Mix Examples - Quadratic Equations Questions in English

(A) $(i)$ True. $A$ quadratic equation is of the form $ax^{2} + bx + c = 0$ where $a \neq 0$. By the Fundamental Theorem of Algebra,a polynomial of degree $n$ has at most $n$ roots. Since a quadratic equation has degree $2$,it can have at most $2$ roots.
$(ii)$ True. For a quadratic equation $ax^{2} + bx + c = 0$,the discriminant is $D = b^{2} - 4ac$. If the coefficient of $x^{2}$ $(a)$ and the constant term $(c)$ have opposite signs,then $ac < 0$. Consequently,$-4ac > 0$. Since $b^{2} \geq 0$,it follows that $D = b^{2} - 4ac > 0$. Because the discriminant is positive,the quadratic equation must have two distinct real roots.

(A) $(i)$ True. For a quadratic equation $ax^{2} + bx + c = 0$,the discriminant is $D = b^{2} - 4ac$. Given $b = 0$,we have $D = -4ac$. If $a$ and $c$ have the same sign,then $ac > 0$,which implies $D = -4ac < 0$. Since the discriminant is negative,the equation has no real roots.
$(ii)$ False. $A$ quadratic equation has exactly two roots (which may be real and distinct,real and equal,or complex/imaginary). It does not necessarily have "at least" two roots in the context of real numbers,and it cannot have more than two roots.

(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 $x = \frac{-48 + 60}{2} = \frac{12}{2} = 6\, km/h$.

(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) Given,$AC \perp CB$,$AC = 2x \, km$,$CB = 2(x+7) \, km$ and $AB = 26 \, km$.
From the figure,we have a right-angled $\triangle ACB$ with the right angle at $C$.
In $\triangle ACB$,by the Pythagoras theorem:
$AB^2 = AC^2 + BC^2$
$(26)^2 = (2x)^2 + \{2(x+7)\}^2$
$676 = 4x^2 + 4(x^2 + 49 + 14x)$
$676 = 4x^2 + 4x^2 + 196 + 56x$
$676 = 8x^2 + 56x + 196$
$8x^2 + 56x - 480 = 0$
Dividing by $8$,we get:
$x^2 + 7x - 60 = 0$
$x^2 + 12x - 5x - 60 = 0$
$x(x + 12) - 5(x + 12) = 0$
$(x + 12)(x - 5) = 0$
$x = -12$ or $x = 5$.
Since distance cannot be negative,$x = 5$ (as $x \neq -12$).
Now,$AC = 2x = 2(5) = 10 \, km$ and $BC = 2(x+7) = 2(5+7) = 24 \, km$.
The distance covered to reach city $B$ from city $A$ via city $C$ is $AC + BC = 10 + 24 = 34 \, km$.
The distance covered to reach city $B$ from city $A$ after the construction of the highway is $AB = 26 \, km$.
Therefore,the distance saved is $34 - 26 = 8 \, km$.

(A) Let the width of the grass strip around the pond be $x\, m$.
Given,the dimensions of the rectangular lawn are $50\, m \times 40\, m$.
Therefore,the length of the pond is $(50 - 2x)\, m$ and the breadth of the pond is $(40 - 2x)\, m$.
The area of the grass surrounding the pond is given by the difference between the area of the lawn and the area of the pond.
Area of grass = (Area of lawn) - (Area of pond)
$1184 = (50 \times 40) - (50 - 2x)(40 - 2x)$
$1184 = 2000 - (2000 - 100x - 80x + 4x^{2})$
$1184 = 2000 - 2000 + 180x - 4x^{2}$
$4x^{2} - 180x + 1184 = 0$
Dividing by $4$:
$x^{2} - 45x + 296 = 0$
$x^{2} - 37x - 8x + 296 = 0$
$x(x - 37) - 8(x - 37) = 0$
$(x - 37)(x - 8) = 0$
So,$x = 37$ or $x = 8$.
If $x = 37$,the dimensions of the pond would be negative,which is not possible. Thus,$x = 8$.
Length of pond = $50 - 2(8) = 50 - 16 = 34\, m$.
Breadth of pond = $40 - 2(8) = 40 - 16 = 24\, m$.
Therefore,the length and breadth of the pond are $34\, m$ and $24\, m$ respectively.

