Prove that x = 1 is a root of (2x² − 4x + 3)(x² − 2x + 2) = 1 and that the equation has no other roots.

Step-by-step solution

1. 1

   Verify the given root by substitution. At $x=1$:

   $2(1)-4(1)+3=1,\qquad 1-2(1)+2=1$

   so the left-hand side is $1\cdot 1=1$ ✓. That settles existence; the real work is uniqueness.

2. 2

   Complete the square in the first factor. Bounding each factor from below is the key idea, because a product of two quantities that are each at least $1$ can only equal $1$ if both are exactly $1$:

   $2x^{2}-4x+3=2\left(x^{2}-2x\right)+3=2\left[(x-1)^{2}-1\right]+3=2(x-1)^{2}+1$

3. 3

   Complete the square in the second factor.

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

   Both factors are now visibly of the form "non-negative square plus $1$".

4. 4

   Bound each factor. For every real $x$, $(x-1)^{2}\ge 0$, so

   $2x^{2}-4x+3\ge 1\qquad\text{and}\qquad x^{2}-2x+2\ge 1$

   with equality in each case only at $x=1$.

5. 5

   Conclude by the product argument. Multiplying two numbers each $\ge 1$ gives a result $\ge 1$, and the product equals $1$ only when both factors equal $1$. So

   $\left[2(x-1)^{2}+1\right]\left[(x-1)^{2}+1\right]=1\ \Longleftrightarrow\ (x-1)^{2}=0\ \Longleftrightarrow\ x=1$

   $\boxed{x=1\text{ is the unique real root}}$

6. 6

   Cross-check by expanding. Setting $u=(x-1)^{2}\ge 0$, the equation becomes $(2u+1)(u+1)=1$, i.e. $2u^{2}+3u=0$, i.e. $u(2u+3)=0$. The roots are $u=0$ and $u=-\tfrac32$; the second is impossible for a square, leaving $u=0$ and $x=1$ ✓ — the same conclusion reached algebraically.