Question

solve for x: 1≤ 1+x^2/2x ≤1

Answer 1

x=1

Answer 2

The given inequality is $1 \leq \frac{1+x^2}{2x} \leq 1$.

For an expression $A$ to satisfy $1 \leq A \leq 1$, it must be that $A=1$. Therefore, the given inequality simplifies to a single equation: $\frac{1+x^2}{2x} = 1$

First, we must note that the denominator $2x$ cannot be zero, which implies $x \neq 0$.

Now, we solve the equation: Multiply both sides by $2x$: $1+x^2 = 2x$ Rearrange the terms to form a standard quadratic equation: $x^2 - 2x + 1 = 0$ This is a perfect square trinomial, which can be factored as: $(x-1)^2 = 0$ Take the square root of both sides: $x-1 = 0$ Solve for $x$: $x = 1$ This value $x=1$ does not violate the condition $x \neq 0$.

To verify, substitute $x=1$ back into the original expression: $\frac{1+(1)^2}{2(1)} = \frac{1+1}{2} = \frac{2}{2} = 1$ Since $1 \leq 1 \leq 1$ is true, $x=1$ is the correct solution.