(C) Let the speed of the stream be $v \, km/h$.
Given that,the speed of the person in still water is $5 \, km/h$.
The speed of the person downstream is $(5 + v) \, km/h$.
The speed of the person upstream is $(5 - v) \, km/h$.
Time taken to cover $40 \, km$ downstream is $t_1 = \frac{40}{5 + v} \, h$.
Time taken to cover $40 \, km$ upstream is $t_2 = \frac{40}{5 - v} \, h$.
According to the problem,the time taken upstream is thrice the time taken downstream,so $t_2 = 3 \times t_1$.
Substituting the values: $\frac{40}{5 - v} = 3 \times \frac{40}{5 + v}$.
Dividing both sides by $40$: $\frac{1}{5 - v} = \frac{3}{5 + v}$.
Cross-multiplying: $5 + v = 3(5 - v)$.
$5 + v = 15 - 3v$.
$4v = 10$.
$v = 2.5 \, km/h$.
Thus,the speed of the stream is $2.5 \, km/h$.

(D) Let the two-digit number be represented as $10x + y$,where $x$ is the tens digit and $y$ is the units digit.
According to the problem,the sum of the digits is equal to the product of the digits:
$x + y = xy$
Rearranging the equation:
$xy - x - y = 0$
Adding $1$ to both sides to factorize:
$xy - x - y + 1 = 1$
$(x - 1)(y - 1) = 1$
Since $x$ and $y$ are digits ($1$ to $9$ for $x$,$0$ to $9$ for $y$),the only integer solution for $(x - 1)(y - 1) = 1$ is:
$x - 1 = 1 \implies x = 2$
$y - 1 = 1 \implies y = 2$
Thus,the number is $10(2) + 2 = 22$.

(A) In $\Delta ABC$,since $m\angle B = 90^{\circ}$,by the Pythagorean theorem:
$AC^2 = AB^2 + BC^2$
$(3x - 1)^2 = (2x + 3)^2 + (x + 2)^2$
$9x^2 - 6x + 1 = (4x^2 + 12x + 9) + (x^2 + 4x + 4)$
$9x^2 - 6x + 1 = 5x^2 + 16x + 13$
$4x^2 - 22x - 12 = 0$
Dividing by $2$:
$2x^2 - 11x - 6 = 0$
$2x^2 - 12x + x - 6 = 0$
$2x(x - 6) + 1(x - 6) = 0$
$(x - 6)(2x + 1) = 0$
Thus,$x = 6$ or $x = -\frac{1}{2}$.
Since side lengths must be positive,$x = -\frac{1}{2}$ is rejected because it results in negative side lengths.
Therefore,$x = 6$.

(A) In a rectangle,the diagonals are equal in length,so $XZ = YW$. Given $XZ + YW = 26$,we have $2XZ = 26$,which implies $XZ = 13$.
In the right-angled triangle $\triangle XYZ$,by the Pythagorean theorem,$XY^2 + YZ^2 = XZ^2 = 13^2 = 169$.
We are given $XY + YZ = 17$. Squaring both sides,we get $(XY + YZ)^2 = 17^2 = 289$.
Expanding this,$XY^2 + YZ^2 + 2(XY \cdot YZ) = 289$.
Substituting $XY^2 + YZ^2 = 169$,we get $169 + 2(XY \cdot YZ) = 289$,so $2(XY \cdot YZ) = 120$,which means $XY \cdot YZ = 60$.
We need two numbers that add to $17$ and multiply to $60$. These are $12$ and $5$.
Since $XY > YZ$,we have $XY = 12$ and $YZ = 5$.

(B) Since $\Delta ABC$ is a right-angled triangle with $m \angle A = 90^{\circ}$,by the Pythagorean theorem,we have:
$AB^2 + AC^2 = BC^2$
Substituting the given expressions:
$(3x - 2)^2 + (5x + 4)^2 = (6x + 2)^2$
Expanding the squares:
$(9x^2 - 12x + 4) + (25x^2 + 40x + 16) = (36x^2 + 24x + 4)$
Combining like terms on the left side:
$34x^2 + 28x + 20 = 36x^2 + 24x + 4$
Rearranging the terms to one side:
$36x^2 - 34x^2 + 24x - 28x + 4 - 20 = 0$
$2x^2 - 4x - 16 = 0$
Dividing the entire equation by $2$:
$x^2 - 2x - 8 = 0$
Factoring the quadratic equation:
$(x - 4)(x + 2) = 0$
This gives $x = 4$ or $x = -2$.
Since lengths must be positive,$3x - 2 > 0$ implies $x > 2/3$. Thus,$x = 4$ is the only valid solution.

