QUADRATIC EQUATIONS
From the same place at 7 am ‘A’ started walking in the north at the
speed of 5 km/hr. After 1 hour B started cycling in the east at a speed of 16 km/hr.
At what time they will be at distance of 52 km apart from each other.
N
E
S
W
Sol.
5 km/hr
16 km/hr
Lets consider direction over here
O
At 8 am
Lets see the animation
At 7 am
?
52 km
B
A
From the same place at 7 am ‘A’ started walking in the north at the
speed of 5 km/hr. After 1 hour B started cycling in the east at a speed of 16 km/hr.
At what time they will be at distance of 52 km apart from each other.
N
E
S
W
Sol.
5 km/hr
B
16 km/hr
O
At 8 am
At 7 am
52 km
A
52 km
A
B
O
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A
B
Speed
Time
Distance =
Speed × Time
5 km/hr
16 km/hr
At 8 am
At 7 am
At what time A started walking?
At what time B started cycling?
That means A travelled 1 hour more than B
If B travelled for 5 hrs,
then A travelled for 6 hrs.
If B travelled for x hrs,
then A travelled for (x + 1)hrs.
x hrs
(x + 1)hrs
5(x + 1)kms
16x kms
From the same place at 7 am ‘A’ started walking in the north at the
speed of 5 km/hr. After 1 hour B started cycling in the east at a speed of 16 km/hr.
At what time they will be at distance of 52 km apart from each other.
Sol.
A
B
O
5(x + 1)kms
16x kms
52 km
In right angled Δ AOB,
(OA)2
By Pythagoras theorem,
+ (OB)2
= (AB)2
∴ [5(x + 1)]2
+ (16x)2
= (52)2
∴ [5x + 5]2
+ (16x)2
= (52)2
Which identity to be used?
(a + b)2 =
a2 + 2ab + b2
∴ 25x2
+ 50x
+ 25
+ 256x2
= 2704
∴ 281x2
+ 50x
+ 25
– 2704
= 0
∴ 281x2
+ 50x
– 2679
= 0
281
2679
281
3
×
893
843
893
×
–
+
∴ 281x2 – 843x + 893x – 2679 = 0
∴ 281x
(x – 3)
+ 893
(x – 3) = 0
∴ (x – 3)
(281x + 893) = 0
∴ x – 3 = 0
or 281x + 893 = 0
∴ x = 3
or x =
– 893
281
x is the time taken by B
∴
∴ x ≠ -893
Hence, x = 3
281
∴ B did the cycling for 3hrs.
∴ B started his journey at 8 a.m.
∴ He will be at point B at 11 a.m.
A and B will be at distance of 52 kms apart from each other at 11 a.m.
Δ AOB is a right angled Δ