what is an extraneous solution

- purplemexican

what is an extraneous solution

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- Nnesha

do you have a equation or just need an example ?

- purplemexican

i have an equation

- madhu.mukherjee.946

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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- Nnesha

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

- purplemexican

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

- madhu.mukherjee.946

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.

- Nnesha

can you solve for x ?

- mathstudent55

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.

- purplemexican

the is no solutions that i have found

- Nnesha

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

- Nnesha

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

- madhu.mukherjee.946

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

- madhu.mukherjee.946

x=12 0r 3

- madhu.mukherjee.946

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

- purplemexican

my answer choices are
x = –12
x = –3
x = 3
x = 12

- madhu.mukherjee.946

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

- Nnesha

sigh. :=)

- purplemexican

i have to pick one of them i cant have both

- madhu.mukherjee.946

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

- Nnesha

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

- madhu.mukherjee.946

take 12

- Nnesha

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 ?

- Nnesha

sides*

- mathstudent55

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.

- madhu.mukherjee.946

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

- purplemexican

##### 1 Attachment

- purplemexican

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

- mathstudent55

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

- madhu.mukherjee.946

okay then its 3

- madhu.mukherjee.946

choose solutions as per your requirement

- purplemexican

thank you to all who helped i appreciate it.

- madhu.mukherjee.946

your welcome

- madhu.mukherjee.946

:)))))))

- Nnesha

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

- Nnesha

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

- madhu.mukherjee.946

@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:)))

- Nnesha

corrected.

- Nnesha

that was a typo

- madhu.mukherjee.946

okay:)))

Looking for something else?

Not the answer you are looking for? Search for more explanations.