(A) Let the sides of the rectangle be $AB = x$ and $BC = y$.
The perimeter of a rectangle is given by $2(x + y) = 28$,which simplifies to $x + y = 14$.
In a rectangle,the diagonal $AC$ forms a right-angled triangle $ABC$ with sides $AB$ and $BC$.
By the Pythagorean theorem,$x^2 + y^2 = AC^2 = 10^2 = 100$.
We have the system of equations:
$1) x + y = 14$
$2) x^2 + y^2 = 100$
From $(1)$,$y = 14 - x$. Substituting this into $(2)$:
$x^2 + (14 - x)^2 = 100$
$x^2 + 196 - 28x + x^2 = 100$
$2x^2 - 28x + 96 = 0$
$x^2 - 14x + 48 = 0$
$(x - 6)(x - 8) = 0$
So,$x = 6$ or $x = 8$.
If $x = 6$,then $y = 14 - 6 = 8$. Since $AB < BC$ $(6 < 8)$,this is a valid solution.
If $x = 8$,then $y = 14 - 8 = 6$. This contradicts $AB < BC$.
Therefore,the sides are $AB = 6$ and $BC = 8$.

(A) In rectangle $HIJK$,$HJ$ is the diagonal and $HI$ is one of the sides.
Let the sides be $HI = l = 5$ and $IJ = w$. The diagonal $HJ$ forms a right-angled triangle $\triangle HIJ$ with sides $HI$ and $IJ$.
By the Pythagorean theorem,$HI^2 + IJ^2 = HJ^2$.
$5^2 + IJ^2 = 13^2$.
$25 + IJ^2 = 169$.
$IJ^2 = 169 - 25 = 144$.
$IJ = \sqrt{144} = 12$.
Now,the perimeter of the rectangle is given by $P = 2(l + w)$.
$P = 2(5 + 12) = 2(17) = 34$.

(D) Let the speed of the motorboat in still water be $x \, km/hr$.
Therefore,the speed of the motorboat going downstream is $(x+3) \, km/hr$ and its speed going upstream is $(x-3) \, km/hr$.
Hence,the time taken to cover $12 \, km$ downstream is $\frac{12}{x+3}$ hours and the time taken to cover the same distance,i.e.,$12 \, km$ to come back upstream is $\frac{12}{x-3}$ hours.
According to the problem,the total time taken is $3$ hours.
$\frac{12}{x+3} + \frac{12}{x-3} = 3$
$12(x-3) + 12(x+3) = 3(x+3)(x-3)$
$12x - 36 + 12x + 36 = 3(x^2 - 9)$
$24x = 3x^2 - 27$
$3x^2 - 24x - 27 = 0$
Dividing by $3$,we get $x^2 - 8x - 9 = 0$.
$(x-9)(x+1) = 0$
$x = 9$ or $x = -1$.
Since the speed cannot be negative,$x = 9 \, km/hr$.

(D) Let the shorter side among the sides containing the right angle (i.e.,base) be $x$ metres.
Then,the other side containing the right angle (i.e.,altitude) is $(x+7)$ metres and the hypotenuse is $(2x+1)$ metres.
By Pythagoras' theorem,$(\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Altitude})^2$.
Therefore,$(2x+1)^2 = x^2 + (x+7)^2$.
$4x^2 + 4x + 1 = x^2 + x^2 + 14x + 49$.
$4x^2 + 4x + 1 = 2x^2 + 14x + 49$.
$2x^2 - 10x - 48 = 0$.
Dividing by $2$,we get $x^2 - 5x - 24 = 0$.
$(x-8)(x+3) = 0$.
So,$x=8$ or $x=-3$.
Since the side of a triangle cannot be negative,$x=8$.
Thus,the base is $8\,m$,the altitude is $8+7=15\,m$,and the hypotenuse is $2(8)+1=17\,m$.
The sides of the garden are $8\,m, 15\,m$ and $17\,m$.

