There are three consecutive positive integers such that the sum of the square of the first and the product of father to is 154 find the integers? Denote the situation in the form of the quadratic equation
Answers
Let the three consecutive positive integers be x,x+1, x+2.
Given that square of the first integer when added to the product of other two integers the sum is 154.
x^2 + (x + 1)(x + 2) = 154
x^2 + x^2 + 2x + x + 2 = 154
x^2 + x^2 + 3x + 2 = 154
x^2 + x^2 + 3x = 152
2x^2 + 3x - 152 = 0
2x^2 - 16x + 19x - 162 = 0
2x(x - 8) + 19(x - 8) = 0
(2x + 19)(x - 8) = 0
2x = -19 and x = 8
x = -19/2.
Since x cannot be negative,so x = 8.
Then,
x = 8
x + 1 = 9
x + 2 = 10.
Therefore the integers are 8,9,10.
Verification:
8^2 + (9 * 10) = 154
64 + 90 = 154
154 = 154.
Answer:
Solution
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Let (x−1),x and (x+1) be the three positive integers.
It is given that the sum of the square of the first integer and the product of other two is 154, therefore,
(x−1)
2
+x(x+1)=154
⇒x
2
+1−2x+x
2
+x=154(∵(a−b)
2
=a
2
+b
2
−2ab)
⇒x
2
+1−2x+x
2
+x−154=0
⇒2x
2
−x−153=0
⇒2x
2
−18x+17x−153=0
⇒2x(x−9)+17(x−9)=0
⇒(2x+17)=0,(x−9)=0
⇒2x=−17,x=9
⇒x=−
2
17
,x=9
Since x is a positive integer thus, x=9.
Now, x−1=9−1=8 and x+1=9+1=10
Since the unit place of the digit is 6 and the tens place is 2
Hence, the three consecutive positive integers are 8,9 and 10.
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