How to solve the equation $x^2-2x=\sqrt{ 2x^2-4x+3} $
Will squaring both sides introduce a $x^4$ term that may complicate the equation?
$\endgroup$ 54 Answers
$\begingroup$Hint:
$$\sqrt{2x^2 - 4x + 3} = \sqrt{2(x^2 - 2x) + 3}$$
Let $v = x^2 - 2x$.
$$\implies x^2 - 2x = \sqrt{2(x^2 - 2x) + 3}\longrightarrow v = \sqrt{2v + 3}$$
Square both sides to get
$$v^2 = 2v + 3 \\ v^2 - 2v - 3 = 0$$
Solve the quadratic equation to find the roots of $v$ and then undo the substitution to find the values of $x$ when $v > 0$.
$\endgroup$ $\begingroup$Hint
Let $\sqrt{2x^2-4x+3}=a\ge0$
$2(x^2-2x)=a^2-3$
$\endgroup$ $\begingroup$Squaring both sides we have $đť‘Ą^4-4x^3+4x^2$=$2x^2-4x+3$ and collecting like terms we have $x^4-4x^3+2x^2+4x-3=0$. By inspection x=-1 and x=1 is a solution hence we have $x^4-4x^3+2x^2+4x-3$=$(x^2-1)(x^2-4x+3)$=$(x-1)(x+1)(x-1)(x-3)$=0. So we have found the four roots, but remember since we have squared both sides we have introduced two new solutions. Substituting back into the original equation we see that x=1 (repeated root) is not the correct solution since it gives a negative value under the square root.
$\endgroup$ $\begingroup$To answer your specific question: yes, squaring introduces additional "sol.utions" which you do not want. So you have to check at the end.
Squaring gives $(x^2-2x)^2-2(x^2-2x)-3=0$ or $$(x^2-2x-3)(x^2-2x+1)=0$$ so $x=1,-1$ or $3$.
But checking we find that $x=1$ is not a solution to the original equation.
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