Question

Solve the radical equation. square root 6n-11= n – 3 Which solution is extraneous? A.)2

Answer 1

D 10 right????????????:

Answer 2

ANSWER


D) 10

EXPLANATION

We want to solve the radical equation,

$\sqrt{6n - 11} = n - 3$

Square both sides,



$6n - 11 = (n - 3) ^{2}$

Expand brackets on right hand side,

$6n - 11 = {n}^{2} - 6n + 9$

Rewrite in general quadratic equation form,

${n}^{2} - 12n + 20 = 0$

Factor to obtain,

$(n - 2)(n - 10) = 0$

Apply the zero product principle to get,

$n = 2 \: or \: n = 10$


We put n=2 into the equation to get,

$\sqrt{6(2) - 11} = 2 - 3$


$1 = - 1$


This statement is false, hence 2 is an extraneous solution


We put n=10, to get

$\sqrt{6(10) - 11} = 10 - 3$


$\sqrt{49} = 7$


$7 = 7$


This is true, hence the only solution is
$n = 10$