Textbook - Quadratic Equations Questions in English

(D) Given quadratic equation: $4 x^{2}+4 \sqrt{3} x+3=0$.
Comparing this equation with the standard form $a x^{2}+b x+c=0$,we get $a=4, b=4 \sqrt{3}, c=3$.
The quadratic formula is given by $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$.
First,calculate the discriminant $D = b^{2}-4 a c = (4 \sqrt{3})^{2} - 4(4)(3) = 48 - 48 = 0$.
Since $D=0$,the roots are real and equal.
Substituting the values into the formula:
$x = \frac{-4 \sqrt{3} \pm \sqrt{0}}{2(4)}$
$x = \frac{-4 \sqrt{3}}{8}$
$x = \frac{-\sqrt{3}}{2}$.
Thus,the roots are $x = \frac{-\sqrt{3}}{2}, \frac{-\sqrt{3}}{2}$.

(NONE) The given quadratic equation is $2x^{2} + x + 4 = 0$.
On comparing this equation with the standard form $ax^{2} + bx + c = 0$,we obtain:
$a = 2, b = 1, c = 4$.
By using the quadratic formula,$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$,we substitute the values:
$x = \frac{-1 \pm \sqrt{(1)^{2} - 4(2)(4)}}{2(2)}$
$x = \frac{-1 \pm \sqrt{1 - 32}}{4}$
$x = \frac{-1 \pm \sqrt{-31}}{4}$
Since the discriminant $D = b^{2} - 4ac = -31$,which is less than $0$,the square root of a negative number is not a real number.
Therefore,there are no real roots for the given quadratic equation.

(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 $x = \frac{3 - \sqrt{13}}{2}$.

(A) Given equation: $\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}$
Taking the common denominator on the left side:
$\frac{(x-7)-(x+4)}{(x+4)(x-7)}=\frac{11}{30}$
Simplifying the numerator:
$\frac{x-7-x-4}{x^2-7x+4x-28}=\frac{11}{30}$
$\frac{-11}{x^2-3x-28}=\frac{11}{30}$
Dividing both sides by $11$:
$\frac{-1}{x^2-3x-28}=\frac{1}{30}$
Cross-multiplying:
$x^2-3x-28 = -30$
$x^2-3x-28+30 = 0$
$x^2-3x+2 = 0$
Factoring the quadratic equation:
$x^2-2x-x+2 = 0$
$x(x-2)-1(x-2) = 0$
$(x-2)(x-1) = 0$
Therefore,the roots are $x = 1$ or $x = 2$.

(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 $30-13 = 17$.

(C) Let the shorter side of the rectangle be $x\, m$.
Then,the longer side of the rectangle is $(x+30)\, m$.
According to the Pythagorean theorem,the diagonal of the rectangle is $\sqrt{x^2 + (x+30)^2}$.
It is given that the diagonal is $60\, m$ more than the shorter side,so $\sqrt{x^2 + (x+30)^2} = x + 60$.
Squaring both sides,we get $x^2 + (x+30)^2 = (x+60)^2$.
Expanding the terms: $x^2 + x^2 + 60x + 900 = x^2 + 120x + 3600$.
Simplifying the equation: $x^2 - 60x - 2700 = 0$.
Factoring the quadratic equation: $x^2 - 90x + 30x - 2700 = 0$,which gives $x(x-90) + 30(x-90) = 0$.
So,$(x-90)(x+30) = 0$.
This gives $x = 90$ or $x = -30$.
Since the length cannot be negative,the shorter side is $90\, m$.
The longer side is $x + 30 = 90 + 30 = 120\, m$.

(A) Let the larger number be $x$ and the smaller number be $y$.
According to the given conditions:
$x^2 - y^2 = 180$ --- $(1)$
$y^2 = 8x$ --- $(2)$
Substituting equation $(2)$ into equation $(1)$:
$x^2 - 8x = 180$
$x^2 - 8x - 180 = 0$
Factoring the quadratic equation:
$x^2 - 18x + 10x - 180 = 0$
$x(x - 18) + 10(x - 18) = 0$
$(x - 18)(x + 10) = 0$
So,$x = 18$ or $x = -10$.
Since $y^2 = 8x$,if $x = -10$,then $y^2 = -80$,which is impossible for real numbers. Thus,$x = 18$.
Now,find $y$:
$y^2 = 8(18) = 144$
$y = \pm \sqrt{144} = \pm 12$.
Therefore,the two numbers are $(18, 12)$ or $(18, -12)$.

