Finding X

[jasm223]

I know this isn't specifically calculus but it is on my calculus review sheet so I would really appreciate your help.
I was given two equations: f(x)=x+6 and g(x)=3?(x+10)
I have to find the two intersection points. I know to set them equal to each other and I have gotten down to
-96=x(x+19)
But I dont know how to get x by itself or if I have even set it up the right way. Any help would be appreciated. Thanks!

 

[Deleted member 4993]

I know this isn't specifically calculus but it is on my calculus review sheet so I would really appreciate your help.
I was given two equations: f(x)=x+6 and g(x)=3?(x+10)
I have to find the two intersection points. I know to set them equal to each other and I have gotten down to
-96=x(x+19)

\(\displaystyle x^2 \ + \ 19x \ + \ 96 \ = \ 0\)

This is a quadratic equation - what method have you been taught to solve such equations?

However, the equation you found [-96=x(x+19) ] is not correct. Please show your work.


But I dont know how to get x by itself or if I have even set it up the right way. Any help would be appreciated. Thanks!

 

[BigGlenntheHeavy]

\(\displaystyle y \ = \ f(x) \ = \ x+6 \ and \ y \ = \ g(x) \ = \ 3\sqrt{x+10}.\)

\(\displaystyle Hence, \ x+6 \ = \ 3\sqrt{x+10}, \ \implies \ (x+6)^2 \ = \ 3^2[\sqrt{x+10}]^2,\)

\(\displaystyle x^2+12x+36 \ = \ 9(x+10) \ = \ 9x+90\)

\(\displaystyle Ergo, \ x^2+3x-54 \ = \ 0, \ \implies \ (x+9)(x-6) \ = \ 0. \ \implies \ x \ = \ -9 \ or \ x \ = \ 6.\)

\(\displaystyle Now, \ when \ x \ = \ -9, \ we \ have \ -3 \ = \ 3, \ no \ good, \ but \ when \ x \ = \ 6,\)

\(\displaystyle we \ have \ 12 \ = \ 12, \ OK.\)

\(\displaystyle Therefore, \ for \ the \ two \ functions, \ there \ is \ only \ one \ point \ of \ intersection, \ namely \ (6,12).\)

\(\displaystyle See \ graph.\)

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