Question

Solve the equation. Identify any extraneous solutions.

Answer 1

squaring on both sides
\[x^2 =-3x+40\]

Answer 2

\[x^2+3x-40=0\]

Answer 3

now solve it with the quadratic formula

Answer 4

You can also take the approach of trying the two proposed solutions and see which, if any, of them work. Any solution that does not work is extraneous.

You have to do this same test even if you go through the work of solving it as suggested, because we have squared both sides of the equation. Whenever you do that in the course of solving, you need to test all the solutions to see if they are true solutions or extraneous ones.

Answer 5

\[x^2+8x-5x-40=0\]
\[x(x+8)-5(x+8)\]\[(x+8)(x-5)=0\]

Answer 6

Actually I see there are more than 2 proposed solutions in the answer choices, but other than changing "two" to "all of" the statement stands :-)

Answer 7

x=-8 and 5

Answer 8

@DHASHNI will now ably illustrate the process of checking the solutions in the original equation. Watch her hands carefully :-)

Answer 9

so 8,-5 are extraneous as u squared

Answer 10

No, they aren't both extraneous, you have to test them to see if they work. They *may* be extraneous.

Answer 11

ohhh i made a mistake wait a minute

Answer 12

i'll do the verification!!

Answer 13

as from the equation, the x values are -8,5 ,,,u got to put this answer in the original question .
\[x=\sqrt{-3x+40}\]

Answer 14

when x= -8
\[x=\sqrt{-3(-8)+40}=\sqrt{24+40}=\sqrt{64}=8\]
so when u put x= -8 in ur left hand side u get x=8 in ur right hand side
its a valid answer , therefore its not extraneous
when x =5
\[x=\sqrt{-3(5)+40}=\sqrt{25}=5\]
the same thing occurs here , so , this is also not extraneous
==> the last option in the answer

Answer 15

@DHASHNI check that first one again please :-)

Answer 16

when \(x=-8\) isn't it
\[-8 = \sqrt{-3(-8)+40} = \sqrt{24+40} = \sqrt{64} = 8\]\[-8=8\]???

Answer 17

ohh yeah!!! sorry for the silly mistakes

Answer 18

\[x=5\]is the only correct solution; \(x=-8\) is extraneous. Therefore, C is the correct answer choice.

Answer 19

The reason this happens is that our original equation has a square root sign, and specifies one of the two possible square roots.

If we had done the equation obtained by squaring each side via the quadratic or completing the square, we would end up with

\[x=\frac{-3\pm\sqrt{3^2-4(1)(-40)}}{2(1)} = \frac{-3\pm\sqrt{169}}{2} =-\frac{1}{2}(3\pm13)\]
That \(\pm13\) at the end is the key to this extraneous solution business. It contributes a second "solution", but the original equation only had one solution because the right hand side only had one value. By squaring both sides, we allowed that second, extraneous solution to worm its way into our problem, and in this case, it was unwelcome. If we had a perfect square (\(b^2-4ac=0\)) the two solutions would have been identical (\(+\sqrt{0} = -\sqrt{0})\) and we would not have had an extraneous solution.

Answer 20

Note that if the original equation had been \[x = -\sqrt{-3x+40}\] we would end up with the same solution candidates of \(x = -8, x = 5\) but this time \(x=5\) would have been extraneous and \(x=8\) the correct solution.

Answer 21

so the answer is A?