Can anyone help me with this ? (1 Viewer)

[captain awesome]

lx^2 + 1l = 17

i don't understand how to do absolute values if there is a squared number :/

 

[SeCKSiiMiNh]

lx^2 + 1l = 17

i don't understand how to do absolute values if there is a squared number :/

lx^2 + 1l = 17


[ (x^2 + 1)^2 ]^1/2 = 17

square both sides

(x^2 + 1)^2 = 289
x^4 + 2x^2 + 1 = 289

x^4 + 2x^2 - 288 = 0

use ur quadratic formula now and u should end up with

x^2 = -1 +- 17

ignore the -1-17 since x wont be real when you root it

your answer should be +4 and -4

 

[thongetsu]

x^2 + 1 = 17
x^2 + 1 = 16
x= -+ 4

x^2 + 1 = -17
x = -+ square root 18

you have to check these answers to mke sure, im not bothered

 

[blackratpoo]

 

[SeCKSiiMiNh]

awww my method takes the longest

 

[captain awesome]

ahh i see thank you !!!!!

one more question tho :
lx^2 + 3xl = x^2 + 6x +4


the answere is -1/2

 

[hscishard]

Yes. After substituting the possible values of x, only -1/2 will remain.

 

[blackratpoo]

 

[captain awesome]

i see, but how did you check it ??

 

[Drongoski]

with this question you guys are all doing too much work.
there is no need to take the negative case of the absolute value.
because x^2 +1 cant be negative anyway. at least in this context.

Xcelz is right. Since (for reals) x^2 + 1 is always > 0, abs(x^2 + 1) = x^2 + 1.

.: eqn is simply: x^2 + 1 = 17 ==> x = -+ 4.

 

[captain awesome]

how do you sub it in ??
thats what i am having trouble with.

 

[alcalder]

Take the Left hand side of the equation and make x = to your possible answers.

Eg let x = -1/2

LHS = |x^2 + 3x|
= |(-1/2)^2 + 3 x -1/2|
= |1/4-3/2|
= |-5/4|
= 5/4 = 1 1/4

Then so the same with the Right Hand Side.
RHS = x^2 + 6x +4
= (-1/2)^2 + 6 x -1/2 + 4
= 1/4 - 3 + 4
= 1 1/4
= RHS

If RHS = LHS then the substitution is valid and therefore an answer.

 

[captain awesome]

Thanks everyone