Question

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

Answer 1

Complete the square to rewrite it as

`2x^2+7x+7=2\left(x^2+\frac72x\right)+7=2\left(x^2+\frac72x+\left(\frac74\right)^2-\left(\frac74\right)^2\right)+7`

`\implies2x^2+7x+7=2\left(x+\frac74\right)+\frac78`

The minimum value of this expression occurs at
`x=-\frac74` because
`\left(x+\frac74\right)^2` will always be non-negative. Then at this point we get a value of
`\frac78`, which means

`2x^2+7x+7\ge\frac78`

and so is always positive for any real value of
`x`