(A) Let the two sides forming the right angle be $x \, cm$ and $(x + 7) \, cm$.
By the Pythagorean theorem,the hypotenuse is $\sqrt{x^2 + (x + 7)^2} = \sqrt{x^2 + x^2 + 14x + 49} = \sqrt{2x^2 + 14x + 49}$.
The perimeter is given as $30 \, cm$,so $x + (x + 7) + \sqrt{2x^2 + 14x + 49} = 30$.
$2x + 7 + \sqrt{2x^2 + 14x + 49} = 30$.
$\sqrt{2x^2 + 14x + 49} = 23 - 2x$.
Squaring both sides: $2x^2 + 14x + 49 = (23 - 2x)^2$.
$2x^2 + 14x + 49 = 529 - 92x + 4x^2$.
$2x^2 - 106x + 480 = 0$.
$x^2 - 53x + 240 = 0$.
$(x - 5)(x - 48) = 0$.
Since $x$ cannot be $48$ (as the perimeter is $30$),$x = 5$.
The sides are $5 \, cm$,$5 + 7 = 12 \, cm$,and the hypotenuse is $\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, cm$.

(A) Let the numerator be $x$. Then the denominator is $2x - 1$. The ratio is $\frac{x}{2x - 1}$.
According to the problem,the sum of the ratio and its reciprocal is $2 \frac{4}{15} = \frac{34}{15}$.
So,$\frac{x}{2x - 1} + \frac{2x - 1}{x} = \frac{34}{15}$.
Let $y = \frac{x}{2x - 1}$. Then $y + \frac{1}{y} = \frac{34}{15}$.
$15y^2 - 34y + 15 = 0$.
$15y^2 - 25y - 9y + 15 = 0$.
$5y(3y - 5) - 3(3y - 5) = 0$.
$(5y - 3)(3y - 5) = 0$.
So,$y = 3/5$ or $y = 5/3$.
If $y = 3/5$,then $\frac{x}{2x - 1} = \frac{3}{5} \implies 5x = 6x - 3 \implies x = 3$. The ratio is $\frac{3}{2(3) - 1} = \frac{3}{5}$.
If $y = 5/3$,then $\frac{x}{2x - 1} = \frac{5}{3} \implies 3x = 10x - 5 \implies 7x = 5 \implies x = 5/7$. The ratio is $\frac{5/7}{2(5/7) - 1} = \frac{5/7}{3/7} = \frac{5}{3}$.

(A) Let the cost price of the calculator be $x$ rupees.
Given that the profit percentage is equal to the cost price,the profit percentage is $x\%$.
Profit = $\text{Cost Price} \times \frac{\text{Profit Percentage}}{100} = x \times \frac{x}{100} = \frac{x^2}{100}$.
Selling Price = $\text{Cost Price} + \text{Profit} = x + \frac{x^2}{100}$.
Given that the selling price is Rs. $56$,we have the equation: $x + \frac{x^2}{100} = 56$.
Multiplying by $100$,we get $100x + x^2 = 5600$,which simplifies to $x^2 + 100x - 5600 = 0$.
Factoring the quadratic equation: $x^2 + 140x - 40x - 5600 = 0$.
$x(x + 140) - 40(x + 140) = 0$.
$(x - 40)(x + 140) = 0$.
Since the cost price cannot be negative,$x = 40$.
Therefore,the cost price of the calculator is Rs. $40$.

(C) Let the two numbers be $x$ and $y$.
Given: $x + y = 15$ (Equation $1$).
Given: $\frac{1}{x} + \frac{1}{y} = \frac{3}{10}$ (Equation $2$).
From Equation $2$,we get $\frac{x + y}{xy} = \frac{3}{10}$.
Substituting the value of $x + y$ from Equation $1$ into this expression: $\frac{15}{xy} = \frac{3}{10}$.
Solving for $xy$: $3xy = 150$,so $xy = 50$.
Now we have $x + y = 15$ and $xy = 50$.
These are the roots of the quadratic equation $t^2 - (x + y)t + xy = 0$.
Substituting the values: $t^2 - 15t + 50 = 0$.
Factoring the quadratic: $(t - 5)(t - 10) = 0$.
Thus,$t = 5$ or $t = 10$.
The two numbers are $5$ and $10$.

