(i)\((x + 1)^2 = 2(x - 3)\)
Using identity \((a + b)^2 = a^2 + 2ab + b^2\)
=>\(x^2 + 1 + 2x = 2x - 6\)
=>\(x^2 + 7 = 0\)
Hence it is a quadratic equation since the degree is degree 2.
(ii)\(x^2 - 2x = (?2) (3 ? x)\)
=>\(x^2 - 2x = -6 + 2x\)
=>\((x^2 - 2x - 2x + 6 = 0\)
=>\(x^2 - 4x + 6 = 0\)
Hence it is a quadratic equation with degree 2.
(iii)\((x ? 2) (x + 1) = (x ? 1) (x + 3)\)
=>\(x^2 + x - 2x - 2 = x^2 + 3x - x - 3\)
=>\(x^2 + x - 2x - 2 - x^2 + 3x - x - 3 = 0\)
=>\((x^2 - 2x - 2 -3x + x + 3 = 0\)
=>\(-3x + 1 = 0\)
Hence it is a not quadratic equation since the degree is 1.
(iv)\((x ? 3) (2x + 1) = x (x + 5)\)
=>\(2x^2 + x - 6x - 3 = x^2 + 5x\)
=>\((2x^2 + x - 6x - 3 - x^2 + 5x = 0\)
=>\(x^2 - 10x - 3 = 0\)
Hence it is a quadratic equation since the degree ish degree 2.
(v)\((2x ? 1) (x ? 3) = (x + 5) (x ? 1)\)
=>\(2x^2 - x - 6x + 3 = x^2 - x + 5x - 5\)
=>\((2x^2 - 7x + 3 - x^2 + x - 5x + 5 = 0\)
=>\(x^2 - 10x - 3 = 0\)
Hence it is a quadratic equation since the degree is degree 2.
(vi)\(x^2 + 3x + 1 = (x - 2)^2\)
Using identity \((a - b)^2 = a^2 - 2ab + b^2\)
=>\(x^2 + 3x + 1 = x^2 - 4x + 4 \)
=>\((x^2 + 3x + 1 - x^2 + 4x - 4 = 0\)
=>\(7x - 3 = 0\)
Hence it is a not quadratic equation since the degree is 1
(vii)\((x + 2)^3 = 2x(x^2 - 1)\)
Using identity \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\)
=>\(x^3 + 2^3 + 3(x)(2)(x + 2) = 2x(x^2 - 1)\)
=>\(x^3 + 8 + 6x(x + 2) = 2x^2 - 2x\)
=>\(2x^3 - 2x - x^3 - 8 - 6x^2 - 12x = 0\)
=>\(x^3 - 6x^2 - 14x - 8 = 0\)
Hence it is a not quadratic equation since the degree is 3
(viii)\(x^3 - 4x^2 - x + 1 = (x - 2)^3\)
Using identity \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\)
=>\(x^3 - 4x^2 - x + 1 = x^3 - 2^3 - 3(x)(2)(x-2)\)
=>\(-4x^2 - x + 1 = -8 - 6x^2 + 12x\)
=>\(2x^2 - 13x + 9 = 0\)
Hence it is a quadratic equation since the degree is 2
(i) A rectangular plot with area 528 \(m^2\) is given.
Let breadth of rectangular plot be x metres
Length is one more than twice its breadth.
Therefore, length of rectangular plot is \(2x + 1\) metres
Area of rectangle = length × breadth
=> \(528 = x (2x + 1)\)
=> \(528 = 2x^2 + x\)
=>\( 2x^2 + x – 528 = 0\)
This is a Quadratic Equation.
(ii) Let two consecutive numbers be x and (x + 1).
It is given that \(x (x + 1) = 306\)
=>\(x^2 + x = 306\)
=>\( x^2 + x – 306 = 0\)
This is a Quadratic Equation.
(iii) Let present age of Rohan = x years
Let present age of Rohan’s mother = \(x + 26\) years
Age of Rohan after 3 years = \(x + 3\) years
Age of Rohan’s mother after 3 years = \(x + 26 + 3 = (x + 29)\) years
According to given condition:
\((x + 3) (x + 29) = 360\)
=>\(x^2 + 29x + 3x + 87 = 360\)
=>\(x^2 + 32x - 273 = 0\)
This is a Quadratic Equation.
(iv) Let speed of train be x km/h
Time taken by train to cover 480 km = 480x hours
If, speed had been 8km/h less then time taken would be \((480x?8)\) hours
According to given condition, if speed had been 8km/h less then time taken is 3 hours less.
