Textbook - 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 $3$:
$x^2 - 8x - 1280 = 0$

(A) To find the roots of the quadratic equation $2x^{2}-5x+3=0$ by factorisation,we split the middle term $-5x$ into $-2x$ and $-3x$,because $(-2x) \times (-3x) = 6x^{2}$,which is the product of the coefficient of $x^{2}$ $(2)$ and the constant term $(3)$.
Now,rewrite the equation as:
$2x^{2}-2x-3x+3=0$
Factor by grouping:
$2x(x-1)-3(x-1)=0$
$(2x-3)(x-1)=0$
Setting each factor to zero:
$2x-3=0 \implies 2x=3 \implies x=\frac{3}{2}$
$x-1=0 \implies x=1$
Thus,the roots of the equation are $x=1$ and $x=\frac{3}{2}$.

(A) We have the quadratic equation $6x^{2}-x-2=0$.
To find the roots,we factorize the quadratic expression by splitting the middle term:
$6x^{2}-x-2 = 6x^{2}+3x-4x-2$
Grouping the terms,we get:
$= 3x(2x+1)-2(2x+1)$
$= (3x-2)(2x+1)$
The roots of $6x^{2}-x-2=0$ are the values of $x$ for which $(3x-2)(2x+1)=0$.
Setting each factor to zero:
$3x-2=0$ or $2x+1=0$
Solving for $x$:
$3x=2 \implies x=\frac{2}{3}$
$2x=-1 \implies x=-\frac{1}{2}$
Therefore,the roots of the quadratic equation are $\frac{2}{3}$ and $-\frac{1}{2}$.

(N/A) The given quadratic equation is $3x^{2}-2\sqrt{6}x+2=0$.
We can factorize the middle term $-2\sqrt{6}x$ as $-\sqrt{6}x - \sqrt{6}x$.
So,$3x^{2}-\sqrt{6}x-\sqrt{6}x+2=0$.
Taking common terms from the first two and last two terms:
$\sqrt{3}x(\sqrt{3}x-\sqrt{2})-\sqrt{2}(\sqrt{3}x-\sqrt{2})=0$.
This simplifies to $(\sqrt{3}x-\sqrt{2})(\sqrt{3}x-\sqrt{2})=0$.
Setting each factor to zero,we get $\sqrt{3}x-\sqrt{2}=0$,which implies $x = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}}$.
Since both factors are identical,the roots are equal.
Therefore,the roots of the equation are $\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}$.

(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 $2(12) + 1 = 25 \ m$.

(A) To find the roots of the quadratic equation $x^{2}-3x-10=0$ by factorisation,we split the middle term.
We need two numbers whose product is $-10$ and whose sum is $-3$. These numbers are $-5$ and $2$.
$x^{2}-5x+2x-10=0$
Factor by grouping:
$x(x-5)+2(x-5)=0$
$(x-5)(x+2)=0$
Setting each factor to zero:
$x-5=0$ or $x+2=0$
Therefore,the roots are $x=5$ or $x=-2$.

(A) To find the roots of the quadratic equation $2x^{2} + x - 6 = 0$ by factorisation,we split the middle term.
We need two numbers whose product is $2 \times (-6) = -12$ and whose sum is $1$.
These numbers are $4$ and $-3$.
So,$2x^{2} + 4x - 3x - 6 = 0$.
Grouping the terms,we get $2x(x + 2) - 3(x + 2) = 0$.
Factoring out $(x + 2)$,we have $(x + 2)(2x - 3) = 0$.
Setting each factor to zero:
$x + 2 = 0 \implies x = -2$
$2x - 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2}$.
Thus,the roots are $x = -2$ and $x = \frac{3}{2}$.

