Class10 NCRT Quadratic Equations Exercise – 4.3 pdf || UP Board
Quadratic
Equations
Exercise
– 4.3
1. Find
the roots of the following quadratic equations, if they exist, by the method of
completing the square:
(i) 2x2
– 7x + 3 = 0
dividing
by 2,
x2
– 7/2 x + 3/2 = 0
x2
– 7/2 x = - 3/2
Adding
on both side,
[a2 + b2 – 2ab = (a + b)2]
Either,
x – 7/4 = 5/4, x – 7/4 =
-5/4
x = 7/4 +
5/4, x = 7/4 – 5/4
x = 12/4 =
3, x = 2/4 = 1/2
Hence, the
roots of quadratic equations are 3 and 1/2.
(ii) 2x2
+ x - 4 = 0
dividing
by 2,
x2
+ 1/2 x - 2 = 0
x2
+ 1/2 x = 2
Adding
on both side,
Either,
x + ¼ = √33/4, x + ¼ =
-√33/4
x = (√33 –
1)/4, x = (-√33 – 1)/4
Hence, the
roots of the quadratic equations are (√33 – 1)/4
and (-√33
– 1)/4.
(iii) 4x2
+ 4√3 x + 3 = 0
dividing
by 4,
x2
+ √3 x + 3/4 = 0
x2
+ √3 x = - 3/4
Adding
on both side,
Either, x
= -√3/2
Hence, the
roots of the quadratic equations are -√3/2 and -√3/2.
(iv) 2x2
+ x + 4 = 0
dividing
by 2,
x2
+ 1/2 x + 4 = 0
x2
+ 1/2 x = - 2
Adding
on both side,
But, we
know that (x
+
1/4)2 cannot be negative for any real value
of x, so no real value of x can satisfy the given equation. Hence the root of
the given equation does not exist.
2. Find
the roots of the quadratic equations gives in Q. 1 above by applying the
quadratic formula.
(i) 2x2
– 7x + 3 =0
2x2 – 7x + 3 =0
Comparing with the general quadratic equation ax2
+ bx + c = 0,
Here, a = 2, b = -7, c = 3
Therefore, b2 – 4ac = (-7)2 – 4 × 2 × 3 =
49 - 24 = 25 > 0
So,
Or x = (7 + 5)/4, x = (7 – 5)/4
x = 12/4, x = 2/4
x = 3, x = ½
Hence, the roots of quadratic equations are 3 and ½.
(ii) 2x2 + x – 4 = 0
2x2 + x – 4 = 0, Here a = 2, b = 1, c = -4
Therfore, b2 – 4ac = (1)2 – 4 × 2 × -4 = 1 + 32 = 33 > 0
Hence, the roots of the quadratic equations are (√33 – 1)/4 and (-√33 – 1)/4.
(iii) 4x2 + 4√3 x + 3 = 0
4x2 + 4√3 x + 3 = 0, Here a = 4, b = 4√3, c = 3
Therefore b2 – 4ac = (4√3)2 – 4 × 4 × 3 =
48 - 48 = 0
So, 
Or x = -(4√3)/8
x = -√3/2
Hence, the roots of the quadratic equations are -√3/2 and -√3/2.
(iv) 2x2 + x + 4 = 0
2x2 + x + 4 = 0, Here a = 2, b = 1, c = 4
Therefore, b2 – 4ac = (1)2 – 4 × 2 × 4 = 1
- 32 = -31 < 0
However, the square of a real number cannot be
negative, so the value of
is not real.
Hence, the root of the given equation does not exist.