Therefore, \(480x – 8 = 480x + 3\)
=>\( 480 (1x – 8 ? 1x) = 3\)
=>\( 480 (x – x + 8) (x) (x - 8) = 3\)
=>\( 480 × 8 = 3 (x) (x - 8)\)
=>\( 3840 = 3x^2 - 24x\)
=>\(3x^2 - 24x - 3840 = 0\)
Dividing equation by 3, we get
=>\(x^2 - 8x - 1280 = 0\)
This is a Quadratic Equation.
1. \(x^2 - 3x - 10 = 0\)
=> \(x^2 - 3x - 10 = 0\)
=> \(x^2 - 5x + 2x - 10 = 0\)
=> \(x(x - 5) + 2 (x - 5) = 0\)
=> \((x - 5) (x + 2) = 0\)
=> \(x = 5, -2\)
2. \(2x^2 + x - 6 = 0\)
=> \(2x^2 + x - 6 = 0\)
=> \(2x^2 + 4x - 3x - 6 = 0\)
=> \(2x(x + 2) - 3(x + 2) = 0\)
=> \((2x - 3) (x + 2) = 0\)
=> \(x = {{3} \over 2}, -2\)
3. \(\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0\)
=> \(\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0\)
=> \(\sqrt{2}x^2 + 2x + 5x + 5\sqrt{2} = 0\)
=> \(\sqrt{2}x^2(x + \sqrt{2}) + 5(x + \sqrt{2}) = 0\)
=> \((\sqrt{2}x + 5) (x + \sqrt{2}) = 0\)
=> \(x = {{-5} \over \sqrt{2}}, -\sqrt{2}\)
=> \(x = {{-5} \over \sqrt{2}} × {{\sqrt{2}} \over \sqrt{2}}, -\sqrt{2}\)
=> \(x = {{-5\sqrt{2}} \over 2}, -\sqrt{2}\)
4. \(2x^2 - x + {{1} \over 8} = 0\)
=> \({{16x^2 - 8x + 1} \over 8} = 0\)
=> \(16x^2 - 8x + 1 = 0\)
=> \(16x^2 - 4x - 4x + 1 = 0\)
=> \(4x(4x - 1) - 1(4x - 1) = 0\)
=> \((4x - 1) (4x - 1) = 0\)
=> \(x = {{1} \over 4}, {{1} \over 4} \)
5. \(100x^2 - 20x + 1 = 0\)
=> \(100x^2 - 20x + 1 = 0\)
=> \(100x^2 - 10x - 10x + 1 = 0\)
=> \(10x(10x - 1) - 1(10x - 1) = 0\)
=> \((10x - 1) (10x - 1) = 0\)
=> \(x = {{1} \over 10}, {{1} \over 10}\)
1. \(x^2 - 45x + 324 = 0\)
=> \(x^2 - 45x + 324 = 0\)
=> \(x^2 - 36x - 9x - 324 = 0\)
=> \(x(x - 36) - 9 (x - 36) = 0\)
=> \((x - 9) (x - 36) = 0\)
=> \(x = 9, 36\)
2. \(x^2 - 55x + 750 = 0\)
=> \(x^2 - 55x + 750 = 0\)
=> \(x^2 - 25x - 30x + 750 = 0\)
=> \(x(x - 25) - 30 (x - 25) = 0\)
=> \((x - 25) (x - 30) = 0\)
=> \(x = 25, 30\)
Ans.Let first number be x and let second number be (27 - x)
According to given condition, the product of two numbers is 182.
Therefore,
\(x(27 - x) = 182\)
=> \(27 - x^2 = 182\)
=> \(x^2 - 27x + 182 = 0\)
=> \(x^2 - 14x + 13x + 182 = 0\)
=> \(x(x - 14) - 13(x - 14) = 0\)
=> \((x - 14) (x - 13) = 0\)
=> \(x = 14, 13\)
Therefore, the first number is equal to 14 or 13
And, second number is = 27 – x = 27 – 14 = 13 or Second number = 27 – 13 = 14
Therefore, two numbers are 13 and 14.
Ans.Let first number be x and let second number be (x + 1)
According to given condition,
\(x^2 + (x + 1)^2 = 365\)
We know, \((a+b)^2 = a^2 + b^2 + 2ab\)
=> \(x^2 + x^2 + 1 + 2x = 365\)
=> \(2x^2 + 2x -364 = 0\)
Dividing equation by 2
=> \(x^2 + x - 182 = 0\)
=> \(x^2 + 14x - 13x - 182 = 0\)
=> \(x(x + 14) - 13(x + 14) = 0\)
=> \((x + 14) (x - 13) = 0\)
=> \(x = -14, 13\)
Therefore, first number = 13 (We discard -14 because it is negative number)
Second number = x + 1 = 13 + 1 = 14
Therefore, two consecutive positive integers are 13 and 14 whose sum of squares is equal to 365.
