Question

-x" alt="\sqrt{2x^{2} -1} \geq 1-x" align="absmiddle" class="latex-formula">

Answer 1

Answer:

$x\geq 2$

Step-by-step explanation:

by squaring both sides:

$2x^2 - 1 \geq x^2 + 2x - 1$

canceling out 1

$2x^2 \geq x^2 + 2x$

subtract x^2 on both sides

$x^2\geq 2x$

divide by x on both sides

$x\geq 2$