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Quadratic Equations
A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0, where a, b, c are real numbers, a ⧣ 0.
Standard form of quadratic equation
The equation of the form ax² +bx+ c= 0, a⧣0 is known as standard quadratic equation.
For example,
5x² + 3x + 6 = 0 is a quadratic equation in the standard form.
Roots of the quadratic equation
If p(x) = 0 is a quadratic equation, then the zeroes of the polynomial p(x) are called the roots of the quadratic equation p(x) = 0.
Thus, x = l is a root of p(x) = x² + x - 2 = 0, since p(l ) = (1)² + 1- 2 = 0
x = - 2 is also a root of x² + x - 2 = 0, since p(- 2) = (- 2)² + (- 2) - 2 = 4 - 2 - 2 = 0
Notes 1 . x = α i s a root of p(x) = 0, then p(α ) = 0.
2 . Every quadratic equation can have at most two real roots.
Nature of Roots
The roots of a general quadratic equation ax²+bx+c=0 are given by the standard quadratic equation formula,
In this formula, the key component is b²−4ac . This is called the discriminant {D} of this quadratic equation.
[1] If b² – 4ac > 0, [D= positive] we get two distinct real roots
EXERCISE 4.1
- Check whether the following are quadratic equations :
(i) (x + 1) ² = 2(x- 3)
⇒ x² + 1 + 2x = 2(x - 3)
⇒ x² + 1 + 2x = 2x - 6
⇒ x² + 1 + 2x - 2x + 6 = 0 =>
⇒ x² + 7 = 0
I t is of the form ax² + bx + c = 0
Therefore, it is a quadratic equation.
(ii) x² - 2x = (-2) (3- x)
⇒ x² - 2x = - 6 + 2x
⇒ x² - 2x - 2x + 6 = 0
⇒ x² - 4x + 6 = 0
lt is of the form ax 2 + bx + c = 0,
where a = 1, b = - 4 and c = 6. Hence, it is a quadratic equation.
(iii) (x - 2 ) (x + l) = (x - l) (x + 3)
⇒ x (x +1)- 2(x + 1 ) = x(x + 3) - 1(x + 3)
⇒ x² + x - 2x - 2 = x² + 3x - x - 3
⇒ x² - x - 2 = x² + 2x -3
=> x²+ 2x- 3- x²+x+2 =0
=> 3x - 1 = 0
It is not of the form ax 2 + bx + c = 0.
Therefore, it is not a quadratic equation.
(iv) (x - 3) (2x + l ) = x(x + 5)
⇒ x(2x + 1) - 3(2x + l ) = x² + 5x
⇒ 2x² + x - 6x - 3 = x² + 5x
=> 2x² - 5x - 3 - x² - 5x = 0
⇒ x 2 - 10x - 3 = 0
It is of the form ax² + bx + c = 0
Therefore, it is a quadratic equation.
(v) (2x - 1)(x - 3) = (x + 5) (x - 1)
⇒ 2x(x - 3) - l(x - 3) = x (x - l) + 5(x - 1)
⇒ 2x - 6x - x + 3 = x² - x + 5x - 5
⇒ 2x² - 7x + 3 = x² + 4x- 5
⇒ 2x² - 7x + 3- x² - 4x+5 = 0
⇒ x² -11x + 8 = 0
It is of the form ax² + bx + c = 0 ,
Therefore, it is a quadratic equation.
(vi) x² + 3x + l = (x - 2)²
⇒ x² + 3x + I = x² + 4 - 4x
⇒ x² + 3x + l - x² - 4 + 4x = 0
⇒ 7x - 3 = 0
It is not of the form ax² + bx + c = 0.
Therefore , the given equation is not a quadratic equation.
(vii) (x + 2)³ = 2x( x² - 1)
⇒ x³ + (2)³ + 3 x 2 x x(x + 2) = 2x 3 - 2x
⇒ x³ + 8 + 6x(x + 2)= 2x³ - 2x
⇒ x³ + 8 + 6x² + I 2x = 2x³ - 2x
⇒ x³ + 8 + 6x 2 + l 2x - 2x³ + 2x = 0
⇒ - x³ + 6x² + 14x + 8 = 0
It is not of the form ax² + bx + c = 0. Hence , the given equation is not a quadratic equation.