Ans.Let base of triangle be x cm and let altitude of triangle be (x ? 7) cm
It is given that hypotenuse of triangle is 13 cm
According to Pythagoras Theorem,
\(13^2 = x^2 + (x - 7)^2 \)
We know, \((a+b)^2 = a^2 + b^2 + 2ab\)
=> \(169 = x^2 + x^2 + 49 - 14x\)
=> \(169 = 2x^2 - 14x + 49 \)
=> \(2x^2 - 14x - 120 = 0\)
Dividing equation by 2
=> \(x^2 - 7x - 60 = 0\)
=> \(x^2 - 12x + 5x - 60 = 0\)
=> \(x(x - 12) + 5(x - 12) = 0\)
=> \((x - 12) (x + 5) = 0\)
=> \(x = -5, 12\)
We discard x = -5 because length of side of triangle cannot be negative.
Therefore, base of triangle = 12 cm
Altitude of triangle = (x -7) = 12 – 7 = 5 cm.
Ans. Let cost of production of each article be Rs x
We are given total cost of production on that particular day = Rs 90
Therefore, total number of articles produced that day = \({{90} \over x}\)
According to the given conditions,
(i) \(2x^2 - 7x + 3 = 0\)
First step is to make coefficient of \(x^2 = 1\)
So on dividing equation by 2
\(x^2 - {{7}\over{2}}x + {{3}\over{2}} = 0\)
On dividing the middle term of the equation by 2x:
\(({{7}\over{2}}x)({{1}\over{2}}x) = {{7}\over{4}}\)
On adding and subtracting square of \({{7}\over{4}}\) from the equation \(x^2 - {{7}\over{2}}x + {{3}\over{2}} = 0\)
\(x^2 - {{7}\over{2}}x + {{3}\over{2}} + ({{7}\over{4}})^2 - ({{7}\over{4}})^2 = 0\)
\(x^2 + ({{7}\over{4}})^2 - {{7}\over{2}}x + {{3}\over{2}} - ({{7}\over{4}})^2 = 0\)
Using identity {\((a-b)^2 = a^2 + b^2 -2ab\)}
\((x - {{7}\over{4}})^2 + {{24 - 49}\over{16}} = 0\)
\((x - {{7}\over{4}})^2 = {{49 - 24}\over{16}}\)
On taking square root both sides:
\(x - {{7}\over{4}} = {{5}\over{4}}\) and \({{-5}\over{4}}\)
\(x = {{5}\over{4}} + {{7}\over{4}} = {{12}\over{4}} = 3\) and \({{-5}\over{4}} - {{7}\over{4}} = {{2}\over{4}} = {{1}\over{2}}\)
Hence, \(x = {{1} \over {2}} , 3\)
(ii) \(2x^2 + x - 4 = 0\)
First step is to make coefficient of \(x^2 = 1\)
So on dividing equation by 2
\(x^2 + {{x}\over{2}} - 2 = 0\)
On dividing the middle term of the equation by 2x:
\(({{x}\over{2}})({{1}\over{2}}x) = {{1}\over{4}}\)
On adding and subtracting square of \({{1}\over{4}}\) from the equation \(x^2 + {{x}\over{2}} - 2 = 0\)
\(x^2 + {{x}\over{2}} - 2 + ({{1}\over{4}})^2 - ({{1}\over{4}})^2 = 0\)
\(x^2 + ({{1}\over{4}})^2 + {{x}\over{2}} - 2 - ({{1}\over{16}}) = 0\)
Using identity {\((a+b)^2 = a^2 + b^2 + 2ab\)}
\((x + {{1}\over{4}})^2 - {{33}\over{16}} = 0\)
\((x + {{1}\over{4}})^2 = {{33}\over{16}}\)
On taking square root both sides:
\(x + {{1}\over{4}} = {{\sqrt{33}}\over{4}}\) and \({{-\sqrt{33}}\over{4}}\)
\(x = {{\sqrt{33}}\over{4}} - {{1}\over{4}} = {{\sqrt{33} - 1}\over{4}}\) and \({{-\sqrt{33}}\over{4}} - {{1}\over{4}} = {{2}\over{4}} = {{-\sqrt{33} - 1}\over{4}}\)
Hence, \(x = {{\sqrt{33} - 1}\over{4}} , {{-\sqrt{33} - 1}\over{4}}\)
(iii) \(4x^2 + 4\sqrt{3}x + 3 = 0\)
First step is to make coefficient of \(x^2 = 1\)
So on dividing equation by 4
\(x^2 + \sqrt{3}x + {{3} \over {4}} = 0\)
On adding and subtracting square of \({{\sqrt{3}}\over{2}}\) (half of the middle term) from the equation \(x^2 + {{x}\over{2}} - 2 = 0\)
\(x^2 + \sqrt{3}x + {{3} \over {4}} + ({{\sqrt{3}}\over{2}})^2 - ({{\sqrt{3}}\over{2}})^2 = 0\)
\(x^2 + ({{3}\over{4}})^2 - \sqrt{3}x + {{3} \over {4}} - ({{3}\over{4}}) = 0\)
Using identity {\((a+b)^2 = a^2 + b^2 + 2ab\)}
\((x + {{\sqrt{3}}\over{2}})^2 = 0\)
