QUADRATIC �EQUATIONS
+ 4x
– 285
= 0
+ 1
∴ x + 2
∴ x + 3
∴ x + 4
= 15 + 1
= 16
= 15 + 2
= 17
= 15 + 3
= 18
= 15 + 4
= 19
+ 4x
+ 4
∴
x2
+ x2
+ x2
+ x2
+ x2
+ 2x
+ 6x
+ 9
+ 8x
+ 16
– 1455
= 0
(Q) The sum of squares of five natural consecutive numbers is 1455. Find the numbers.
Sol.
Let the five required consecutive natural numbers be x , x + 1, x + 2, x + 3 and x + 4.
As per the given condition,
+
+
+
+
( )2 ( )2 ( )2 ( )2 ( )2
x
x + 1
x + 2
x + 4
x + 3
=
1455
∴
x2
+
x2 + 2x + 1
+
x2 + 4x + 4
+
x2 + 6x + 9
+
x2 + 8x + 16
– 1455 = 0
∴ 5x2
(Dividing throughout by 5)
∴ x2
∴ x2
∴ x
∴ (x – 15)
∴ x – 15 = 0
∴ x = 15
x is a natural number
∴
x ≠ –19
Hence, x = 15
∴
∴ x + 1
How many numbers?
Means +
Means (something)2
Square of what ?
Means =
x, x + 1, x + 2, x + 3 and x + 4
1
Let’s arrange all the like terms together
285
57
19
5
3
19
1
19
Find two factors of 285 in such a way that by subtracting factors we get middle number 4
19
15
285
+
–
15
–
= 4
– 15x
+ 19x
– 285
= 0
(x – 15)
+ 19
(x – 15)
= 0
(x + 19)
= 0
or x + 19 = 0
or x = – 19
The required five consecutive natural numbers are 15, 16, 17, 18 and 19.
∴
What we need to find ?
Calculation
+ 20x
– 1425
= 0
Since we are subtracting the factors give middle term sign to the bigger factor and the opposite sign to the smaller factor
Let’s do the prime factorization of 285
How to assume five consecutive natural numbers ?
(a+b)2 = a2 + 2ab + b2
homework