(viii) x³ - 4x² - x+ l = (x - 2)³
⇒ x³ - 4x² - x + l = x³ - 3x²(2) +3x(2)² - (2)³ [(a – b)³ = a³ – 3a²b + 3ab² – b³]
⇒ x³- 4x ² -x+ 1= x³ - 6x² +12x - 8
⇒ x³ - 4x² - x + 1 - x³ + 6x² - 12x +8 = 0
⇒ 2x² -13x+ 9 = 0
lt is of the form ax²+ bx+ c = 0,
Therefore, the given equation is a quadratic equation. - Represent the following situations in the form of quadratic equations :
(i) The area of a rectangular plot is 528 m2 The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Let the breadth of rectangular plot be x metres.
Then, length of the rectangular plot = (2x + 1) metres.
Area of rectangular plot = [x (2x + l)]
= (2x² + x)
According to question
⇒ 2x² + x = 528
Hence , given problem in the form of quadratic equation is 2x² + x - 528 = 0
Hence , given problem in the form of quadratic equation is 2x² + x - 528 = 0
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Let first positive integers be x
Then second positive integers = x + l.
Product of integers = x (x + 1)
= x² + x
According to question
x² + x = 306
x² + x - 306 =0
Hence, given problem in the form of quadratic equation is x² + x - 306 = 0.
(iii) Rohan's mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan's present age.
Let the present age of Rohan be x years
Then Present age of Rohan's mother = (x + 26) years
After 3 years, Rohan's age = (x + 3) years
After 3 years, Rohan's mother's age =(x + 26 + 3) years
= (x + 29) years
The product of their ages = (x + 3) (x + 29)
= x² + 29x + 3x + 87
= x² + 32x + 87
According to question
⇒ x² + 32x + 87 = 360
⇒ x² + 32x + 87 - 360 = 0
⇒ x² + 32.x - 273 = 0
Hence, given problem in the form of quadratic equation is x² + 32.x - 273 = 0.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have take 3 hours more to cover the same distance. We need to find the speed of the train.
Let the speed of train be x km/h
Total distance 480 km
Time taken by train = Distance/Speed
If speed of train is decreased by 8 km/h
Then, decreased speed of train= (x - 8) km/h
Then, Time taken by train =Distance/Speed
According to question
480x - 480x - 3840 = 3x² - 24x
3x² - 24x -3840 = 0
x² - 8x - 1280 =0
Hence , given problem in the form of quadratic equation is x² - 8x -1280 =0
EXERCISE 4.2
- Find the roots of the following quadratic equations by factrisation :
(i) x² - 3x - 10 = 0
x² - 5x + 2x - 10 = 0
x(x - 5) + 2(x - 5) = 0
(x - 5) (x + 2) = 0
x - 5 = 0 or x + 2 = 0
x = 5 or x = - 2
Hence , 5 and -2 are the roots of the given quadratic equation.
(ii ) 2x² + X - 6 = 0
2x² + 4x - 3x - 6 = 0
2.x(x + 2) - 3(x + 2) = 0
(x + 2) (2x - 3) = 0
x + 2 = 0 or 2x - 3 = 0,
x = - 2 or x = 3/2
Hence , - 2 and 3/2 are the roots of the given quadratic equation.
(iii) √2x² + 7x + 5√2 = 0
√2x² + 2x+ 5x +5√2= 0
√2x(x+√2)+5( x+ √2)= 0
(x + √2)(√2x+5) = 0
x + √2=0 or √2x+5 = 0
x = -√2 or x = - 5/√2
Hence, -√2 and -5/√2 are the roots of the given quadratic equation
(iv) 2x -x+½= 0 [Multipy By 8]
16x² - 8x + 1 = 0
l6x² - 4x - 4x + l = 0
4x(4 x - 1) - 1 (4 x - 1) = 0
(4x - 1) (4x - 1) = 0
(4x - 1) = 0 or (4x - 1) = 0
4x = ¼ or 4x = ¼
Hence, ¼ and ¼ are the roots of the given quadratic equation.
