Question

4x^2+1 greater than or equal to 4x

Answer 1

\[4x^2 + 1 \ge 0\]
first step is to subtract 1 from both sides

Answer 2

The original question was not either of the above responses. The question is\[4 x ^{2} + 1 \ge 4 x\]

Answer 3

The best way to solve inequalities like this is to solve the equality \[4 x^{2} - 4 x + 1 = 0 \] for \[ x \] and then to determine values satisfy the inequality.

Answer 4

So you could use the quadratic formula, or you could factor by hand.

Answer 5

\[4x^2-2x-2x+1 \ge 0\]\[2x(2x-1)-(2x-1) \ge 0\]\[(2x-1)(2x-1) \ge0\]\[(2x-1)^2 \ge 0\]\[2x-1 \ge 0\]\[2x \ge 1\]\[x \ge \frac{1}{2}\]

Answer 6

This happens to factorize into \[ (2 x -1 )^2 = 0 \] which means that \[ x =\frac{1}{2} \] is a double root. So the original problem will be true on either side of \[ x =\frac{1}{2} \] or on both sides. Find a convenient x point and check it out on either side of 1/2.

Answer 7

but u can do it as