(A) To find the roots of the quadratic equation $\sqrt{2} x^{2}+7 x+5 \sqrt{2}=0$ by factorisation:
$1$. Split the middle term $7x$ into two parts such that their sum is $7x$ and their product is $(\sqrt{2} x^{2}) \times (5 \sqrt{2}) = 10x^{2}$.
$2$. The two parts are $5x$ and $2x$ because $5x + 2x = 7x$ and $5x \times 2x = 10x^{2}$.
$3$. Rewrite the equation: $\sqrt{2} x^{2} + 2x + 5x + 5 \sqrt{2} = 0$.
$4$. Factor by grouping: $\sqrt{2} x(x + \sqrt{2}) + 5(x + \sqrt{2}) = 0$.
$5$. Take out the common factor $(x + \sqrt{2})$: $(x + \sqrt{2})(\sqrt{2} x + 5) = 0$.
$6$. Set each factor to zero:
- $x + \sqrt{2} = 0 \implies x = -\sqrt{2}$
- $\sqrt{2} x + 5 = 0 \implies \sqrt{2} x = -5 \implies x = -\frac{5}{\sqrt{2}}$
Thus,the roots are $x = -\sqrt{2}$ and $x = -\frac{5}{\sqrt{2}}$.

(A) Given equation: $2x^{2} - x + \frac{1}{8} = 0$
Multiply the entire equation by $8$ to remove the fraction:
$16x^{2} - 8x + 1 = 0$
Now,factorise the quadratic expression by splitting the middle term:
$16x^{2} - 4x - 4x + 1 = 0$
Group the terms:
$4x(4x - 1) - 1(4x - 1) = 0$
$(4x - 1)(4x - 1) = 0$
$(4x - 1)^{2} = 0$
Setting the factor to zero:
$4x - 1 = 0$
$4x = 1$
$x = \frac{1}{4}$
Thus,the roots of the equation are $x = \frac{1}{4}, \frac{1}{4}$.

(A) Given quadratic equation: $100 x^{2}-20 x+1=0$
To factorise,we split the middle term $-20x$ into $-10x - 10x$:
$100 x^{2}-10 x-10 x+1=0$
Group the terms:
$10 x(10 x-1)-1(10 x-1)=0$
Factor out the common term $(10 x-1)$:
$(10 x-1)(10 x-1)=0$
$(10 x-1)^{2}=0$
For the roots,set the factor to zero:
$10 x-1=0$
$10 x=1$
$x=\frac{1}{10}$
Since the factor is squared,the roots are equal: $x = \frac{1}{10}, \frac{1}{10}$.

(A) Let the number of John's marbles be $x$.
Therefore,the number of Jivanti's marbles $= 45 - x$.
After losing $5$ marbles each:
Number of John's marbles $= x - 5$.
Number of Jivanti's marbles $= 45 - x - 5 = 40 - x$.
It is given that the product of their remaining marbles is $124$.
So,$(x - 5)(40 - x) = 124$.
Expanding the equation: $40x - x^2 - 200 + 5x = 124$.
Rearranging terms: $-x^2 + 45x - 200 = 124$.
$x^2 - 45x + 324 = 0$.
Factoring the quadratic equation: $x^2 - 36x - 9x + 324 = 0$.
$x(x - 36) - 9(x - 36) = 0$.
$(x - 36)(x - 9) = 0$.
Thus,$x = 36$ or $x = 9$.
If John had $36$ marbles,Jivanti had $45 - 36 = 9$ marbles.
If John had $9$ marbles,Jivanti had $45 - 9 = 36$ marbles.

(A) Let the number of toys produced in a day be $x$.
The cost of production of each toy (in rupees) is $(55 - x)$.
The total cost of production is given by the product of the number of toys and the cost per toy:
$x(55 - x) = 750$
Expanding the equation:
$55x - x^2 = 750$
Rearranging into the standard quadratic form $ax^2 + bx + c = 0$:
$x^2 - 55x + 750 = 0$
To solve this,we factorize the quadratic equation by splitting the middle term:
$x^2 - 25x - 30x + 750 = 0$
Grouping the terms:
$x(x - 25) - 30(x - 25) = 0$
$(x - 25)(x - 30) = 0$
Setting each factor to zero:
$x - 25 = 0 \Rightarrow x = 25$
$x - 30 = 0 \Rightarrow x = 30$
Therefore,the number of toys produced on that day was either $25$ or $30$.

