Non Linear equation w/ square roots on both sides of = sign

[tklopfstein]

(√3(x+4))=1+(√2x+7)

(√3(x+4))²=(1+(√2x+7))²
3x + 12 = (1+(√2x+7)) * (1+(√2x+7))
3x +12 = 1 + 2√2x+7 + 2x +7
-2x -7
(X+4)²=(2√2x+7)²
X²+16=4(2x+7)
(X+4) (X+4) = 8x + 28
x² +8x + 16 = 8x + 28
-8x -28 -8x -28
X²-12=0
X²-12+36= 0 +36
(X-6)²=36
X-6=√36
X= 6+ √36 x= 6 + 2√3

Answer should be +- 2√3

Where did I go wrong. I have done this problem five times over now, still can’t get it right. In the original problem 3(x+4) is all under the square root and only 2x+7 is under the square root on the other side of the equal sign.

 

[stapel]

Is the equation, as posted:

. . . . .sqrt(3)×(x + 4) = 1 + sqrt(2)×(x) + 7

...or did you mean this:

. . . . .sqrt(3x + 4) = 1 + sqrt(2x + 7)

...or something else?

Eliz.

 

[tklopfstein]

It is sqrt 3(x+4)=1 + sqrt(2x + 7)

 

[Guest]

"
X²-12=0
X²-12+36= 0 +36
(X-6)²=36
X-6=√36
X= 6+ √36 x= 6 + 2√3

Answer should be +- 2√3 "

HERE:

You have everything right up x^2-12 = 0

I will continue it for you right here:

x^2=12
x = +-sqrt[12]
x = +-2sqrt[3]

because the square root of 12 reduces to sqrt[4]sqrt[3]
which reduces to 2sqrt[3]

I hope this helps.

Take care,
Beckie

 

[stapel]

Using grouping symbols can be quite helpful. Do you mean:

. . . . .sqrt[3]×(x + 4) = 1 + sqrt[2x + 7]

...as you have written, or do you mean:

. . . . .sqrt[3(x + 4)] = 1 + sqrt[2x + 7]

...or something else?

Eliz.

 

[soroban]

Re: Non Linear equation w/ square roots on both sides of = s

Hello, tklopfstein!

You did fine . . . up to the last few steps . . . a silly error.

√[3(x + 4)] = 1 + √(2x+7)

[√3(x+4)]² = [1 + √(2x + 7)]²

3x + 12 = [1 + √(2x+7)] * [1 + √(2x+7))]

3x + 12 = 1 + 2√(2x+7) + 2x +7

x + 4 = 2/(2x + 7)

(x + 4)² = [2√(2x+7)]² . . . . then you squared incorrectly **

. . x² + 8x + 16 .= . 4(2x + 7)

. . x² + 8x + 16 .= .8x + 28

. . . . . . . . . . x² .= .12

. . . . . . . . . . .x .= .±√12 .= . ±2√3

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

** And <u>obviously</u> ... you know better . . .

After all, you squared that ugly radical-thing in steps 2, 3, and 4.