NCERT CHAPTER 4 Ex 4.1 Solution
Exercise 4.1
Form of a quadratic equation:
A quadratic equation can be written in the form ax2
+bx+ c = 0
|
Variable |
Coefficient |
|
X2 |
a |
|
x |
b |
“c” is constant in this equation.
Q1. Check whether the following are quadratic equations
or not.
(i)
(x+1)2
= 2(x-3)
Solution: To solve this equation, we start by
expanding the left-hand side and opening the bracketed term on the right-hand
side
We get
X2 + 2x + 1 = 2x - 6
x2+2x+1 -2x + 6 =0
2x will cancel out
Now remaining terms are
x2+7 = 0; it has the x2 term,
hence it is quadratic.
(ii) x2-2x
= (-2) (3-x)
x2 - 2x = -6+2x (simplify the RHS)
x2² - 4x + 6 = 0
it is in the
form of ax2 +bx+ c = 0; hence, it is quadratic.
(iii)
(x-2)(x+1)
=(x-1)(x+3)
x2-2x+x
-2 = x2 +2x-3 (use identity: (a+b)2 = a2+2ab+b2 )
X2
will cancel out
The remaining
terms are
-2x-2x+x -2+3
= 0
-3x+1= 0
This is not of the
form ax2 +bx + c = 0
Hence, it is not quadratic.
(iv)
(x-3)(2x+1)
= x(x+5)
Multiply the
bracketed terms and rewrite as
2x2+x-6x-3
= x2+5x
2x2
- 5x -3 –x2-5x =0 (take all terms in the left-hand side.)
x2-10x-3
=0
It is of the form
ax2 + bx + c = 0
So, it is quadratic.
(v)
(2x-1) (x-3)
=(x + 5) (x-1)
Multiply the
bracketed terms
2x2-6x-x+3
= x2-x+5x-5
2x2-x2-7x+3-4x
+ 5 =0
X2-11x+8
=0
This is of
the form ax² + bx c = 0, hence quadratic.
(vi)
X2+3x+1
= (x-2)2
X2
+ 3x + 1 = x2- 4x +4
X2 +3x+1
–x2+4x--4=
7x-3 =0
Not of the
form ax2 + bx + c = 0; hence, it is not quadratic
(vii)
(x+2)3
= 2x(x2-1)
Identity
(a+b)3 = a3+3a2b+3ab2+b3
X3 + 8 +
6x2 + 12x = 2x3 -2x
-x3+6x2+14x
+ 8 =0
Not of the
form ax2 + bx + c = 0
Hence, not
quadratic.
(viii)
X3-4x2-x+1
= (x-2)3
x3-4x2-x+1
= x3 -8 -6x2 +12x (using identity (a-b)3= a3-b3-3a2b+3ab2)
ð
-4x2+6x2-x-12x+1+8
= 0
ð
2x2
-13x +9 = 0
It is of the
form ax2 + bx + c = 0
So this equation
is quadratic.
2. 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.
(ii) The product of two consecutive positive integers
is 306. We need to find the integers.
(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.
(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 taken
Solution:
(i)
Given
that the area of a rectangular plot is 528 m2.
The length of the plot (in meters) is one more than
twice its breadth
Let the breadth be x meters; then the length is 1 + 2x
Then area,
Length x breadth = 528
ð
x(2x+1)
= 528
ð
2x2
+ x - 528 = 0
We need to find the value of x.
(ii)
The
product of two consecutive positive integers is 306
Consecutive
numbers are those that comes one after another
Like 1, 2,3,4 ….
Now, come to
the solution
Let the two
consecutive numbers be x and x+1
According to the question
ð
x(x+1) = 306
ð
x2
+ x - 306 = 0
ð
This
is the required quadratic equation where we need to accept the positive value
of x.
(iii) Rohan’s
mother is 26 years older than he is. The product of their ages (in years) 3 years
from now will be 360
In the
question, it is given that his mom is 26 years older than him, so we can assume
that his current age is = x years
Rohan’s
mother’s age in the present = x+26 years
According to the question
ð
x*(x+26)
= 360
ð
x2 + 26x - 360 = 0
Here, we need
to find Rohan’s age which is “x,” and then we can find his mother’s age by adding
26 to x.
(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 taken
Let the speed
of the train is x km/hr
Distance covered = 480 Km
Condition 1
When train is
moving with normal speed ‘x’
Time =
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