Radical Problem (Algebra)?

Question

√(x+15) - √(2x+7) = 1

I need to know how to solve this problem for tommorrow's test.

Here's what I have:

x+ 15 = 2x + 7+1
15 = x+8
7=x

I squared both sides but I don't know what to do with the one. I'm pretty sure I got this problem wrong, can someone explain this to me? Thanks

Please I have a 72% in Algebra, I need to know how to solve this problem. I typically score in Cs range in tests. I need a B to stay afloat

Raj said:

"squaring again
4(x+15)(2x+7)=(3x+21)^2"

Where did you get the 4 from?

Answer 1

ok to begin we should start by breaking down what squaring actually accomplishes. So lets square both sides

sqrt(x+15) = (sqrt(2x+7)+1)

So square

x+15 = (sqrt(2x+7) +1)(sqrt(2x+7)+1)

It is not the simple square you did as when you square, you actually have to foil since squaring does not distribute over addition. Therefore we get

x+15 = sqrt(2x+7))^2 +2*(sqrt(2x+7)) +1

that leaves us with:

x+15 = 2x+7+1+2*sqrt(2x+7)

So simplify

-x+7 = 2*sqrt(2x+7)

So lets take out the negative from the right side:

-(x-7) = 2*sqrt(2x+7)

So now we are ready to square both sides again and get

(x-7)^2 = 4*(2x+7)
x^2-14x+49 = 8x+28

Now simplifty

x^2 -22x+21 = 0

we can then foil to see:

(x-1)(x-21)= 0

so x = 1 or x=21.

Hope this helps.

Answer 2

√(x+15) - √(2x+7) = 1
<=> √(x+15) = √(2x+7) + 1
<=> x + 15 = 2x + 8 + 2√(2x+7)
<=> 7 - x = 2√(2x+7)
<=> 4(2x + 7) = (x-7)^2 and x <= 7
<=> 8x + 28 = x^2 - 14x + 49 and x <= 7
<=> x^2 - 22x + 21 = 0 and x <= 7
<=> x = 1 or x = 21 and x <= 7
<=> x =1

note that A = B <=> A^2 = B^2 when and only when A and B > 0
sqrt(A) = B <=> A = B^2 and B>=0

Answer 3

First rearrange the terms:

√(x+15) = 1 + √(2x+7)

Now square both sides:

x+15 = 1 + 2√(2x+7) + 2x+7

Collect terms:

0 = x - 7 +2√(2x+7)

Now subsitute √(2x+7 = r;

The equation becomes

0 = x - 7 + 2r

If √(2x+7) = r, then 2x+7 = r^2 and x = (r^2 - 7)/2; put that in for x:

0 = (r^2-7)/2 - 7 + 2r and rearrange, collecting terms to

r^2 + 4*r -21 = 0. solve this quadratic for r

(r-3)*(r+7) so r = 3 or r = -7

then from x = (r^2 - 7)/2, x = 1 or x = 21

Putting both terms back into the orginal expression shows that x=1 gives the right result (+1). x = 21 gives -1

Answer 4

[√(x+15)]² = [1² + √(2x+7)]²
x + 15 = 1+ 2x + 7
x - 2x = 8 - 15
-x = -7
x = 7
<><>

Answer 5

First seem for any factors that are suited squares: that is not clean, yet i'm assuming the x's are below the sq. root sign one million. ?144x + ?36x - ?25x one hundred forty four = 12² 36 = 6² 25 = 5² So each and all of the numbers below the unconventional are suited squares. because of the fact which you already understand that the sq. root of a variety squared is purely the variety: ?144x + ?36x - ?25x = ?12²x + ?6²x - ?5²x = 12?x + 6?x - 5?x = 17?x 2. ?27x^3 * ?3x 27 = 9*3 = 3²*3 ?27x^3 * ?3x = ?(3² * 3)x³ + ?3x = 3?3x³ * ?3x = 3(?3)(?3)?(x³)(x) = 3*3*?x^4 = 9x² 3. ?18x³/?3xy 18/3 = 6 ?18x³/?3xy = ?(18/3)*x³/xy = ?6x²/y = x?(6/y)

Answer 6

squaring
x+15+2x+7-2rt(x+15)(2x+7)=1
3x+22-2rt(x+15)(2x+7)=1
2rt(x+15)(2x+7)=3x+21
squaring again
4(x+15)(2x+7)=(3x+21)^2
4(2x^2+37x+105)=9x^2+126x+441
8x^2+148x+420=9x^2+126x+441
=>x^2-22x+21=0
=(x-21)(x-1)=0
x=21 or 1
21 is not admissible as it doesn't check
so x=1 is the solution

Answer 7

√(x+15) - √(2x+7) = 1

ok it's:

(x+15)-(2x+7)=(1)^2

x+15-2x-7 (remember...the second part is distributive) = 1

-x+8=1

subtract 8

-x= -7

divide by -1, b/c x is NEVER negative.

x=7

and there's your answer.

Hope this helped. :)