Originally Posted by

**flora33**
So I would do, for your example 1 - 25 = 24 because 5 squared is 25, right?

Um... what...? Where is "25" coming from? And how is 1 - 25 not negative?

You have "x[sup:28epy69h]2[/sup:28epy69h] - 2x - 4 = 0", and a proposed solution value of x = 1 - sqrt[5]. To "plug in for x", you would take the "1 - sqrt[5]" and use this to

*replace* (plug in for) every instance of "x" in the

equation:

. . . . .(1 - sqrt[5])[sup:28epy69h]2[/sup:28epy69h] - 2(1 - sqrt[5]) - 4 ?=? 0

You would then square out the first term:

. . . . .(1 - 2 sqrt[5] + 5) - 2(1 - sqrt[5]) - 4 ?=? 0

...and simplify:

. . . . .1 + 5 - 2 sqrt[5] - 2 + 2 sqrt[5] - 4 ?=? 0

. . . . .1 + 5 - 2 - 4 - 2 sqrt[5] + 2 sqrt[5] ?=? 0

. . . . .6 - 6 + 2 sqrt[5] - 2 sqrt[5] ?=? 0

. . . . .0 + 0 = 0

See? It works just like every other plug-n-chug or check-the-solution you've done. No new rules; no new formulas or processes; just the exact same plugging in and checking as always.

Eliz.

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