(v) 100x² - 20x + 1 = 0
100x² - 10x - 10x + l = 0
10x (10x - 1) - l (10x - 1) = 0
(10x - l ) (10 x - 1) = 0
(10x - l ) = 0 or (10 x - 1) = 0
x = ⅒ or x = ⅒
Hence, ⅒ and ⅒ are the roots of given quadratic equation. - Solve the problems :
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
Let the number of marbles John had be x.
Then, Number of marbles Jivanti had = 45 - x.
The number of marbles left with John, when he lost 5 marbles = x - 5
The number of marbles left with Jivanti, when she lost 5 marbles= 45 - x - 5
= 40 - x
According to question
The product of the number of marbles =124
(x - 5) (40 - x) = 124
40x - x² - 200 + 5x = 124
- x² + 40x + 5x - 200 - 124 = 0
- x² + 45x - 324 =0
x² - 45x + 324 =0
x² - 36x - 9x + 324 =0
x(x - 36) - 9 (x - 36) = 0
(x - 36) = 0 or (x - 9) = 0
x = 36 or x = 9
Hence, they had started with 36 and 9 marbles.
(ii) A cottage industry produces a ce11ain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day. the total cost of production was Rs 750. We would like to find out the number of toys produced on that day.
Let the number of toys produced on a particular day be x.
Then, the cost of production of each toy on that day = (55 - x)
The total cost of production on that day = x(55 - x)
According to question
x(55 -x) = 750
55x - x2 = 750
x² - 55x + 750 = 0
x² - 25x - 30x + 750 = 0
x(x - 25) - 30(x - 25) = 0
(x - 25) (x - 30) = 0
(x - 25) = 0 or (x - 30) = 0
x = 25 or x = 30
Hence, the number of toys produced on a particular day can be 25 or 30 - Find two numbers whose sum is 27 and product is 182.
Let the first number be x.
Then second number = 27 -x
Product of two numbers = x(27 - x)
As per condition
x(27 - x) = 182
27x - x² = 182
-x² + 27x - 182 = 0
x² - 27x + 182 = 0
x² - 13x - 14x + 182 = 0
x( x - 13) - 14(x - 13) =0
(x - 13) (x - 14) = 0
(x - 13) = 0 or (x - 14) = 0
x = 13 or x = 14
Hence, the required numbers are 13 and 14. - Find two consecutive positive integers, sum of whose squares is 365.
Let the first integer be x
Then second consecutive integer = x + 1
Sum of their squares = x² + (x + 1 )²
According to question
=x² +x² +2x + 1 =365
=2x² + 2x + l =365
2x² + 2x = 364 [Divide by 2]
x² + x = 182
x² + X - 182 = 0
x² + 14x - 13x - 182 = 0
x(x + 14) - 13 (x + 14) = 0
(x - 13) (x + 14) = 0
(x - 13) =0 or (x + 14) = 0
x = 13 or x = - I 4
Since, x is a positive integer, therefore x ⧣ -14
If x = 13 then, 2nd number = 13 + I = 14
Hence, the required two positive consecutive integers are 13 and 14. - The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Let the base of the right triangle = x cm
Height (altitude) of the right triangle = (x - 7) cm
Hypotenuse = 13
By Pythagoras theorem,
(Hyp.)² = (Base)² + (Height)²
(13)² = x² + (x- 7)²
l69 = x² + x² +49 - I4x
2x² - 14x + 169 - 49 = 0
x² - 14x - 120 = 0 [ Divide by 2]
x² - 7x - 60 = 0
x² - 12x + 5x - 60 = 0
x(x - 12) + 5(x - 12) = 0
(x - 12) (x + 5) = 0
(x - 12) = 0 or (x + 5) = 0
x= 12 or x = - 5
Side cannot be negative, so
Base of the right triangle (x) = 12 cm
altitude of the right triangle = x - 7
= 12 - 7
= 5 cm.