3. Find
the roots of the following equations:
(i) x – 1/x = 3, x ≠ 0
x2 – 1 = 3x
x2 – 3x – 1 = 0
x2 – 3x – 1 = 0, Here a = 1, b = -3, c = -1
Therefore, b2 – 4ac = (-3)2 – 4 × 1 × -1 =
9 + 4 = 13 > 0
So,
Or x = (3 + √13)/2, x = (3 - √13)/2
Hence, the roots of the quadratic equations are (3 +
√13)/2 and (3 - √13)/2.
x2 – 3x – 28 = -30
x2 – 3x + 2 = 0
x2 – 3x + 2 = 0, Here a = 1, b = -3, c = 2
Therefore, b2 – 4ac = (-3)2 – 4 × 1 × 2 = 9
- 8 = 1 > 0
So, 
Or x = (3 + 1)/2, x = (3 – 1)/2
x = 4/2, x = 2/2
x = 2, x = 1
Hence, the roots of quadratic equations are 2 and 1.
4. The sum of the reciprocals of Rehman’s ages, (in
years) 3 years ago and 5 years from now is 1/3. Find his present age.
Let Rehman's present age = x years
Hence, age 3 years ago = x – 3 years
Hence, age after 5 years = x + 5 years
According to Question,
x2 + 2x – 15 = 6x +6
x2 – 4x – 21 = 0
x2 – 4x – 21 = 0, Here a = 1, b = -4, c = -21
Therefore, b2 – 4ac = (-4)2 – 4 × 1 × -21 =
16 + 84 = 100 > 0
So,
Or, x = (4 + 10)/2, x = (4 – 10)/2
x = 14/2, x = -6/2
x = 7, x = -3
Since age cannot be negative, Rehman's present age is 7 years.
5. In
a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had
she got 2 marks more in Mathematics and 3 marks less in English, the product of
their marks would have been 210. Find her marks in the two subjects.
Let Shefali's marks in Mathematics = x
Hence, Shefali's marks in English = 30 – x
If he had got 2 marks more in Mathematics and 3 marks less in English, then
Marks in Mathematics = x + 2
Marks in English = 30 – x – 3
According to Question,
Product = (x + 2)(27 – x)
210 = (x + 2)(27 – x)
27x – x2 + 54 – 2x = 210
– x2 – 25x – 156 = 0
x2 – 25x + 156 = 0
x2 – 12x – 13x + 156 = 0
x(x – 12) – 13(x – 12) = 0
(x – 12)(x – 13) = 0
Or, x – 12 = 0, x – 13 = 0
x = 12, x = 13
If x = 12, then marks in maths = 12, marks in English = 30 – 12 = 18
If x = 13, then marks in maths = 13, marks in English = 30 – 3 = 17
6. The
diagonal of a rectangular field is 60 metres more than the shorter side. If the
longer side is 30 metres more than the shorter side, find the sides of the
field.
Let short side = x m
Hence, diagonal = x + 60 m
and longer side = x + 30 m
According to Question,
(x + 60)2 = x2 + (x + 30)2
x2 + 120x + 3600 = x2 + x2
+ 60x + 900
-x2 + 60x + 2700 = 0
x2 – 60x – 2700 = 0
x2 – 90x + 30x – 2700 = 0
x(x – 90) + 30(x – 90) = 0
(x – 90)(x + 30) = 0
x – 90 = 0, x + 30 = 0
x = 90, x = -30
But x ≠ -30, since x is the side of the field and will
not be negative.
So, x = 90
So, shorter side = 90 m
and longer side = 90 + 30 = 120 m
7. The
difference of squares of two numbers is 180. The square of the smaller number
is 8 times the larger number. Find the two numbers.
Let the smaller number = y
and larger number = x
So,
y2 = 8x
According to Question,
x2 – y2 = 180
x2 – 8x = 180 [y2 = 8x]
x2 – 8x – 180 = 0
x2 – 18x + 10x -180 = 0
x(x – 18) + 10(x – 18) = 0
(x – 18)(x + 10) = 0
x – 18 = 0, x + 10 = 0
x = 18, x = -10
But x ≠ -10, because x is a large number and will not
be negative.
So, x = 18
So, smaller side = y ⇒ y2 = √8x
⇒ y = √8x
⇒ y = √8×18
⇒ y = √144
⇒ y = 12
And larger side = 18
8. A train travels 360 km at a uniform speed. If the
speed had been 5 km/h more, it would have taken 1 hour less for the same
journey. Find the speed of the train.
