Prove that x = 2 is a root of (x² − 4x + 5)(2x² − 8x + 9) = 1 and that there are no other roots.

Step-by-step solution

1. 1

   Check x = 2 directly.

   $4-8+5=1,\qquad 8-16+9=1$

   so the product is $1\cdot 1=1$ ✓.

2. 2

   Complete the square in the first factor. Both factors have their vertex at the same place, $x=2$, which is what makes the argument work:

   $x^{2}-4x+5=(x-2)^{2}+1$

3. 3

   Complete the square in the second factor.

   $2x^{2}-8x+9=2\left(x^{2}-4x\right)+9=2\left[(x-2)^{2}-4\right]+9=2(x-2)^{2}+1$

4. 4

   Bound both factors below by 1. Since $(x-2)^{2}\ge 0$ for all real $x$,

   $x^{2}-4x+5\ge 1,\qquad 2x^{2}-8x+9\ge 1$

   with equality in each case only at $x=2$.

5. 5

   Finish with the product argument. A product of two reals each $\ge 1$ is $\ge 1$, and equals $1$ only if both equal $1$. Hence the equation forces $(x-2)^{2}=0$:

   $\boxed{x=2\text{ is the unique real root}}$

6. 6

   Verify with a substitution. Let $u=(x-2)^{2}\ge 0$. The equation becomes $(u+1)(2u+1)=1$, i.e. $2u^{2}+3u=0$, i.e. $u(2u+3)=0$. Only $u=0$ is admissible, giving $x=2$ ✓. Numerically, at $x=3$ the left side is $(2)(3)=6>1$ and at $x=1$ it is $(2)(3)=6>1$, confirming the minimum sits at $x=2$.