\((x + {{\sqrt{3}}\over{2}}) (x + {{\sqrt{3}}\over{2}}) = {{33}\over{16}}\)
On taking square root both sides:
\(x + {{\sqrt{3}}\over{2}} = 0\) and \(x + {{\sqrt{3}}\over{2}} = 0\)
\(x = -{{\sqrt{3}}\over{2}}\) and \(x = -{{\sqrt{3}}\over{2}}\)
(iv) \(2x^2 + x + 4 = 0\)
First step is to make coefficient of \(x^2 = 1\)
So on dividing equation by 2
\(x^2 + {{x}\over{2}} + 2 = 0\)
On dividing the middle term of the equation by 2x:
\(({{x}\over{2}})({{1}\over{2}}x) = {{1}\over{4}}\)
On adding and subtracting square of \({{1}\over{4}}\) from the equation \(x^2 - {{7}\over{2}}x + {{3}\over{2}} = 0\)
\(x^2 + {{x}\over{2}}x + 2 + ({{1}\over{4}})^2 - ({{1}\over{4}})^2 = 0\)
\(x^2 + ({{1}\over{4}})^2 + {{x}\over{2}} + 2 - ({{1}\over{4}})^2 = 0\)
Using identity {\((a + b)^2 = a^2 + b^2 -2ab\)}
\((x + {{1}\over{4}})^2 + 2 - {{1}\over{16}} = 0\)
\((x + {{1}\over{4}})^2 = {{1}\over{16}} - 2 = {{1 - 32}\over{16}}\)
On taking square root both sides:
Right hand side does not exist since square root of negative number does not exist.
Therefore, there is no solution.
Ans.(i) \(2x^2 - 7x + 3 = 0\)
The general form of equation is \(ax^2 + bx + c = 0\)
Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{7 ± \sqrt{(-7)^2 - 4(2)(3)}} \over {(2)(2)}}\)
=>\(x = {{7 ± \sqrt{49 - 24}} \over {4}}\)
=>\(x = {{7 ± 5} \over {4}}\)
=>\(x = {{7 + 5} \over {4}} , {{7 - 5} \over {4}}\)
=>\(x = 3 , {{1} \over {2}}\)
(ii) \(2x^2 + x - 4 = 0\)
The general form of equation is \(ax^2 + bx + c = 0\)
Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{-1 ± \sqrt{(1)^2 - 4(2)(-4)}} \over {(2)(2)}}\)
=>\(x = {{-1 ± \sqrt{33}} \over {4}}\)
=>\(x = {{-1 + \sqrt{33}} \over {4}} , {{-1 - \sqrt{33}} \over {4}}\)
(iii) \(4x^2 + 4\sqrt{3}x + 3 = 0\)
The general form of equation is \(ax^2 + bx + c = 0\)
Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{-4\sqrt{3} ± \sqrt{(4\sqrt{3})^2 - 4(4)(3)}} \over {(2)(4)}}\)
=>\(x = {{-4\sqrt{3} ± \sqrt{0}} \over {8}}\)
=>\(x = {{-\sqrt{3}} \over {2}} , {{-\sqrt{3}} \over {2}}\)
(iv) \(2x^2 + x + 4 = 0\)
The general form of equation is \(ax^2 + bx + c = 0\)
Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{-1 ± \sqrt{(1)^2 - 4(2)(-4)}} \over {(2)(2)}}\)
=>\(x = {{-1 ± \sqrt{-31}} \over {4}}\)
since square root of negative number does not exist.
Therefore, there is no solution.
Ans.(i)\({{x} - {{1}\over {x}}} = 3 \) where x is not equal to 0
=>\({{x^2 - 1} \over {x}} = 3\)
=>\(x^2 - 1 = 3x\)
=>\(x^2 -3x - 1 = 0\)
The general form of equation is \(ax^2 + bx + c = 0\)
Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{3 ± \sqrt{(3)^2 - 4(1)(-1)}} \over {(2)(1)}}\)
=>\(x = {{3 ± \sqrt{13}} \over {2}}\)
=>\(x = {{3 + \sqrt{13}} \over {2}} , {{3 - \sqrt{13}} \over {2}}\)
(ii)\({{1} \over {x + 4}} - {{1} \over {x - 7}} = {{11} \over {30}}, x ? -4,7\)
=>\({{(x - 7) - (x + 4)} \over {(x - 4)(x - 7)}} = {{11} \over {30}}\)
=>\({{-11} \over {(x - 4)(x - 7)}} = {{11} \over {30}}\)
=>\(-30 = x^2 - 7x + 4x -28\)
=>\(x^2 -3x + 2 = 0\)
The general form of equation is \(ax^2 + bx + c = 0\)
Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{3 ± \sqrt{(3)^2 - 4(1)(2)}} \over {(2)(1)}}\)
=>\(x = {{3 ± \sqrt{1}} \over {2}}\)
=>\(x = {{3 + \sqrt{1}} \over {2}} , {{3 - \sqrt{1}} \over {2}}\)
=>\(x = 2,1\)
Ans.Let present age of Rehman = x years
Age of Rehman 3 years ago = (x - 3) years.
