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Fynjy0 [20]
3 years ago
6

Show that 2x^2 + 7x + 7 is positive for all real values of x. PLEASE HELP

Mathematics
1 answer:
Gre4nikov [31]3 years ago
5 0

Complete the square to rewrite it as

2x^2+7x+7=2\left(x^2+\dfrac72x\right)+7=2\left(x^2+\dfrac72x+\left(\dfrac74\right)^2-\left(\dfrac74\right)^2\right)+7

\implies2x^2+7x+7=2\left(x+\dfrac74\right)+\dfrac78

The minimum value of this expression occurs at x=-\dfrac74 because \left(x+\dfrac74\right)^2 will always be non-negative. Then at this point we get a value of \dfrac78, which means

2x^2+7x+7\ge\dfrac78

and so is always positive for any real value of x

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Hello!

\large\boxed{4 : 3 \text{ and } 16 : 12}

We can go through each answer to verify whether each is equivalent to 8 : 6.

4 : 3 is equivalent to 8 : 6 because reducing the ratio 8 : 6  by a factor of 2 yields 4 : 3.

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7 : 5 is not equal to 8 : 6 for the same reason.

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