Question

Find 3 consecutive integers so that 4 times the square of the third decreased by 3, times the square of the first is 41 more than twice the square of the second.​

Answer 1

Answer:

3, 4, 5 or 9, 10, 11

Step-by-step explanation:

Let the first no. be x

.: No. are x, x+1, x+2

ATQ,

$4(x+2)^{2} - 3x^{2} = 41 + 2(x+1)^{2} \\4(x^{2} +4+4x)-3x^{2} = 41 + 2(x^{2}+1+2x)\\4x^{2} + 16+16x -3x^{2}=41 +2x^{2}+2+4x\\-x^{2}+12x-27=0\\x^{2}-12x+27=0\\x^{2}-9x-3x+27=0\\x(x-9)-3(x-9)=0\\(x-3)(x-9)=0$

.: x = 3 or 9

When x = 3,

First No. - 3

Second No. - 4

Third No. - 5

When x = 9,

First No. - 9

Second No. - 10

Third No. - 11

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