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purplemexican
 one year ago
what is an extraneous solution
purplemexican
 one year ago
what is an extraneous solution

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Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1do you have a equation or just need an example ?

Purplemexican
 one year ago
Best ResponseYou've already chosen the best response.1i have an equation

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.1an 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.

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1post 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
 one year ago
Best ResponseYou've already chosen the best response.1(453x)^1/2=x9 is the equation i have

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.1Extraneous 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.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1An 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
 one year ago
Best ResponseYou've already chosen the best response.1the is no solutions that i have found

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1you can convert 1/2 exponent to square root \[\sqrt{453x}=x9\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1okay but you need x value x=?? no solution means the value of xx is an extraneous solution

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.1hey its coming to x^215x+36=0

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.1if two sides of the equation can be squared then its coming

Purplemexican
 one year ago
Best ResponseYou've already chosen the best response.1my answer choices are x = –12 x = –3 x = 3 x = 12

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.1its 12 and 3 .i can show you my workings if you really want

Purplemexican
 one year ago
Best ResponseYou've already chosen the best response.1i have to pick one of them i cant have both

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.1dear its a quadratic equation so you'll always have two answers

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1plugin the x value into the equation which one give you the equal sides ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1you need an*extraneous solution* \[\huge\rm \sqrt{453\color{reD}{x}}=\color{reD}{x}9\] first replace x by 12 do you get equal sides and then replace by 3 do you get equal sdes ?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Square both sides of your equation to get rid of the 1/2 exponent. \(\large (453x)^{\frac{1}{2}}=x9\) \(\large [(453x)^{\frac{1}{2}}]^2=(x9)^2\) \(\large 453x=x^2  18x + 81\) \(\large x^2  15x + 36 = 0\) \(\large(x  12)(x  3) = 0\) \(\large x = 12\) 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
 one year ago
Best ResponseYou've already chosen the best response.1i told you if you want a proper solution then take 12 otherwise take 3

Purplemexican
 one year ago
Best ResponseYou've already chosen the best response.1it cant be 12 it has to be 3 if n extraneous solution is a invalid answer

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1The extraneous solution is 3. The real solution is 12.

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.1okay then its 3

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.1choose solutions as per your requirement

Purplemexican
 one year ago
Best ResponseYou've already chosen the best response.1thank you to all who helped i appreciate it.

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.1your welcome

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1now substitute 3 \[\huge\rm \sqrt{453\color{reD}{(3)}}=\color{reD}{3}9\] \[\huge\rm \sqrt{36} =6\] square root of 36 =6 but \[6\cancel{=}6 \] both sides are not equal so 3 is ane extraneous

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1\[\huge\rm \sqrt{453\color{reD}{(12)}}=\color{reD}{12}9\] \[\huge\rm \sqrt{9} =3\] now take square root of 9 both sides are equal then 12 is not an extraneous solution

madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.1@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:)))
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