Question

what is an extraneous solution

Answer 1

do you have a equation or just need an example ?

Answer 2

i have an equation

Answer 3

an extraneous solution represents a solution, such as that to an equation, that emerges from the process of solving the problem but is not a valid solution to the original problem.

Answer 4

post that
you have to plug in the x value into the equation to check if is it extraneous or not
if the value of x give you equal sides then yes
if not then no that's not an extraneous solution

Answer 5

(45-3x)^1/2=x-9 is the equation i have

Answer 6

Extraneous solutions can arise naturally in problems involving fractions with variables in the denominator. For example, consider this equation:

\frac{1}{x - 2} = \frac{3}{x + 2} - \frac{6x}{(x - 2)(x + 2)}\,.
To begin solving, we multiply each side of the equation by the least common denominator of all the fractions contained in the equation. In this case, the LCD is (x - 2)(x + 2). After performing these operations, the fractions are eliminated, and the equation becomes:

x + 2 = 3(x - 2) - 6x\,.
Solving this yields the single solution x = −2. However, when we substitute the solution back into the original equation, we obtain:

\frac{1}{-2 - 2} = \frac{3}{-2 + 2} - \frac{6(-2)}{(-2 - 2)(-2 + 2)}\,.
The equation then becomes:

\frac{1}{-4} = \frac{3}{0} + \frac{12}{0}\,.
This equation is not valid, since one cannot divide by zero.

Answer 7

can you solve for x ?

Answer 8

An extraneous solution is a solution that appears in the solving of an equation that is not a solution to the original equation.

Sometimes, you need to square both sides of an equation to solve the equation. The process of squaring both sides creates a new equation that is similar but not necessarily completely equivalent to the original equation. Sometimes there is a solution to the squared equation that is not a solution to the original equation. This is an extraneous solution, and it must be discarded.

Answer 9

the is no solutions that i have found

Answer 10

you can convert 1/2 exponent to square root \[\sqrt{45-3x}=x-9\]

Answer 11

okay but you need x value
x=??
no solution means the value of xx is an extraneous solution

Answer 12

hey its coming to x^2-15x+36=0

Answer 13

x=12 0r 3

Answer 14

if two sides of the equation can be squared then its coming

Answer 15

my answer choices are

x = –12
x = –3
x = 3
x = 12

Answer 16

its 12 and 3 .i can show you my workings if you really want

Answer 17

sigh. :=)

Answer 18

i have to pick one of them i cant have both

Answer 19

dear its a quadratic equation so you'll always have two answers

Answer 20

plugin the x value into the equation which one give you the equal sides ?

Answer 21

take 12

Answer 22

you need an*extraneous solution*
\[\huge\rm \sqrt{45-3\color{reD}{x}}=\color{reD}{x}-9\]
first replace x by 12 do you get equal sides
and then replace by 3 do you get equal sdes ?

Answer 23

sides*

Answer 24

Square both sides of your equation to get rid of the 1/2 exponent.

\(\large (45-3x)^{\frac{1}{2}}=x-9\)

\(\large [(45-3x)^{\frac{1}{2}}]^2=(x-9)^2\)

\(\large 45-3x=x^2 - 18x + 81\)

\(\large x^2 - 15x + 36 = 0\)

\(\large(x - 12)(x - 3) = 0\)

\(\large x = 12\) or \(\large x = 3\)

12 and 3 are the solutions to the quadratic equation we got after squaring both sides. The squaring of both sides can introduce extraneous solution, so both solutions to the quadratic, 3 and 12, must be checked in the original equation.

When you check x = 12 in the original equation, it does work, so the only solution to the original equation is x = 12.

When you use x = 3 in the original equation, you see that the right side becomes -6. A square root cannot be a negative number, so x = 3 must be discarded since it is an extraneous solution.

Answer 25

i told you if you want a proper solution then take 12 otherwise take 3

Answer 26

it cant be 12 it has to be 3 if n extraneous solution is a invalid answer

Answer 27

The extraneous solution is 3.
The real solution is 12.

Answer 28

okay then its 3

Answer 29

choose solutions as per your requirement

Answer 30

thank you to all who helped i appreciate it.

Answer 31

your welcome

Answer 32

:)))))))

Answer 33

now substitute 3 \[\huge\rm \sqrt{45-3\color{reD}{(3)}}=\color{reD}{3}-9\]
\[\huge\rm \sqrt{36} =-6\]
square root of 36 =6
but \[6\cancel{=}-6 \] both sides are not equal so 3 is ane extraneous

Answer 34

\[\huge\rm \sqrt{45-3\color{reD}{(12)}}=\color{reD}{12}-9\]
\[\huge\rm \sqrt{9} =3\]
now take square root of 9
both sides are equal then 12 is not an extraneous solution

Answer 35

@Nnesha an extraneous solution is an invalid one so how can you think of 12 .coz 12 is a perfect solution .3 is an extraneous solution coz its giving you an invalid answer:)))

Answer 36

corrected.

Answer 37

that was a typo

Answer 38

okay:)))