Solve the equation $2^{x+1}+2^{-x}=3$. The equation of a curve is $y=2x^2-4ax+b$ where $a$ and $b$ are non-zero constants. Explain why $y>0$ if $b>2a^2$.

1. Rewrite the equation using laws of indices to isolate $2^x$

$2(2^x) + \frac{1}{2^x} = 3$

2. Use a substitution to form a quadratic equation

Let $u = 2^x$.
$2u + \frac{1}{u} = 3$
$2u^2 - 3u + 1 = 0$

3. Solve the quadratic equation for $u$

$(2u - 1)(u - 1) = 0$
$u = \frac{1}{2}$ or $u = 1$

4. Solve for $x$

When $u = \frac{1}{2}$:
$2^x = 2^{-1} \implies x = -1$

When $u = 1$:
$2^x = 2^0 \implies x = 0$

1. Express the quadratic function in completed square form.

$y = 2x^2 - 4ax + b$
$y = 2(x^2 - 2ax) + b$
$y = 2[(x - a)^2 - a^2] + b$
$y = 2(x - a)^2 - 2a^2 + b$

2. Identify the minimum value of the function.

Since $2(x - a)^2 \geqslant 0$ for all real $x$, the minimum value of $y$ is $-2a^2 + b$.

3. Use the given condition to explain why $y > 0$.

Given that $b > 2a^2$, we have $b - 2a^2 > 0$.
Since the minimum value of $y$ is strictly greater than $0$, it follows that $y > 0$ for all real values of $x$.