Age of Rehman after 5 years = (x + 5) years
According to the given condition:
\({{1} \over {x - 3}} + {{1} \over {x + 5}} = {{1} \over {3}}\)
=>\({{(x + 5) + (x - 3)} \over {(x - 3)(x + 5)}} = {{1} \over {3}}\)
=>\( 3 (2x + 2) = (x - 3) (x + 5)\)
=>\( 6x + 6 = x^2 - 3x + 5x -15\)
=>\(x^2 - 4x - 15 - 6 = 0\)
=>\(x^2 - 4x -21 = 0\)
Comparing quadratic equation \(x^2 - 4x -21 = 0\) with general form \(ax^2 + bx + c = 0\),
We get a = 1, b = ?4 and c = ?21
Using quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{4 ± \sqrt{(4)^2 - 4(1)(-21)}} \over {(2)(1)}}\)
=>\(x = {{4 ± \sqrt{16 + 84}} \over {2}}\)
=>\(x = {{4 + 10} \over {2}} , {{4 - 10} \over {2}}\)
=>\(x = 7, -3\)
We discard x=-3. Since age cannot be in negative.
Therefore, present age of Rehman is 7 years.
Ans.Let Shefali’s marks in Mathematics = x
Let Shefali’s marks in English = 30 - x
If, she had got 2 marks more in Mathematics, her marks would be = x + 2
If, she had got 3 marks less in English, her marks in English would be = 30 – x - 3 = 27 - x
According to given condition:
\((x + 2) (27 - x) = 210\)
=>\(27x - x^2 + 54 - 2x = 210\)
=>\(x^2 - 25x + 156 = 0\)
Comparing quadratic equation \(x^2 - 25x + 156 = 0\) with general form \(ax^2 + bx + c = 0\),
We get a = 1, b = ?25 and c = 156
Applying Quadratic Formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{25 ± \sqrt{(25)^2 - 4(1)(156)}} \over {(2)(1)}}\)
=>\(x = {{25 ± \sqrt{625 - 624}} \over {2}}\)
=>\(x = {{25 + 1} \over {2}} , {{25 - 1} \over {2}}\)
=>\(x = 13, 12\)
Therefore, Shefali’s marks in Mathematics = 13 or 12
Shefali’s marks in English = 30 – x = 30 – 13 = 17
Or Shefali’s marks in English = 30 – x = 30 – 12 = 18
Therefore, her marks in Mathematics and English are (13, 17) or (12, 18).
Ans.Let shorter side of rectangle = x metres
Let diagonal of rectangle = (x + 60) metres
Let longer side of rectangle = (x + 30) metres
According to pythagoras theorem,
\((x + 60)^2 = (x + 30)^2 + x^2\)
=>\( x^2 + 3600 + 120x = x^2 + 900 + 60x + x^2\)
=>\(x^2 - 60x - 2700 = 0\)
Comparing equation with \(x^2 - 60x - 2700 = 0\) standard form \(ax^2 + bx + c = 0\),
We get a = 1, b = ?60 and c = ?2700
Applying quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{60 ± \sqrt{(60)^2 - 4(1)(-2700)}} \over {(2)(1)}}\)
=>\(x = {{60 ± \sqrt{3600 + 10800}} \over {2}} = {{60 ± \sqrt{14400}} \over {2}}\)
=>\(x = {{60 + 120} \over {2}} , {{60 - 120} \over {2}}\)
=>\(x = 90,-30\)
We ignore –30. Since length cannot be in negative.
Therefore, x = 90 which means length of shorter side = 90 metres
And length of longer side = x + 30 = 90 + 30 = 120 metres
Therefore, length of sides are 90 and 120 in metres.
Ans.Let smaller number = x and let larger number = y
According to condition:
=>\(y^2 - x^2 = 180\)… (1)
Also, we are given that square of smaller number is 8 times the larger number.