(A) Let the uniform speed of the train be $x\, km/h$.
Time taken to cover $360\, km$ at speed $x$ is $t_1 = \frac{360}{x}\, h$.
If the speed is increased by $5\, km/h$,the new speed is $(x + 5)\, km/h$.
Time taken at the new speed is $t_2 = \frac{360}{x + 5}\, h$.
According to the problem,the difference in time is $1\, hour$:
$\frac{360}{x} - \frac{360}{x + 5} = 1$
$360 \left( \frac{1}{x} - \frac{1}{x + 5} \right) = 1$
$360 \left( \frac{x + 5 - x}{x(x + 5)} \right) = 1$
$360 \left( \frac{5}{x^2 + 5x} \right) = 1$
$1800 = x^2 + 5x$
$x^2 + 5x - 1800 = 0$
Factoring the quadratic equation:
$x^2 + 45x - 40x - 1800 = 0$
$x(x + 45) - 40(x + 45) = 0$
$(x + 45)(x - 40) = 0$
This gives $x = -45$ or $x = 40$.
Since speed cannot be negative,the speed of the train is $40\, km/h$.

(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 the average speed of the passenger train be $x\, km/h$.
Average speed of the express train $= (x + 11)\, km/h$.
Time taken by the passenger train to cover $132\, km = \frac{132}{x}\, hours$.
Time taken by the express train to cover $132\, km = \frac{132}{x + 11}\, hours$.
According to the problem,the express train takes $1\, hour$ less than the passenger train:
$\frac{132}{x} - \frac{132}{x + 11} = 1$
$132 \left[ \frac{x + 11 - x}{x(x + 11)} \right] = 1$
$\frac{132 \times 11}{x^2 + 11x} = 1$
$x^2 + 11x = 1452$
$x^2 + 11x - 1452 = 0$
Solving the quadratic equation by splitting the middle term:
$x^2 + 44x - 33x - 1452 = 0$
$x(x + 44) - 33(x + 44) = 0$
$(x + 44)(x - 33) = 0$
$x = -44$ or $x = 33$.
Since speed cannot be negative,$x = 33\, km/h$.
Therefore,the speed of the passenger train is $33\, km/h$ and the speed of the express train is $33 + 11 = 44\, km/h$.

(A) Let the sides of the two squares be $x \, m$ and $y \, m$. Therefore,their perimeters are $4x$ and $4y$ respectively,and their areas are $x^2$ and $y^2$ respectively.
It is given that the difference of their perimeters is $24 \, m$:
$4x - 4y = 24$
$x - y = 6$
$x = y + 6$
It is also given that the sum of their areas is $468 \, m^2$:
$x^2 + y^2 = 468$
Substituting $x = y + 6$ into the area equation:
$(y + 6)^2 + y^2 = 468$
$y^2 + 12y + 36 + y^2 = 468$
$2y^2 + 12y - 432 = 0$
Dividing by $2$:
$y^2 + 6y - 216 = 0$
Factoring the quadratic equation:
$y^2 + 18y - 12y - 216 = 0$
$y(y + 18) - 12(y + 18) = 0$
$(y + 18)(y - 12) = 0$
This gives $y = -18$ or $y = 12$. Since the side of a square cannot be negative,we take $y = 12 \, m$.
Then,$x = 12 + 6 = 18 \, m$.
Thus,the sides of the two squares are $18 \, m$ and $12 \, m$.

(D) The given quadratic equation is in the form $ax^{2}+bx+c=0$,where $a=2$,$b=-4$,and $c=3$.
The discriminant $D$ is given by the formula $D = b^{2}-4ac$.
Substituting the values,we get:
$D = (-4)^{2} - (4 \times 2 \times 3)$
$D = 16 - 24$
$D = -8$
Since the discriminant $D < 0$,the quadratic equation has no real roots.

(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 $A$.

(D) Given equation: $3x^{2}-2x+\frac{1}{3}=0$.
Comparing with $ax^{2}+bx+c=0$,we get $a=3, b=-2, c=\frac{1}{3}$.
Discriminant $D = b^{2}-4ac = (-2)^{2}-4(3)(\frac{1}{3}) = 4-4 = 0$.
Since $D=0$,the equation has two equal real roots.
The roots are given by $x = \frac{-b}{2a}$.
$x = \frac{-(-2)}{2(3)} = \frac{2}{6} = \frac{1}{3}$.
Thus,the roots are $\frac{1}{3}, \frac{1}{3}$.

(C) For a quadratic equation $ax^2 + bx + c = 0$,the discriminant $D$ is given by $D = b^2 - 4ac$.
$(A)$ If $D > 0$,there are two distinct real roots.
$(B)$ If $D = 0$,there are two equal real roots.
$(C)$ If $D < 0$,there are no real roots.
Given the equation $2x^2 - 3x + 5 = 0$,we compare it with $ax^2 + bx + c = 0$ to get:
$a = 2$,$b = -3$,$c = 5$.
Calculating the discriminant:
$D = b^2 - 4ac = (-3)^2 - 4(2)(5)$
$D = 9 - 40 = -31$.
Since $D < 0$,the given quadratic equation has no real roots.

