1. Check whether the following are quadratic equations:
(i) (x + 1)2 = 2(x – 3)
(ii) x2 – 2x = (–2)(3 – x
(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
(iv) (x – 3)(2x + 1) = x(x + 5)
(v) (2x – 1) (x – 3) – (x + 5) (x – 1)
(vi) x2 + 3x +1 = (x – 2)2
(vii) (x + 2)3 = 2x(x2 – 1)
(viii) x3 – 4x2 – x + 1 = (x – 2)3
Answer:
(x + 1)2 =
2(x – 3)
(x + 1)2 = 2 (x – 3)
x2 + 2x + 1 = 2x – 6
x2 +
2x + 1 – 2x + 6 = 0
x2 + 7=0
This
is of the form ax2 + bx + c = 0
∴ (x + 1)2 = 2(x – 3) is
a quadratic equation.
(ii) x2– 2x = (–2) (3 – x)
x2 – 2x = –6 + 2x
x2 –
2x – 2x + 6 = 0
x2 –
4x + 6 = 0
This is of the form ax2 + bx + c =
0
∴ x2 – 2x = (–2) (3 – x) is a quadratic
equation.
(iii)
(x – 2) (x + 1) = (x – 1) (x + 3)
x2 –
x – 2 = x2 + 2x – 3
x2 –
x – 2 – x2 – 2x + 3 = 0
–3x
+ 1 = 0
This
is not the form of ax2 + bx + c = 0
∴ (x – 2) (x + 1) = (x – 1) (x + 3) is
not quadratic equation.
(iv) (x – 3) (2x
+ 1) = x(x + 5)
2x2 +
x – 6x – 3 = x2 + 5x
2x2 – 5x – 3 – x2 –
5x – 0
x2 +
10x – 3 = 0
This
is of the form ax2 + bx + c = 0
∴
(x – 3) (2x + 1) = x(x + 5) is a quadratic equation.
(v) (2x – 1) (x
– 3) = (x + 5) (x – 1)
2x2 –
6x – x + 3 = x2 – x + 5x – 5
2x2 –
x2 – 6x – x + x – 5x + 3 + 5 = 0
x2 –
11x + 8 = 0
This
is of the form ax2 + bx + c = 0
∴ (2x – 1) (x – 3) = (x + 5) (x – 1) is a
quadratic equation.
(vi) x2 +
3x + 1 = (x – 2
x2 +
3x + 1 = (x – 2
x2 +
3x + 1 = x2 – 4x + 4
x2 +
3x + 1 – x2 + 4x – 4 =0
7x
– 3 = 0
This
is not the form of ax2 + bx + c = 0
∴
x2 + 3x + 1 = (x – 2 is not a quadratic equation.
(vii) (x + 2)3 = 2x(x2 – 1)
(a+b)3 = a3 + 3a2b + 3ab2 + b3
x3 +
3x2(2) + 3x(2)2 + (2)3 = 2x3 –
2x
x3 +
6x2 + 12x + 8 = 2x3 – 2x
x3 +
6x2 + 12x + 8 – 2x3 + 2x = 0
–x3 +
6x2 + 14x + 8 = 0
This
is not the form of ax2 + bx + c = 0
∴ (x + 2)3 = 2x(x2 –
1) is not a quadratic equation.
(viii) x3 –
4x2 – x + 1 = (x – 2)3
We
have:
x3 –
4x2 – x + 1 = (x – 2)3
x3 –
4x2 – x + 1 = x3 + 3x2(– 2) + 3x(–
2)2 + (– 2)3
x3 –
4x2 – x + 1 = x3 – 6x2 + 12x –
8
x3 –
4x2 – x – 1 – x3 + 6x2 – 12x +
8 = 0
2x2 –
13x + 9 = 0
This
is of the form of ax2 + bx + c = 0
∴ x3 – 4x2 – x + 1 = (x – 2)3 is a quadratic equation.
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 3 hours more to cover the same distance. We need to find the speed of the train.
Answer:
(i) Let
the breadth of the plot be x m.
Hence, the length of the plot is (2x +
1) m.
Area of a rectangle = Length × Breadth
(2x + 1) × x = 528
2x2 +
x = 528
2x2 +
x – 528 = 0
Thus, the required quadratic equation is 2x2 + x – 528 = 0
(ii)
Let the consecutive integers be x and x + 1.
According to the question
x (x +
1) = 306
x2 +
x = 306
x2 +
x – 306 = 0
iii)
Let Rohan’s age be x.
Hence, his mother’s age = x +
26
3 years hence,
Rohan’s age = x + 3
Mother’s age = x + 26 + 3 = x +
29
According
to the question
(x + 3) × (x + 29) = 360
x2 + 29x + 3x + 87 = 360
x2 + 29x + 3x + 87 – 360 = 0
x2 + 32x – 273 = 0
(iv) Let the speed of train be x km/h.
In the second case, speed = x – 8 km/h
According to the question,
T2 - T1 = 3
480x – 480x + 3840 = 3x2 – 24x
3x2 – 24x -3840 = 0
x2 – 8x – 1280 = 0
EXERCISE 4.2
1. Find the roots of the following quadratic
equations by factorisation:
(i)
x2 – 3x – 10 = 0
(ii)
2x2 + x – 6 = 0
(iii) 
X2 +7X + 5 = 0
(iv)
2x2 – x + 8 = 0
(v)
100x2 – 20x + 1 = 0
Answer:
(i) x2 –
3x – 10 = 0
x2 –
3x – 10 = 0
x2 –
5x + 2x – 10 = 0
x (x – 5) + 2(x – 5) = 0
(x – 5) (x + 2) = 0
x – 5 = 0 ⇒ x = 5
or
x + 2 = 0 ⇒ x = –2
Thus,
the required roots are x = 5 and x = –2.
(ii)
2x2 + x – 6 = 0
We
have:
2x2 +
x – 6 = 0
2x2 +
4x – 3x – 6 = 0
2x(x
+ 2) – 3 (x + 2) = 0
(x
+ 2) (2x – 3) = 0
x + 2 = 0 ⇒ x = –2
or
2x – 3 = 0 ⇒ x = 3/2
Thus,
the required roots are x = –2 and 3/2
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