Let speed of train = x km/h
and distance covered = 360 km
so time t1 = 360/x hours [ time =
distance / speed ]
If the speed had been 5 km/h more, then the time t2 = 360/(x + 5) hours
According to Question,
360x + 1800 – 360x = x(x + 5)
1800 = x2 + 5x
x2 + 5x – 1800 = 0
x2 + 45x – 40x – 1800 = 0
x(x + 45) – 40(x + 45) = 0
(x + 45)(x – 40) = 0
x + 45 = 0, x – 40 = 0
x = -45, x = 40
But x ≠ -45, because x is the speed of the train and
will not be negative.
Therefore, x = 40
Hence, the speed of the train is 40 km/h.
9. Two
water taps together can fill a tank in
hours.
The tap of larger diameter takes 10 hours less than the smaller one to fill the
tank separately. Find the time in which each tap can separately fill the tank.
Let the time taken by the larger diameter tap = x
hours
and time taken by the tap of smaller diameter = x + 10 hours
Therefore, the tank filled by the larger diameter tap
in 1
hour = 1/x
And the tank filled by the smaller diameter tap in 1 hour = 1/(x + 10)
According to Question,
75(2x + 10) = 8x(x + 10)
150x + 750 = 8x2 + 80x
8x2 – 70x – 750 = 0
4x2 – 35x – 375 = 0
4x2 – 60x + 25x – 375 = 0
4x(x – 60) + 25(x – 15) = 0
(x – 15)(x + 25) = 0
x – 15 = 0, x + 25 = 0
x = 15, x = -25
But x ≠ -25, because x is the time to fill the tank and it will
not be negative.
Therefore, x = 15
Hence, time taken by the tap of larger diameter = 15 hours
And time taken by smaller diameter tap = 15 + 10 hours
10. An
express train takes 1 hour less than a passenger train to travel 132 km between
Mysore and Bangalore (without taking into consideration the time they stop at
intermediate stations). If the average speed of the express train is 11 Km/h
more than that of the passenger train, find the average speed of the two
trains.
Let the average speed of the passenger train = x km/h
Hence, average speed of express train = x + 11 km/h
Distance covered = 132 km
Hence, the time taken by the passenger train t1 = 132/x hours [ time =
distance / speed ]
and the time taken by the express train t2 = 132/(x + 11) hours
According to Question,
132x + 1452 – 132x = x(x + 11)
1452 = x2 + 11x
x2 + 11x – 1452 = 0
x2 + 44x – 33x – 1452 = 0
x(x + 44) – 33(x + 44) = 0
(x + 44)(x – 33) = 0
x + 44 = 0, x – 33 = 0
x = - 44, x = 33
But x ≠ -44, since x is the speed of the train and will not be
negative.
Therefore, x = 33
Hence, average speed of passenger train = 33 km/h
And, average speed of express train = 33 + 11 = 44 km/h
11. Sum
of the areas of two squares is 468 m2. If the difference of their perimeters
is 24 m, find the sides of the two squares.
Let the side of the larger square = x m
Let the side of the smaller square = y m
According to Question,
x2 + y2 = 468 ……………(i)
difference of dimensions,
4x – 4y = 24
x – y = 6
x = 6 + y ………………(ii)
Substituting the value of x in equation (i),
(y + 6)2 + y2 = 468
y2 + 12y + 36 + y2 = 468
2y2 + 12y – 432 = 0
y2 + 6y – 216 = 0
y2 +18y – 12y – 216 = 0
y(y + 18) – 12 (y + 18) = 0
(y + 18)(y – 12) = 0
y + 18 = 0, y – 12 =0
y = -18, y = 12
But x ≠ -18, since x is the side of the square and will not be
negative.
Therefore, x = 12
Hence, side of smaller square = 12 m
Substituting the value of y in equation (ii),
Side of the larger square = x = y + 6 = 12 + 6 = 18 m
Download Class10 NCRT Quadratic Equations Exercise – 4.3 pdf
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