Hence, the length of the other two sides are 5 cm and 12 cm. - A cottage industry produces a certain number of pottery articles in a day. ft was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.
Let the number of articles produced = x
Then, cost of production of an article = 2x + 3
Total cost of production = x (2x +3)
As per condition :
x(2x + 3) = 90
2x² + 3x - 90 = 0
2x + 15x - 12x - 90 = 0
x(2x + 15) - 6 (2x + 15) = 0
(2, + 15) (x - 6) = 0
(2x + 15) = 0 or (x - 6) = 0
x = - 15/2 or x = 6
Number of articles cannot be a negative number :
∴ x = 6
Cost of each article =2x + 3
= 2(6) + 3
= 15
Hence, Cost of each article is Rs.15 and Number of articles produced is 6.
EXERCISE 4.3
- Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:(i) 2x² – 3x + 5 = 0a=2 , b=−3 , c=5D = b²−4ac= 3²− 4(2( (5)= 9 - 40= -31. D < 0 [Negative]Therefore, there are no real roots.(ii) 3x² – 4 √3 x + 4 = 0a=3 , b=− 4√3 , c=4D = b²−4ac=48−48D =0Therefore, there are two equal real roots.
(iii) 2x² – 6x + 3 = 0a=2 , b = − 6 c=3D = b²−4ac=(6)²−4(2) (3)=36 - 24= 12D > 0 [Positive]Therefore, there are two distinct real roots - Find the values of k for each of the following quadratic equations, so that they have two equal roots.(i) 2x + kx + 3 = 0a=2 , b=k , and c=3D = b²−4ac =0 [ for the roots to be equal]k²−24 =0k² = 24k=±√4 x 6k = ± 2√6(ii) kx (x – 2) + 6 = 0kx(x−2)+6=0kx²− 2kx + 6 = 0a=k, , b=−2k, and c=6D = b² − 4ac(2k)²− 4k(6) = 0 [ for equal roots D =0]4k² − 24k = 0k² − 6k = 0k(k−6)=0The two values of k satisfying quadratic equation are, k=0 and k=6But if k=0 , the given quadratic equation turns out to be invalid.So, k=6 for the two roots to be of equal value.
- Is it possible to design a rectangular mango grove whose length is twice its breadth,and the area is 800 m²? If so, find its length and breadth.Let us the breadth of the mango grove to be xSo its length is 2xArea = Length x breadthAccording to Questionx × 2x = 8002x² = 800x² = 400x= ±20breadth cannot be negative, so its value is, x=20 metersthe length is = 40 metersIt is possible to design a rectangular mango grove with these dimensions that satisfy the problem conditions.
- Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.Let the present age of one friend be x years.So the age of the second friend = (20−x) years.Four years agoFirst friend's age : (x−4) yearsSecond friend age : (20−x−4)=(16−x) years.The product of these age :(x − 4) (16 − x) = 48−x² + 20x − 64 = 48x² − 20x + 112 = 0The coefficients are,a=1, b=−20 and c=112D =b²−4ac=(-20)²−4(1)(112)=400 - 448= -48D< 0 [Negative]Therefore, the quadratic equation does not have any real roots.The given situation is not possible.
- Is it possible to design a rectangular park of perimeter 80 m and area 400 m² ? If so, find its length and breadthLet the breadth of the park be x meters and length yPerimeter = 2(x+y)=80x + y = 40breadth (y) =40−xAlso the area is product of the length and breadth,x(40−x) = 400−x² + 40x = 400x² − 40x+ 400 = 0The coefficients are,a=1, b=−40 and c=400D = b² −4ac=(-40)² - 4 (1) (400)= 1600 −1600D = 0Againx² − 40x+ 400 = 0x² − 20x - 20x + 400 = 0x (x - 20) - 20 (x -20) =0(x - 20) = 0 or (x - 20) = 0x =20 = x =20Breadth x = 40 m,Length y=40−x=20 m.Answer: Yes, it is possible. Both length and breadth are of equal value 20 m.