(N/A) We know that for a quadratic equation $ax^2 + bx + c = 0$,the discriminant is $D = b^2 - 4ac$.
$(A)$ If $D > 0$,there are two distinct real roots.
$(B)$ If $D = 0$,there are two equal real roots.
$(C)$ If $D < 0$,there are no real roots.
Given equation: $3x^2 - 4\sqrt{3}x + 4 = 0$.
Comparing this with $ax^2 + bx + c = 0$,we get $a = 3$,$b = -4\sqrt{3}$,and $c = 4$.
Discriminant $D = b^2 - 4ac = (-4\sqrt{3})^2 - 4(3)(4)$.
$D = 48 - 48 = 0$.
Since $D = 0$,the equation has two equal real roots.
The roots are given by $x = \frac{-b}{2a}$.
$x = \frac{-(-4\sqrt{3})}{2(3)} = \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$.
Thus,the roots are $\frac{2}{\sqrt{3}}$ and $\frac{2}{\sqrt{3}}$.

(A) For a quadratic equation $ax^2 + bx + c = 0$,the discriminant $D$ is given by $D = b^2 - 4ac$.
$(A)$ If $D > 0$,there are two distinct real roots.
$(B)$ If $D = 0$,there are two equal real roots.
$(C)$ If $D < 0$,there are no real roots.
Given the equation $2x^2 - 6x + 3 = 0$,comparing it with $ax^2 + bx + c = 0$,we get:
$a = 2, b = -6, c = 3$.
Discriminant $D = (-6)^2 - 4(2)(3) = 36 - 24 = 12$.
Since $D > 0$,the equation has two distinct real roots.
Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:
$x = \frac{-(-6) \pm \sqrt{12}}{2(2)} = \frac{6 \pm 2\sqrt{3}}{4}$.
Dividing the numerator and denominator by $2$,we get:
$x = \frac{3 \pm \sqrt{3}}{2}$.
Thus,the roots are $\frac{3 + \sqrt{3}}{2}$ and $\frac{3 - \sqrt{3}}{2}$.

(D) We know that if a quadratic equation $a x^{2}+b x+c=0$ has two equal roots,its discriminant $D = (b^{2}-4 a c)$ must be equal to $0$.
Given equation: $2 x^{2}+k x+3=0$
Comparing this with the standard form $a x^{2}+b x+c=0$,we get:
$a=2, b=k, c=3$
The discriminant is given by:
$D = b^{2}-4 a c$
$D = (k)^{2}-4(2)(3)$
$D = k^{2}-24$
For the equation to have two equal roots,we set the discriminant to $0$:
$k^{2}-24 = 0$
$k^{2} = 24$
$k = \pm \sqrt{24}$
$k = \pm 2 \sqrt{6}$

(K=6) Given equation: $k x(x-2)+6=0$
Expanding the equation: $k x^{2}-2 k x+6=0$
Comparing this with the standard quadratic form $a x^{2}+b x+c=0$,we get:
$a=k, b=-2 k, c=6$
For a quadratic equation to have two equal roots,the discriminant $D$ must be equal to $0$:
$D = b^{2}-4 a c = 0$
Substituting the values:
$(-2 k)^{2}-4(k)(6) = 0$
$4 k^{2}-24 k = 0$
Factoring out $4k$:
$4 k(k-6) = 0$
This gives two possibilities: $4 k=0$ or $k-6=0$,which implies $k=0$ or $k=6$.
However,if $k=0$,the equation becomes $6=0$,which is not a quadratic equation. Thus,$k$ cannot be $0$.
Therefore,the only valid value is $k=6$.

(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 $2(20) = 40 \, m$.

(D) Let the age of one friend be $x$ years.
Age of the other friend will be $(20-x)$ years.
Four years ago,the age of the first friend was $(x-4)$ years.
The age of the second friend was $(20-x-4) = (16-x)$ years.
According to the given condition:
$(x-4)(16-x) = 48$
$16x - x^2 - 64 + 4x = 48$
$-x^2 + 20x - 64 = 48$
$-x^2 + 20x - 112 = 0$
$x^2 - 20x + 112 = 0$
Comparing this equation with the standard form $ax^2 + bx + c = 0$,we get $a=1, b=-20, c=112$.
The discriminant $D = b^2 - 4ac = (-20)^2 - 4(1)(112) = 400 - 448 = -48$.
Since the discriminant $D < 0$,there are no real roots for this quadratic equation.
Therefore,the given situation is not possible.

(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 $b = 40 - 20 = 20\, m$.