estadsocm3

Answered

2022-01-21

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

Answer & Explanation

sainareon2

Expert

2022-01-22Added 10 answers

Step 1
We can avoid squaring both sides.
Let
$x-5={u}^{2}$,
where
$u\ge 0$
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
${u}^{2}-u-2=\left(u-2\right)\left(u+1\right)$
Thus the solutions of
${u}^{2}-u-2=0$
are
$u=2$
and  $u=-1$
Since $u\ge 0$, we reject the solution $u=-1$
We conclude that $u=2$, and therefore $x={u}^{2}+5=9$.

Kudusind

Expert

2022-01-23Added 11 answers

Step 1
$x-7=\sqrt{x-5}$
${\left(x-7\right)}^{2}={\sqrt{x-5}}^{2}$
${\left(x-7\right)}^{2}=x-5$
$\left(x-7\right)\left(x-7\right)=x-5$
${x}^{2}-7x-7x+49=x-5$
${x}^{2}-15x+54=0$
$\left(x-9\right)\left(x-6\right)=0$
$x-9=0$ or $x-6=0$
$x=9$ or $x=6$
Now check for extraneous solutions

RizerMix

Expert

2022-01-27Added 437 answers

Step 1 $\left(x-7\right)=\sqrt{x-5}$ Check the domain first $x-5\ge 0\cap x-7\ge 0$ $x\ge 7$ $\left(x-7{\right)}^{2}={\sqrt{x-5}}^{2}$ $\left(x-7{\right)}^{2}=x-5$ $\left(x-7\right)\left(x-7\right)=x-5$ ${x}^{2}-7x-7x+49=x-5$ ${x}^{2}-15x+54=0$ $\left(x-9\right)\left(x-6\right)=0$ $x-9=0$ or $x-6=0$ $x=9$ or $x=6$ checking the domain $x=6$ is an extraneous root. $x=9$

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