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3 years ago

find two consecutive odd integers whose product is 1 less than 6 times their sum. Domain for the smallest : {-1, 1, 11}

1 answer:

6 0

Let x be the 1st odd number, and x+2 the second odd consecutive number:

(x)(x + 2) = 6[((x) + (x+2)] -1
x² + 2x = 6(2x + 2) - 1

x² + 2x = 12x +12 - 1
And x² - 10x - 11=0

Solve this quadratic expression:

x' = [+10 +√(10²- 4.(1)(-11)]/2 and x" = [+10 -√(10²- 4.(1)(-11)]/2

x' = [10 + √144]/2 and x" = [10 - √64]/2

x' = (10+12)/2 and x" = (10-12)/2

x = 11 and x = -1

We have 2 solutions that satisfy the problem:

1st for x = 11, the numbers at 11 and 13
2nd for x = - 1 , the numbers are -1 and +1
If you plug each one in the original equation :(x)(x + 2) = 6[((x) + (x+2)] -1
you will find that both generates an equlity

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Step-by-step explanation:

Part A: Look at the graph! The highest y-value should be the maximum of each graph, which can be located at the vertex of a quadratic function. In this graph, the vertex is (2.5, 2531.25), where which its y-value contains the highest point of a graph, since the graph is concave down.

$y=a(x-2.5)^{2} +2531.25\\$

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$y=-5(x-2.5) ^{2}+2531.25$

$y=-5(0-2.5)^{2} +2531.25$

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