(D) Let the speed of the current be $x\,km/h$.
Speed downstream = $(9 + x)\,km/h$.
Speed upstream = $(9 - x)\,km/h$.
Time taken downstream = $\frac{12}{9 + x}$.
Time taken upstream = $\frac{12}{9 - x}$.
Total time = $3\,hours$,so $\frac{12}{9 + x} + \frac{12}{9 - x} = 3$.
Dividing by $3$: $\frac{4}{9 + x} + \frac{4}{9 - x} = 1$.
$4(9 - x) + 4(9 + x) = (9 + x)(9 - x)$.
$36 - 4x + 36 + 4x = 81 - x^2$.
$72 = 81 - x^2$.
$x^2 = 81 - 72 = 9$.
$x = 3\,km/h$ (since speed cannot be negative).

(B) Let the tens digit be $x$ and the units digit be $y$. The number is $10x + y$.
Given that the product of the digits is $xy = 15$.
When the digits are interchanged,the new number is $10y + x$.
According to the problem,the new number is $18$ more than the original number:
$(10y + x) - (10x + y) = 18$
$9y - 9x = 18$
$y - x = 2$,which implies $y = x + 2$.
Substitute $y = x + 2$ into the product equation $xy = 15$:
$x(x + 2) = 15$
$x^2 + 2x - 15 = 0$
$(x + 5)(x - 3) = 0$
Since $x$ must be a positive digit,$x = 3$.
Then $y = 3 + 2 = 5$.
The original number is $10(3) + 5 = 35$.

(C) Let the side of the smaller square be $x\,m$ and the side of the bigger square be $y\,m$.
The perimeter of the smaller square is $4x$ and the perimeter of the bigger square is $4y$.
According to the problem,$4y = 4x + 12$,which simplifies to $y = x + 3$.
The area of the smaller square is $x^2$ and the area of the bigger square is $y^2$.
According to the problem,$3x^2 = y^2 + 11$.
Substitute $y = x + 3$ into the second equation:
$3x^2 = (x + 3)^2 + 11$
$3x^2 = x^2 + 6x + 9 + 11$
$2x^2 - 6x - 20 = 0$
$x^2 - 3x - 10 = 0$
$(x - 5)(x + 2) = 0$
Since the side length cannot be negative,$x = 5$.
Therefore,the side of the bigger square is $y = x + 3 = 5 + 3 = 8\,m$.

(D) Let the two-digit number be represented as $10x + y$,where $x$ is the tens digit and $y$ is the units digit.
According to the problem,the sum of the digits is equal to the product of the digits:
$x + y = xy$
Rearranging the equation to solve for $y$:
$x = xy - y$
$x = y(x - 1)$
$y = \frac{x}{x - 1}$
Since $x$ and $y$ must be integers between $1$ and $9$ (with $x \neq 0$ for a two-digit number):
If $x = 2$,then $y = \frac{2}{2 - 1} = 2$.
The number is $10(2) + 2 = 22$.
Check: Sum of digits $= 2 + 2 = 4$; Product of digits $= 2 \times 2 = 4$. Both are equal.
Thus,the number is $22$.

(B) First,solve the quadratic equation $x^{2}-5x+6=0$.
Factorizing the equation: $(x-2)(x-3)=0$.
Thus,the roots are $x=2$ or $x=3$.
Case $1$: If $x=2$ is a root of $x^{2}+3x+c=0$,then $(2)^{2}+3(2)+c=0$.
$4+6+c=0 \implies 10+c=0 \implies c=-10$.
Case $2$: If $x=3$ is a root of $x^{2}+3x+c=0$,then $(3)^{2}+3(3)+c=0$.
$9+9+c=0 \implies 18+c=0 \implies c=-18$.
Therefore,the possible values for $c$ are $-10$ or $-18$.

(D) triplet $(a, b, c)$ is a Pythagorean triplet if it satisfies the condition $a^2 + b^2 = c^2$,where $c$ is the largest number.
For option $A$: $7^2 + 24^2 = 49 + 576 = 625 = 25^2$. This is a Pythagorean triplet.
For option $B$: $20^2 + 21^2 = 400 + 441 = 841 = 29^2$. This is a Pythagorean triplet.
For option $C$: $11^2 + 60^2 = 121 + 3600 = 3721 = 61^2$. This is a Pythagorean triplet.
For option $D$: $13^2 + 35^2 = 169 + 1225 = 1394$,while $37^2 = 1369$. Since $1394 \neq 1369$,this is not a Pythagorean triplet.