## anonymous one year ago what is an extraneous solution

1. Nnesha

do you have a equation or just need an example ?

2. anonymous

i have an equation

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.

4. 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

5. anonymous

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

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.

7. Nnesha

can you solve for x ?

8. 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.

9. anonymous

the is no solutions that i have found

10. Nnesha

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

11. Nnesha

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

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

x=12 0r 3

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

15. anonymous

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

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

17. Nnesha

sigh. :=)

18. anonymous

i have to pick one of them i cant have both

20. Nnesha

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

take 12

22. 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 ?

23. Nnesha

sides*

24. 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.

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

26. anonymous

27. anonymous

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

28. mathstudent55

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

okay then its 3

choose solutions as per your requirement

31. anonymous

thank you to all who helped i appreciate it.

:)))))))

34. 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

35. 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

@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:)))

37. Nnesha

corrected.

38. Nnesha

that was a typo