=>\(x^2 = 8y\) … (2)
Putting equation (2) in (1), we get
\(y^2 - 8y = 180\)
=>\(y^2 - 8y - 180 = 0\)
Comparing equation \(y^2 - 8y - 180 = 0\) with general form \(ax^2 + bx + c = 0\),
We get a = 1, b = ?8 and c = ?180
Using quadratic formula = \(y = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(y = {{8 ± \sqrt{(-8)^2 - 4(1)(-180)}} \over {(2)(1)}}\)
=>\(y = {{8 ± \sqrt{64 + 720}} \over {2}} = {{60 ± \sqrt{784}} \over {2}}\)
=>\(y = {{8 + 28} \over {2}} , {{8 - 28} \over {2}}\)
=>\(y = 18,-10\)
Using equation (2) to find smaller number:
\(x^2 = 8y\)
=>\(x^2 = 8y = (8)(18) = 144\)
=>\( x = ±12\)
And, = \(x^2 = 8y = (8)(-10) = -80\) {No real solution for x}
Therefore, two numbers are (12, 18) or (-12, 18)
Ans.Let the speed of the train = x km/hr
If, speed had been 5 km/hr more, train would have taken 1 hour less.
So, according to this condition
\({{360} \over {x}} = {{360} \over {x + 5}} + 1\)
=>\(360 ( {{1}\over{x}} - {{1}\over {x + 5}}) = 1\)
=> \(360 ( {{x + 5 - x}\over{x(x + 5)}}) = 1\)
=>\((360)(5) = x^2 + 5x\)
=>\(x^2 + 5x - 1800 = 0\)
Comparing equation \(x^2 + 5x - 1800 = 0\) with general equation \(ax^2 + bx + c = 0\),
We get a = 1, b = 5 and c = -1800
Applying quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{-5 ± \sqrt{(5)^2 - 4(1)(-1800)}} \over {(2)(1)}}\)
=>\(x = {{-5 ± \sqrt{25 + 7200}} \over {2}} = {{-5 ± \sqrt{7225}} \over {2}}\)
=>\(x = {{-5 + 85} \over {2}} , {{-5 - 85} \over {2}}\)
=>\(x = 40,-45\)
Since speed of train cannot be in negative. Therefore, we discard x = -45
Therefore, speed of train = 40 km/hr
Ans.Let time taken by tap of smaller diameter to fill the tank = x hours
Let time taken by tap of larger diameter to fill the tank = (x – 10) hours
It means that tap of smaller diameter fills \({{1} \over {x}} th\) part of tank in 1 hour.… (1)
And, tap of larger diameter fills \({{1} \over {x - 10}} th\) part of tank in 1 hour. … (2)
When two taps are used together, they fill tank in 758 hours.
In 1 hour \({{8} \over {75}} th\), they fill part of tank \({{1} \over {{{75} \over {8}}}} = {{8} \over {75}}\)… (3)
From (1), (2) and (3),
=>\( {{1}\over{x}} + {{1}\over {x - 10}} = {{8} \over {75}}\)
=> \( {{x - 10 + x}\over{x(x - 10)}} = {{8} \over {75}}\)
=>\( 75 (2x - 10) = 8 (x^2 - 10x)\)
=> \(150x – 750 = 8x^2 - 80x\)
=>\(8x^2 - 80x - 150x +750 = 0\)
=>\(4x^2 - 115x + 375 = 0\)
Comparing equation \(4x^2 - 115x + 375 = 0\) with general equation \(ax^2 + bx + c = 0\),
We get a = 4, b = -115 and c = 375
Applying quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{115 ± \sqrt{(-115)^2 - 4(4)(375)}} \over {(2)(4)}}\)
=>\(x = {{115 ± \sqrt{13225 - 6000}} \over {8}} = {{115 ± \sqrt{7225}} \over {8}}\)
=>\(x = {{115 + 85} \over {8}} , {{115 - 85} \over {8}}\)
=>\(x = 25,3.75\)
Time taken by larger tap = x – 10 = 3.75 – 10 = -6.25 hours
Time cannot be in negative. Therefore, we ignore this value.
Time taken by larger tap = x – 10 = 25 – 10 = 15 hours
Therefore, time taken by larger tap is 15 hours and time taken by smaller tap is 25 hours.
