Solving Radical Equations x−7=x−5

Question

Solving Radical Equations  x−7=x−5

sainareon2 · Accepted Answer

Step 1
We can avoid squaring both sides.
Let
$x - 5 = u^{2}$ ,
where
$u \geq 0$
Then
$\sqrt{x - 5} = u$
Also,
$x = u^{2} + 5$
so
$x - 7 = u^{2} - 2$
Thus our equation can be rewritten as
$u^{2} - 2 = u$
or equivalently
$u^{2} - u - 2 = 0$
But
$u^{2} - u - 2 = (u - 2) (u + 1)$
Thus the solutions of
$u^{2} - u - 2 = 0$
are
$u = 2$
and $u = - 1$
Since $u \geq 0$ , we reject the solution $u = - 1$
We conclude that $u = 2$ , and therefore $x = u^{2} + 5 = 9$ .

Kudusind · Accepted Answer

Step 1  x−7=x−5  (x−7)2=x−52  (x−7)2=x−5  (x−7)(x−7)=x−5  x2−7x−7x+49=x−5  x2−15x+54=0  (x−9)(x−6)=0  x−9=0 or x−6=0  x=9 or x=6  Now check for extraneous solutions

RizerMix · Accepted Answer

Step 1 (x−7)=x−5 Check the domain first x−5≥0∩x−7≥0 x≥7 (x−7)2=x−52 (x−7)2=x−5 (x−7)(x−7)=x−5 x2−7x−7x+49=x−5 x2−15x+54=0 (x−9)(x−6)=0 x−9=0 or x−6=0 x=9 or x=6 checking the domain x=6 is an extraneous root. x=9

Solving Radical Equations x-7=\sqrt{x-5}

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