urgent need help

[topdawgizalaw]

( parentheses symbolize that the thing inside of them is being square rooted)

1. (x + 10) + (x-6) = 8 remember those first two polynomials are being rooted
2. (k+9) - (k) = (3)
3. (x+2) - 7= (x+9)
4. (4x^2 or 4x squared -3x +2) -2x - 5=0



pls try to explain it clearly so that i can actually understand wtf ur talking about

 

[Guest]

( parentheses symbolize that the thing inside of them is being square rooted)

1. (x + 10) + (x-6) = 8 remember those first two polynomials are being rooted
2. (k+9) - (k) = (3)
3. (x+2) - 7= (x+9)
4. (4x^2 or 4x squared -3x +2) -2x - 5=0


pls try to explain it clearly so that i can actually understand wtf ur talking about

Well, the first thing you need to do is get rid of the square roots in each equation, by squaring everything on both sides (square = opposite of square root). Once they're gone, you can solve like a normal linear equation. For example:

1.
(x+10)^2 + (x-6)^2 = 8^2
x + 10 + x - 6 = 64
2x + 4 = 64
2x = 60
x = 30

 

[topdawgizalaw]

no u square both sides not all 3 terms, so wouldnt u square the polynomial

 

[Guest]

no u square both sides not all 3 terms, so wouldnt u square the polynomial

Huh?

 

[Unco]

\(\displaystyle \sqrt{x + 10} + \sqrt{x-6} = 8\)

\(\displaystyle \sqrt{x+10} = 8 - \sqrt{x-6}\)

Square both sides:

\(\displaystyle x + 10 = 64 - 16\sqrt{x-6} + x-6\)

etc.

 

[Unco]

no u square both sides not all 3 terms, so wouldnt u square the polynomial

Huh?

\(\displaystyle (a + b + c)^2 \neq a^2 + b^2 + c^2\)

 

[stapel]

no u square both sides not all 3 terms, so wouldnt u square the polynomial

This might be why telling other students that they have to help you is not generally the best way to obtain assistance.

Eliz.

 

[Daniel_Feldman]

Woah! Hold the phone!

Yes, you must square both sides.




1. (x + 10) + (x-6) = 8 remember those first two polynomials are being rooted

[(x+10)+(x-6)]^2=8^2

I will use the word "root" before a term to indicate square root, because these parentheses confuse rather easily.


[root(x+10)+root(x-6)][root(x+10)+root(x-6)]=64

(x+10)+2root(x^2+4x-60)+(x-6)=64

2x+4+2root(x^2+4x-60)=64

2x-60=-2root(x^2+4x-60)

x-30=-root(x^2+4x-60)

30-x=root(x^2+4x-60)

Square both sides to get

900-60x+x^2=x^2+4x-60

64x=960

x=15



Check

root(x + 10) + root(x-6) = 8


root(15+10)+root(15-6)=8

root(25)+root(9)=8

5+3=8 Yep, it works. X=15 is your answer.

 

[Denis]

1. (x + 10) + (x-6) = 8 remember those first two polynomials are being rooted

Next time, show these like this (or you get no help!):
sqrt(x + 10) + sqrt(x - 6) = 8

Here's another way:
a = x+10, b = x-6 ; so:
sqrt(a) + sqrt(b) = 8
sqrt(a) = 8 - sqrt(b)
a = 64 - 16sqrt(b) + b : square both sides
x+10 = 64 - 16sqrt(x-6) + x - 6 : substitute back in
16sqrt(x-6) = 48
sqrt(x-6) = 48/16 = 3
x - 6 = 9 : square again
x = 15

ALL your others will work same way...