Question

prove that 2x²+2x>-4 for all real values of x​

Answer 1

Step-by-step explanation:

$2x {}^{2} - 4x + 3 = ( \sqrt{2x) {}^{2} } - 2 \times ( \sqrt{2x)}$

$\times \frac{2}{ \sqrt{2} } + ( \frac{2}{ \sqrt{2} } ) {}^{2} - ( \frac{2}{ \sqrt{2} }) {}^{2} + 3$

$= ( \sqrt{2x} - ( \frac{2}{ \sqrt{2} } ) {}^{2} + 1$