Ans.Let average speed of passenger train = x km/h
Let average speed of express train = (x + 11) km/h
Time taken by passenger train to cover 132 km = \({{132} \over {x}}\) hours
Time taken by express train to cover 132 km = \({{132} \over {x + 11}}\) hours
According to the given condition,
\({{132} \over {x}} = {{132} \over {x + 11}} + 1\)
=>\(132({{1} \over {x}} - {{1} \over {x + 11}}) = 1\)
=>\(132({{x + 11 - x} \over {x(x + 11)}}) = 1\)
=> \(132 (11) = x (x + 11)\)
=>\(1452 = x^2 + 11x\)
=>\(x^2 + 11x -1452 = 0\)
Comparing equation \(x^2 + 11x -1452 = 0\) with general quadratic equation \(ax^2 + bx + c = 0\), we get a = 1, b = 11 and c = ?1452
Applying quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{-11 ± \sqrt{(11)^2 - 4(1)(-1452)}} \over {(2)(1)}}\)
=>\(x = {{-11 ± \sqrt{121 + 5808}} \over {2}} = {{-11 ± \sqrt{5929}} \over {2}}\)
=>\(x = {{-11 + 77} \over {2}} , {{-11 - 77} \over {2}}\)
=>\(x = 33, -44\)
As speed cannot be in negative. Therefore, speed of passenger train = 33 km/h
And, speed of express train = x + 11 = 33 + 11 = 44 km/h
Ans.Let perimeter of first square = x metres
Let perimeter of second square = (x + 24) metres
Length of side of first square = \({{x} \over {4}}\) metres {Perimeter of square = 4 × length of side}
Length of side of second square = \({{x + 24} \over {4}}\) metres
Area of first square = side × side = \({{x} \over {4}} × {{x} \over {4}} = {{x^2} \over {16}} m^2\)
Area of second square = \(({{x + 24} \over {4}})^2\)
According to given condition:
\({{x^2} \over {16}} + ({{x + 24} \over {4}})^2 = 468\)
=>\({{x^2} \over {16}} + {{x^2 + 576 + 48x} \over {16}} = 468\)
=>\({{x^2 + x^2 + 576 + 48x} \over {16}} = 468\)
=> \(2x^2 + 576 + 48x = (468)(16)\)
=> \(2x^2 + 576 + 48x = 7488\)
=>\(2x^2 + 48x - 6912 = 0\)
=>\(x^2 + 24x - 3456 = 0\)
Comparing equation \(x^2 + 24x - 3456 = 0\) with standard form \(ax^2 + bx + c = 0\),
We get a = 1, b =24 and c = -3456
Applying quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{-24 ± \sqrt{(24)^2 - 4(1)(-3456)}} \over {(2)(1)}}\)
=>\(x = {{-24 ± \sqrt{576 + 13824}} \over {2}} = {{-24 ± \sqrt{14400}} \over {2}}\)
=>\(x = {{-24 + 120} \over {2}} , {{-24 - 120} \over {2}}\)
=>\(x = 48, -72\)
Perimeter of square cannot be in negative. Therefore, we discard x=-72.
Therefore, perimeter of first square = 48 metres
And, Perimeter of second square = x + 24 = 48 + 24 = 72 metres
=> Side of First square = \({{Perimeter} \over {4}} = {{48} \over {4}} = 12\) m
And, Side of second Square = \({{Perimeter} \over {4}} = {{72} \over {4}} = 18\) m
Ans.(i)\(2x^2 - 3x + 5 = 0\)
Comparing this equation with general equation \(ax^2 + bx + c\),
We get a = 2, b = ?3 and c = 5
Discriminant = \(b^2 - 4ac = (-3)^2 - 4(2)(5) = 9 – 40 = ?31\)
Discriminant is less than 0 which means equation has no real roots.
(ii)\(3x^2 -4 \sqrt{3} x + 4 = 0\)
Comparing this equation with general equation \(ax^2 + bx + c\),
We get a = 3, b = \(-4 \sqrt{3}\) and c = 4
Discriminant = \(b^2 - 4ac = (-4 \sqrt{3})^2 - 4(3)(4) = 48 – 48 = 0\)
Discriminant is equal to zero which means equations has equal real roots.
Applying quadratic \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\) to find roots,
=>\(x = {{4 \sqrt{3} ± \sqrt{0}} \over {6}} = {{2 \sqrt{3}} \over {3}} \)
Because, equation has two equal roots, it means x = \({{2 \sqrt{3}} \over {3}} ,{{2 \sqrt{3}} \over {3}} \)
(iii)\(2x^2 - 6x + 3 = 0\)
Comparing this equation with general equation \(ax^2 + bx + c\),
We get a = 2, b = ?6, and c = 3
Discriminant =\(b^2 - 4ac = (-6)^2 - 4(2)(3) = 36 – 24 = 12\)
Value of discriminant is greater than zero.
Therefore, equation has distinct and real roots.
Applying quadratic \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\) to find roots,
\(x = {{6 ± \sqrt{12}} \over {4}} = {{6 ± 2 \sqrt{3}} \over {4}} = {{3 ± \sqrt{3}} \over {2}}\)
\(x = {{3 + \sqrt{3}} \over {2}} , {{3 - \sqrt{3}} \over {2}}\)
Ans.(i)\(2x^2 + kx + 3 = 0\)
We know that quadratic equation has two equal roots only when the value of discriminant is equal to zero.
