Solving the quadratic equation by splitting the middle term:
$x^2 - 12x - 13x + 156 = 0$
$x(x-12) - 13(x-12) = 0$
$(x-12)(x-13) = 0$
So,$x = 12$ or $x = 13$.
Case $1$: If marks in Mathematics are $12$,then marks in English are $30-12 = 18$.
Case $2$: If marks in Mathematics are $13$,then marks in English are $30-13 = 17$.

(A) Let the time taken by the smaller pipe to fill the tank be $x \text{ hours}$.
Time taken by the larger pipe $= (x - 10) \text{ hours}$.
Part of the tank filled by the smaller pipe in $1 \text{ hour} = \frac{1}{x}$.
Part of the tank filled by the larger pipe in $1 \text{ hour} = \frac{1}{x - 10}$.
It is given that the tank can be filled in $9 \frac{3}{8} = \frac{75}{8} \text{ hours}$ by both pipes together. Therefore,
$\frac{1}{x} + \frac{1}{x - 10} = \frac{8}{75}$.
$\frac{x - 10 + x}{x(x - 10)} = \frac{8}{75}$.
$\frac{2x - 10}{x^2 - 10x} = \frac{8}{75}$.
$75(2x - 10) = 8(x^2 - 10x)$.
$150x - 750 = 8x^2 - 80x$.
$8x^2 - 230x + 750 = 0$.
Dividing by $2$,we get $4x^2 - 115x + 375 = 0$.
$4x^2 - 100x - 15x + 375 = 0$.
$4x(x - 25) - 15(x - 25) = 0$.
$(x - 25)(4x - 15) = 0$.
Thus,$x = 25$ or $x = \frac{15}{4} = 3.75$.
If $x = 3.75$,the larger pipe takes $3.75 - 10 = -6.25 \text{ hours}$,which is impossible.
Therefore,the smaller pipe takes $25 \text{ hours}$ and the larger pipe takes $25 - 10 = 15 \text{ hours}$.

(A) Let $P$ be the required location of the pole on the boundary of the circular park.
Let the distance of the pole from gate $B$ be $x \, m$,i.e.,$BP = x \, m$.
The difference of the distances of the pole from the two gates is $AP - BP = 7 \, m$.
Therefore,$AP = (x + 7) \, m$.
Since $AB$ is the diameter of the circular park,$AB = 13 \, m$.
In a circle,the angle subtended by the diameter at any point on the boundary is $90^{\circ}$. Thus,$\angle APB = 90^{\circ}$.
By Pythagoras theorem in $\triangle APB$,we have $AP^2 + BP^2 = AB^2$.
Substituting the values,$(x + 7)^2 + x^2 = 13^2$.
Expanding the equation: $x^2 + 14x + 49 + x^2 = 169$.
$2x^2 + 14x - 120 = 0$.
Dividing by $2$,we get the quadratic equation: $x^2 + 7x - 60 = 0$.
To check if this is possible,we calculate the discriminant $D = b^2 - 4ac = 7^2 - 4(1)(-60) = 49 + 240 = 289$.
Since $D > 0$,the equation has two distinct real roots,so it is possible to erect the pole.
Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$,we get $x = \frac{-7 \pm \sqrt{289}}{2} = \frac{-7 \pm 17}{2}$.
This gives $x = \frac{10}{2} = 5$ or $x = \frac{-24}{2} = -12$.
Since distance cannot be negative,we take $x = 5$.
Thus,the pole should be erected at a distance of $5 \, m$ from gate $B$ and $12 \, m$ from gate $A$.

(A) Let the breadth of the mango grove be $x \, m$.
Then,the length of the mango grove will be $2x \, m$.
Area of the rectangular mango grove $= \text{Length} \times \text{Breadth} = (2x)(x) = 2x^2 \, m^2$.
Given that the area is $800 \, m^2$,we have $2x^2 = 800$.
$x^2 = 400$.
$x^2 - 400 = 0$.
Comparing this with the standard quadratic equation $ax^2 + bx + c = 0$,we get $a = 1, b = 0, c = -400$.
The discriminant $D = b^2 - 4ac = (0)^2 - 4(1)(-400) = 1600$.
Since $D > 0$,the equation has two distinct real roots,meaning such a design is possible.
$x^2 = 400 \implies x = \pm 20$.
Since the breadth cannot be negative,we take $x = 20$.
Thus,the breadth is $20 \, m$ and the length is $2(20) = 40 \, m$.