(N/A) Let the first number be $x$. Then the second number is $27-x$.
Given that their product is $182$,we have:
$x(27-x) = 182$
Expanding the equation:
$27x - x^2 = 182$
Rearranging into the standard quadratic form $ax^2 + bx + c = 0$:
$x^2 - 27x + 182 = 0$
Factoring the quadratic equation:
$x^2 - 13x - 14x + 182 = 0$
$x(x - 13) - 14(x - 13) = 0$
$(x - 13)(x - 14) = 0$
Setting each factor to zero:
$x - 13 = 0 \Rightarrow x = 13$
$x - 14 = 0 \Rightarrow x = 14$
If the first number is $13$,the second number is $27 - 13 = 14$.
If the first number is $14$,the second number is $27 - 14 = 13$.
Therefore,the two numbers are $13$ and $14$.

(A) Let the consecutive positive integers be $x$ and $x+1$.
Given that $x^{2} + (x+1)^{2} = 365$.
Expanding the equation: $x^{2} + x^{2} + 2x + 1 = 365$.
Simplifying: $2x^{2} + 2x + 1 = 365$.
$2x^{2} + 2x - 364 = 0$.
Dividing by $2$: $x^{2} + x - 182 = 0$.
Factoring the quadratic equation: $x^{2} + 14x - 13x - 182 = 0$.
$x(x + 14) - 13(x + 14) = 0$.
$(x + 14)(x - 13) = 0$.
This gives $x = -14$ or $x = 13$.
Since the integers must be positive,we take $x = 13$.
Therefore,the next integer is $x + 1 = 13 + 1 = 14$.
The two consecutive positive integers are $13$ and $14$.

(A) Let the base of the right triangle be $x \, cm$.
Its altitude is $(x - 7) \, cm$.
According to the Pythagoras theorem:
$\text{Base}^2 + \text{Altitude}^2 = \text{Hypotenuse}^2$
$x^2 + (x - 7)^2 = 13^2$
$x^2 + x^2 - 14x + 49 = 169$
$2x^2 - 14x - 120 = 0$
Dividing by $2$:
$x^2 - 7x - 60 = 0$
Factoring the quadratic equation:
$x^2 - 12x + 5x - 60 = 0$
$x(x - 12) + 5(x - 12) = 0$
$(x - 12)(x + 5) = 0$
So,$x = 12$ or $x = -5$.
Since the length of a side cannot be negative,we take $x = 12$.
Therefore,the base is $12 \, cm$ and the altitude is $12 - 7 = 5 \, cm$.

(N/A) Let the number of articles produced be $x$.
Therefore,the cost of production of each article $= (2x + 3)$ rupees.
It is given that the total cost of production is ₹ $90$.
Thus,$x(2x + 3) = 90$.
Expanding the equation: $2x^2 + 3x - 90 = 0$.
Factoring the quadratic equation: $2x^2 + 15x - 12x - 90 = 0$.
$x(2x + 15) - 6(2x + 15) = 0$.
$(2x + 15)(x - 6) = 0$.
This gives $x = -15/2$ or $x = 6$.
Since the number of articles produced must be a positive integer,we have $x = 6$.
Therefore,the number of articles produced is $6$,and the cost of each article is $2(6) + 3 = 15$ rupees.

(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 $\frac{3-\sqrt{19}}{5}$.

