Question

Solve the equation. Identify any extraneous solutions.

Answer 1

A. 6 is a solution to the original equation. The value –5 is an extraneous solution.

B. 6 and 5 are both extraneous solutions.

C. 5 is a solution to the original equation. The value –6 is an extraneous solution.

D. 6 and –5 are solutions.

Answer 2

do u know the meaning of extraneous

Answer 3

That it's unrelated, I think.

Answer 4

square both sides

x^2 = x+ 30
x^2 -x - 30 = 0

can u solve this?

Answer 5

Ill try

Answer 6

extraneous means that although the working out gives it as a solution , it is not possible as a root

Answer 7

5 and 6?

Answer 8

not quite
(x - 6)(x + 5) = 0

giving x = 6 , -5

Answer 9

So what do I do with that?

Answer 10

well see if these roots are possible by substituting each into the original equation

Answer 11

eg x = 6

6 = sqrt(6 + 30)

6 = 6 - so that is a genuine root

Answer 12

So I can eliminate B and C

Answer 13

yes

Answer 14

is -5 a solution?

Answer 15

I dont think so.

Answer 16

try plugging in -5 for x in the original equation

Answer 17

- 5 = sqrt (-5 + 30) does that work out?

Answer 18

No

Answer 19

what are the square roots of 25?

Answer 20

5 ANd -5 right?

Answer 21

Yes

Answer 22

so its a genuine root

Answer 23

So they're both genuine making them both solutions.

Answer 24

right

Answer 25

I do believe I'd pick D

Answer 26

correct

Answer 27

Thank you very much.

Answer 28

yw

Answer 29

wrong.

Answer 30

please explain why thats wrong

Answer 31

ask the teacher, I answered D and I got it wrong.

Answer 32

hii