Where did the extra solutions come from?

[Guest]

If we start with the equation [tex](x+1)^{2 } = x^{2} + 2x + 1[/tex], we know it is true for all [tex]x[/tex], then if we wish to solve this simultaneously with [tex]2x^{2} - 5x = x[/tex], we can substitute equation 2 into equation 1 (here specifically I am going to put it into the [tex]x[/tex] which is in the bracket being squared. We would then expect the solutions of [tex](2x^{2} - 5x +1)^{2 } = x^{2} + 2x + 1[/tex] to be the intersection of the solution sets of equation 1 and 2, that being the intersection of all real [tex]x[/tex] and [tex]x = 0, x = 3[/tex], leaving us with [tex]x =0, x=3[/tex], however there is an extraneous solution when we solve the equation. Why is this? Where is the flaw in my thinking? Thank you all n advance.

[Guest]

If we start with the equation [tex](x+1)^{2 } = x^{2} + 2x + 1[/tex], we know it is true for all [tex]x[/tex], then if we wish to solve this simultaneously with [tex]2x^{2} - 5x = x[/tex], we can substitute equation 2 into equation 1 (here specifically I am going to put it into the [tex]x[/tex] which is in the bracket being squared. We would then expect the solutions of [tex](2x^{2} - 5x +1)^{2 } = x^{2} + 2x + 1[/tex] to be the intersection of the solution sets of equation 1 and 2, that being the intersection of all real [tex]x[/tex] and [tex]x = 0, x = 3[/tex], leaving us with [tex]x =0, x=3[/tex], however there is an extraneous solution when we solve the equation. Why is this? Where is the flaw in my thinking? Thank you all n advance.

When you squared [tex]2x^2 - 5x[/tex], you introduced extra solutions. Squaring is an irreversible step, and requires the extra checks.

Note: The equation [tex](x + 1)^2 = x^2 + 2x + 1[/tex] is a tautology, and cannot be "solved" for [tex]x[/tex]. Instead, just solve the equation [tex]2x^2 - 5x = x[/tex]:

[tex]\qquad 2x^2 - 5x = x[/tex]

[tex]\qquad 2x^2 - 6x = 0[/tex]

[tex]\qquad x^2 - 3x = 0[/tex]

[tex]\qquad (x)(x - 3) = 0[/tex]

[tex]\qquad x = 0,\,3[/tex]