(NONE) Given the quadratic equation: $4 x^{2}+3 x+5=0$.
To solve by completing the square,we first divide the entire equation by $4$:
$x^{2} + \frac{3}{4}x + \frac{5}{4} = 0$.
Now,we rewrite the expression by completing the square:
$(x + \frac{3}{8})^{2} - (\frac{3}{8})^{2} + \frac{5}{4} = 0$.
$(x + \frac{3}{8})^{2} - \frac{9}{64} + \frac{80}{64} = 0$.
$(x + \frac{3}{8})^{2} + \frac{71}{64} = 0$.
$(x + \frac{3}{8})^{2} = -\frac{71}{64}$.
Since the square of any real number cannot be negative,there is no real value of $x$ that satisfies this equation.
Therefore,the equation $4 x^{2}+3 x+5=0$ has no real roots.

(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 $\frac{2}{3}$.

(NONE) The given quadratic equation is $x^{2}+4x+5=0$.
Comparing this with the standard form $ax^{2}+bx+c=0$,we get $a=1$,$b=4$,and $c=5$.
The discriminant $D$ is given by $D = b^{2}-4ac$.
Substituting the values,$D = (4)^{2} - 4(1)(5) = 16 - 20 = -4$.
Since the discriminant $D < 0$,the square root of the discriminant $\sqrt{D}$ does not yield a real number.
Therefore,the given quadratic equation has no real roots.

(A) 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 $a = 2, b = -2\sqrt{2}, c = 1$.
First,calculate the discriminant $D = b^2 - 4ac$.
$D = (-2\sqrt{2})^2 - 4(2)(1) = 8 - 8 = 0$.
Since $D = 0$,the equation has two equal real roots.
The quadratic formula is $x = \frac{-b \pm \sqrt{D}}{2a}$.
Substituting the values,$x = \frac{-(-2\sqrt{2}) \pm \sqrt{0}}{2(2)} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Thus,the roots are $\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$.

(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 $\frac{3 - \sqrt{3}}{3}$.

(C) Let the speed of the stream be $x\, km/h$.
Therefore,the speed of the boat upstream is $(18-x)\, km/h$ and the speed of the boat downstream is $(18+x)\, km/h$.
The time taken to go upstream is $\frac{\text{distance}}{\text{speed}} = \frac{24}{18-x}\, \text{hours}$.
Similarly,the time taken to go downstream is $\frac{24}{18+x}\, \text{hours}$.
According to the problem,the difference in time is $1\, \text{hour}$:
$\frac{24}{18-x} - \frac{24}{18+x} = 1$
Multiplying by $(18-x)(18+x)$,we get:
$24(18+x) - 24(18-x) = (18-x)(18+x)$
$432 + 24x - 432 + 24x = 324 - x^2$
$48x = 324 - x^2$
$x^2 + 48x - 324 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-48 \pm \sqrt{48^2 - 4(1)(-324)}}{2(1)}$
$x = \frac{-48 \pm \sqrt{2304 + 1296}}{2} = \frac{-48 \pm \sqrt{3600}}{2}$
$x = \frac{-48 \pm 60}{2}$
Since speed cannot be negative,we take $x = \frac{-48 + 60}{2} = \frac{12}{2} = 6\, km/h$.

(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 $x = -\frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}$.

(D) Given equation: $2x^{2} + x + 4 = 0$.
Step $1$: Divide the entire equation by $2$ to make the coefficient of $x^{2}$ equal to $1$:
$x^{2} + \frac{1}{2}x + 2 = 0$.
Step $2$: Move the constant term to the right side:
$x^{2} + \frac{1}{2}x = -2$.
Step $3$: 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:
$x^{2} + 2(x)(\frac{1}{4}) + (\frac{1}{4})^{2} = -2 + \frac{1}{16}$.
Step $4$: Simplify the equation:
$(x + \frac{1}{4})^{2} = -\frac{32}{16} + \frac{1}{16}$.
$(x + \frac{1}{4})^{2} = -\frac{31}{16}$.
Since the square of any real number cannot be negative,$(x + \frac{1}{4})^{2}$ cannot be equal to $-\frac{31}{16}$.
Therefore,there are no real roots for the given quadratic equation.

(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 $x = \frac{-1 - \sqrt{33}}{4}$.