SOLUTION: the square root of (3x+10) = 1 + the square root of (2x+5)

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Question 31046This question is from textbook Algebra 2, Student Edition
: the square root of (3x+10) = 1 + the square root of (2x+5) This question is from textbook Algebra 2, Student Edition

Found 2 solutions by Cintchr, ikdeep:
Answer by Cintchr(481) About Me  (Show Source):
You can put this solution on YOUR website!
+sqrt%283x%2B10%29+=+1+%2B+sqrt%282x%2B5%29+ square both sides
+3x+%2B+10+=+%281+%2B+sqrt%282x%2B5%29%29%281+%2B+sqrt%282x%2B5%29%29+ foil the right side
F: +1+%2A+1+=+1+
O: +1+%2A+sqrt%282x%2B5%29+=+sqrt%282x%2B5%29+
I: +1+%2A+sqrt%282x%2B5%29+=+sqrt%282x%2B5%29+
L: +sqrt%282x%2B5%29%2Asqrt%282x%2B5%29+=+2x%2B5+ combine like terms
+3x+%2B+10+=+1+%2B+sqrt%282x%2B5%29+%2B+sqrt%282x%2B5%29+%2B+2x+%2B+5+
+3x+%2B+10+=+6+%2B+2%28sqrt%282x%2B5%29%29+%2B+2x+ Move everything AWAY from the sqrt
+x+%2B+4+=+2%28sqrt%282x%2B5%29%29+ divide all parts by 2
+1x%2F2+%2B+2+=+sqrt%282x%2B5%29+ square both sides
+%281%2F4%29x%5E2+%2B+2x+%2B+4+=+2x+%2B+5+ move everything to the left side
+%281%2F4%29x%5E2+-+1+=+0+ This is the difference of two squares.
+%28%281%2F2%29x+-1%29%28%281%2F2%29x%2B1%29+=+0+
1x%2F2+-1+=+0+ and 1x%2F2+%2B1+=+0+ solve for x in both equations
+1x%2F2+=+1+ and +1x%2F2+=+-1+
+x+=+2+ and +x+=+-2+

Answer by ikdeep(226) About Me  (Show Source):
You can put this solution on YOUR website!
My earlier solution is incorrect ,,kindly study this one ,,,
the square root of (3x+10) = 1 + the square root of (2x+5)
on squaring both sdes we get.......
(3x+10) = [1 + the square root of (2x+5)]^2
O
n RHS we apply fromula (a+b)^2 = (a)^2 + (b)^2 + 2(a)(b)
here a = 1 and b = the square root of (2x+5)..
on applying formala we get ...
(3x+10)=(1)^2 + [the square root of (2x+5)]^2 + 2(1)(the square root of (2x+5)
on solving RHS we get...
3x+10 = 1 + 2x + 5 + 2 (1)(the square root of (2x+5)
3x+10 = 2x + 6 + 2 (1)(the square root of (2x+5)
taking together the like variables we get.....
3x - 2x + 10 - 6 = 2(the square root of (2x+5)
x + 4 = 2(the square root of (2x+5)
again squaring both sides we get
(x + 4)^2 = [2(the square root of (2x+5)]^2
now we apply the earlier fromula on LHS i.e
(a+b)^2 = (a)^2 + (b)^2 + 2(a)(b)
here a = x and b = 4
on applying formula we get...
(x)^2 + (4)^2 + 2(x)(4) = [2(the square root of (2x+5)]^2

x^2 + 16 + 8x = 4(2x+5)
x^2 + 16 + 8x = 8x+ 20
subtract 8x from both sides ,,we get
x^2 + 16 = 20
subtract 16 from both sides ,,we get
x^2 = 4
taking square root of both sides we get..
either x = 2 0r x = -2

Now if you want to verify you answer ,,,you can put the values of x in the given equation and if you get LHS=RHS,,,this means that you answer is correct..
hope this will help you
Please feel free to revert back for any further queries.