Three consecutive positive integers are such that the sum of the square of the first and the product of the other two is 46.

Question:

Three consecutive positive integers are such that the sum of the square of the first and the product of the other two is 46. Find the integers

Solution:

Let the three consecutive positive integers be xx + 1 and x + 2.

According to the given condition,

$x^{2}+(x+1)(x+2)=46$

$\Rightarrow x^{2}+x^{2}+3 x+2=46$

$\Rightarrow 2 x^{2}+3 x-44=0$

$\Rightarrow 2 x^{2}+11 x-8 x-44=0$

$\Rightarrow x(2 x+11)-4(2 x+11)=0$

$\Rightarrow(2 x+11)(x-4)=0$

$\Rightarrow 2 x+11=0$ or $x-4=0$

$\Rightarrow x=-\frac{11}{2}$ or $x=4$

∴ x = 4           (x is a positive integer)

When x = 4,
x + 1 = 4 + 1 = 5
x + 2 = 4 + 2 = 6

Hence, the required integers are 4, 5 and 6.

 

 

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Comments

Shifali
June 21, 2023, 6:35 a.m.
Thanks❤ this answer was really helpful
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