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

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

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

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 $x$