I need help on another algebra question

[tboo]

My son threw another question by me that has me stumped. I figure Id start another thread since this equation is different

He is suppose to solve the following quadratic equation by factoring:

This time I used dpaste: link

And to think I used to whiz through these years ago.

Thanks again

 

[Mo0o]

Cant you just plug it in and solve

 

[StevenYoo]

(6x + sqrt(11)) + (2x + sqrt(7))

the equation can be re-arranged to be in standard quadratic form:

ax^2 + bx + c = 0

12x^2 + (6sqrt11 + 2sqrt7)x + sqrt11*sqrt7 = 0

I can do it all out in MS Word + Equation editor, if you want

 

[tboo]

Originally posted by: StevenYoo
(6x + sqrt(11)) + (2x + sqrt(7))

the equation can be re-arranged to be in standard quadratic form:

ax^2 + bx + c = 0

12x^2 + (6sqrt11 + 2sqrt7)x + sqrt11*sqrt7 = 0

I can do it all out in MS Word + Equation editor, if you want

That would be great if you could

Thanks

 

[jlee]

Heh..it's only been a few years since I was doing those and now I've completely forgotten. Shows how often I've used it, I guess.

 

[GodlessAstronomer]

It's not clear if you are writing 6 * sqrt(7) * x or 6 * sqrt(7 * x) so I'll assume the former.

Don't let the square roots scare you, this can be factorised in the same way as most very simple quadratics. This is actually an easy problem designed to look difficult. In fact the very last term (sqrt(7)sqrt(11)) gives the answer away immediately.

(6x + sqrt(11))(2x + sqrt(7)) is the answer.

Edit - just noticed you said "solve". The solutions are the roots of the equation - the additive inverses of the "non-x" terms (zeroth-order terms) in the brackets. So the solutions are -sqrt(11) and -sqrt(7).

 

[tboo]

Originally posted by: GodlessAstronomer
It's not clear if you are writing 6 * sqrt(7) * x or 6 * sqrt(7 * x) so I'll assume the former.

Don't let the square roots scare you, this can be factorised in the same way as most very simple quadratics. This is actually an easy problem designed to look difficult. In fact the very last term (sqrt(7)sqrt(11)) gives the answer away immediately.

(6x + sqrt(11))(2x + sqrt(7)) is the answer.

Thats the answer I came up with . I split the problem in half & factored out 6x & sqrt(11). I wasnt sure if I was supposed to continue though & solve for x.

 

[StevenYoo]

you got email

 

[StevenYoo]

Originally posted by: tboo


Thats the answer I came up with . I split the problem in half & factored out 6x & sqrt(11). I wasnt sure if I was supposed to continue though & solve for x.

Oh,
x = -sqrt11 / 6

and

x = -sqrt7 / 2

 

[GodlessAstronomer]

Originally posted by: StevenYoo

Originally posted by: tboo


Thats the answer I came up with . I split the problem in half & factored out 6x & sqrt(11). I wasnt sure if I was supposed to continue though & solve for x.

Oh,
x = -sqrt11 / 6

and

x = -sqrt7 / 2

Oops, yeah I forgot to divide out those factors in my final x values.

 

[tboo]

Thanks, again for your help guys.

Almost makes me want to audit an algebra class just to refresh my memory