extraneous solutions of Squareroot of x^2-4x-5 = 4
- Thread starter Mrsmac
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Your formatting is ambiguous. Do you mean any of the following?
. . . . .sqrt[x<sup>2</sup>] - 4x - 5 = 4
. . . . .sqrt[x<sup>2</sup> - 4](x - 5) = 4
. . . . .sqrt[x<sup>2</sup> - 4x] - 5 = 4
. . . . .sqrt[x<sup>2</sup> - 4x - 5] = 4
Or did you mean something else?
As your book (and any dictionary) explains, "extraneous" solutions are the "extra" ones that don't actually count. Back when you were doing quadratic word problems, these were, for instance, the negative solutions to equations for which the problem required the answer to be positive. They're mathematical solutions that don't work in the original exercise, and are thus thrown out.
So what solution(s) did you get? And what did you find when you plugged the solution(s) back in to the original equation?
Please be specific. Thank you.
Eliz.
. . . . .sqrt[x<sup>2</sup>] - 4x - 5 = 4
. . . . .sqrt[x<sup>2</sup> - 4](x - 5) = 4
. . . . .sqrt[x<sup>2</sup> - 4x] - 5 = 4
. . . . .sqrt[x<sup>2</sup> - 4x - 5] = 4
Or did you mean something else?
As your book (and any dictionary) explains, "extraneous" solutions are the "extra" ones that don't actually count. Back when you were doing quadratic word problems, these were, for instance, the negative solutions to equations for which the problem required the answer to be positive. They're mathematical solutions that don't work in the original exercise, and are thus thrown out.
So what solution(s) did you get? And what did you find when you plugged the solution(s) back in to the original equation?
Please be specific. Thank you.
Eliz.
Re: extraneous solutions
Hello, Mrsmac!
Square both sides: \(\displaystyle \,\left(\sqrt{x^2\,-\,4x\,-\,5}\right)^2\;=\;4^2\)
and we have: \(\displaystyle \,x^2\,-\,4x\,-\,5\;=\;16\;\;\Rightarrow\;\;x^2\,-\,4x\,-\,21\;=\;0\)
Factor: \(\displaystyle \,(x\,-\,7)(x\,+\,3)\;=\;0\)
And we have two roots: \(\displaystyle \,\begin{array}{cc}x\,-\,7\:=\:0 & \;\Rightarrow\; & x\,=\,7 \\ x\,+\,3\:=\:0 & \;\Rightarrow\; & x\,=\,-3\end{array}\)
Check \(\displaystyle x\,=\,7:\;\;\sqrt{7^2\,-\,4(7)\,-\,5} \:=\:\sqrt{49\,-\,28\,-\,5}\:=\:\sqrt{16}\:=\:4\) . . . Okay!
Check \(\displaystyle x\,=\,-3:\;\;\sqrt{(-3)^2\,-\,4(-3)\,-\,5} \:=\:\sqrt{9\,+\,12\,-\,5} \:=\:\sqrt{16}\:=\:4\) . . . Okay!
Both answers check out . . . there are no extraneous roots.
Hello, Mrsmac!
Square both sides: \(\displaystyle \,\left(\sqrt{x^2\,-\,4x\,-\,5}\right)^2\;=\;4^2\)
and we have: \(\displaystyle \,x^2\,-\,4x\,-\,5\;=\;16\;\;\Rightarrow\;\;x^2\,-\,4x\,-\,21\;=\;0\)
Factor: \(\displaystyle \,(x\,-\,7)(x\,+\,3)\;=\;0\)
And we have two roots: \(\displaystyle \,\begin{array}{cc}x\,-\,7\:=\:0 & \;\Rightarrow\; & x\,=\,7 \\ x\,+\,3\:=\:0 & \;\Rightarrow\; & x\,=\,-3\end{array}\)
Check \(\displaystyle x\,=\,7:\;\;\sqrt{7^2\,-\,4(7)\,-\,5} \:=\:\sqrt{49\,-\,28\,-\,5}\:=\:\sqrt{16}\:=\:4\) . . . Okay!
Check \(\displaystyle x\,=\,-3:\;\;\sqrt{(-3)^2\,-\,4(-3)\,-\,5} \:=\:\sqrt{9\,+\,12\,-\,5} \:=\:\sqrt{16}\:=\:4\) . . . Okay!
Both answers check out . . . there are no extraneous roots.