Comparing equation with \(2x^2 + kx + 3 = 0\) general quadratic equation \(ax^2 + bx + c = 0\), we get a = 2, b = k and c = 3
Discriminant \( b^2 - 4ac = k^2 - 4 (2) (3) = k^2 - 24\)
Putting discriminant equal to zero
\(k^2 - 24= 0\)
\( k^2 = 24 \)
\( k = ± \sqrt{2} = ± 2 \sqrt{6}\)
(ii)\(kx (x - 2) + 6 = 0\)
\( kx^2 - 2kx + 6 = 0\)
Comparing quadratic equation \(kx (x - 2) + 6 = 0\) with general form \(ax^2 + bx + c = 0\), we get a = k, b = ?2k and c = 6
Discriminant \( b^2 - 4ac = (-2k)^2 - 4 (k) (6) = 4k^2 - 24k\)
We know that two roots of quadratic equation are equal only if discriminant is equal to zero.
Putting discriminant equal to zero
\(4k^2 - 24k = 0\)
\(4k (k - 6) = 0=> k = 0, 6\)
The basic definition of quadratic equation says that quadratic equation is the equation of the form \(ax^2 + bx + c = 0\), where a ? 0.
Therefore, in equation , we cannot have k = 0.
Therefore, we discard k = 0.
Hence the answer is k = 6.
Ans.Let breadth of rectangular mango grove = x metres
Let length of rectangular mango grove = 2x metres
Area of rectangle = length × breadth = \(x × 2x = 2x^2 m^2\)
According to given condition:
\(2x^2 = 800\)
\(2x^2 - 800 = 0 => x^2 - 400 = 0\)
Comparing equation \( x^2 - 400 = 0\) with general form of quadratic equation \(ax^2 + bx + c = 0\) ,
we get a = 1, b = 0 and c = -400
Discriminant = \( b^2 -4ac = (0)^2 - 4 (1) (-400) = 1600\)
Discriminant is greater than 0 means that equation has two distinct real roots.
Therefore, it is possible to design a rectangular grove.
Applying quadratic formula,\(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\) to solve equation,
\(x = {{0 ± \sqrt{1600}} \over {2(1)}}\) = ± {{40} \over {2} } = ± 20\)
x = 20, -20
We discard negative value of x because breadth of rectangle cannot be in negative.
Therefore, x = breadth of rectangle = 20 metres
Length of rectangle = 2x = 40 metres
Ans. Let age of first friend = x years and let age of second friend = (20 - x) years
Four years ago, age of first friend = (x - 4) years
Four years ago, age of second friend = (20 - x) - 4 = (16 - x) years
According to given condition,
\((x - 4) (16 - x) = 48\)
=>\(16x - x^2 - 64 + 4x = 48\)
=>\(20x - x^2 - 112 = 0\)
=>\(x^2 - 20x + 112 = 0\)
Comparing equation , \(x^2 - 20x + 112 = 0\) with general quadratic equation \(ax^2 + bx + c = 0\),
we get a = 1, b = -20 and c = 112
Discriminant = \( b^2 - 4ac = (-20)^2 - 4 (1) (112) = 400 – 448 = -48 < 0\)
Discriminant is less than zero which means we have no real roots for this equation.
Therefore, the give situation is not possible.
Ans. Let length of park = x metres
We are given area of rectangular park = 400 \(m^2\)
Therefore, breadth of park = \({{400} \over {x}}\) metres{Area of rectangle = length × breadth}
Perimeter of rectangular park = 2 (length + breath) = \( 2 ( x + {{400} \over {x}})\) metres
We are given perimeter of rectangle = 80 metres
According to condition:
\( 2 ( x + {{400} \over {x}}) = 80 \)
\( 2 ( {{x^2 + 400} \over {x}})\)
\(2x^2 + 800 = 80x\)
\(2x^2 - 80x + 800 =0\)
\( x^2 - 40x + 400 = 0\)
Comparing equation, \( x^2 - 40x + 400 = 0\) with general quadratic equation \( ax^2 + bx + c = 0\),
we get a = 1, b = -40 and c = 400
Discriminant = \( b^2 - 4ac = (-40)^2 - 4 (1) (400) = 1600 – 1600 = 0\)
Discriminant is equal to 0.
Therefore, two roots of equation are real and equal which means that it is possible to design a rectangular park of perimeter 80 metres and area 400 \(m^2\).
Using quadratic formula \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\) to solve equation,
\(x = {{40 ± \sqrt{0}} \over {2}} = {{40} \over {2}} = 20\)
Here, both the roots are equal to 20.
Therefore, length of rectangular park = 20 metres
Breadth of rectangular park = \({{400} \over {x}} = {{400} \over {20}} = 20 m \)