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mixer [17]
2 years ago
7

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

Mathematics
2 answers:
Mandarinka [93]2 years ago
6 0
D 10 right????????????:
Anvisha [2.4K]2 years ago
3 0
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
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