How do you solve $\sqrt{x - 5} = x - 7$ $sqrt( x-5) = x-7$ ?

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Answer 1 · 2016-04-18T12:22:20

$x = {9}$ $x={9}$

square both sides.
x-5=(x-7)².
x-5=x²-14x+49.
0=x²-15x+54.

$Delta=sqrt((-15)^2-4*1*54)$
$Delta=sqrt(225-216)$
$Delta=sqrt(9)$
$Delta=±3$

$x_1=(15-3)/2$

$x_1=6$

$x_2=(15+3)/2$

$x_2=9$
$x={9}$

Answer 2 · 2016-04-18T12:22:55

$x = {9}$

$sqrt(x-5)=x-7$
$(sqrt(x-5))^2=(x-7)^2$

$x-5=(x-7)^2$

$x-5=x^2-14x+49$

$x^2-14x-x+49+5=0$

$x^2-15x+54=0$

$(x-6)(x-9)=0$

$if (x-6)=0" "then" "x=6$

$if (x-9)=0" "then" "x=9$

Watch out for extraneous solutions! 6 does not work in the original equation, however, 9 does. Thus, ${x =9}$

How do you solve sqrt( x-5) = x-7√x−5=x−7sqrt( x-5) = x-7?