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scoundrel [369]
3 years ago
5

Find three consecutive odd positive integers such that 5 times the sum of all three is 72more than the product of the first and

second integers.
Mathematics
1 answer:
Alenkasestr [34]3 years ago
8 0
Let the positive odd numbers be x,x+2,x+4.
According to the question,
5(3x+30)=72+x^2+4x
x^2-13x+42=0
(x-6) and (x-7)=0
x=6 or x=7
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Step-by-step explanation:

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4 years ago
How many solutions are there in the given equation?<br> y = 2x² + 6x + 4
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Step-by-step explanation:

Use the quadratic formula

=

−

±

2

−

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√

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x=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}

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Once in standard form, identify a, b and c from the original equation and plug them into the quadratic formula.

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a={\color{#c92786}{2}}

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6

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⋅

2

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x=\frac{-{\color{#e8710a}{6}} \pm \sqrt{{\color{#e8710a}{6}}^{2}-4 \cdot {\color{#c92786}{2}} \cdot {\color{#129eaf}{4}}}}{2 \cdot {\color{#c92786}{2}}}

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3 years ago
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