document.write( "Question 31046This question is from textbook Algebra 2, Student Edition
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document.write( ": the square root of (3x+10) = 1 + the square root of (2x+5) \n" );
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Algebra.Com's Answer #17818 by ikdeep(226) You can put this solution on YOUR website! My earlier solution is incorrect ,,kindly study this one ,,,\r \n" ); document.write( "\n" ); document.write( "the square root of (3x+10) = 1 + the square root of (2x+5)\r \n" ); document.write( "\n" ); document.write( "on squaring both sdes we get.......\r \n" ); document.write( "\n" ); document.write( "(3x+10) = [1 + the square root of (2x+5)]^2 \n" ); document.write( "O \n" ); document.write( "n RHS we apply fromula (a+b)^2 = (a)^2 + (b)^2 + 2(a)(b) \n" ); document.write( "here a = 1 and b = the square root of (2x+5).. \n" ); document.write( "on applying formala we get ...\r \n" ); document.write( "\n" ); document.write( "(3x+10)=(1)^2 + [the square root of (2x+5)]^2 + 2(1)(the square root of (2x+5)\r \n" ); document.write( "\n" ); document.write( "on solving RHS we get...\r \n" ); document.write( "\n" ); document.write( "3x+10 = 1 + 2x + 5 + 2 (1)(the square root of (2x+5)\r \n" ); document.write( "\n" ); document.write( "3x+10 = 2x + 6 + 2 (1)(the square root of (2x+5)\r \n" ); document.write( "\n" ); document.write( "taking together the like variables we get.....\r \n" ); document.write( "\n" ); document.write( "3x - 2x + 10 - 6 = 2(the square root of (2x+5)\r \n" ); document.write( "\n" ); document.write( "x + 4 = 2(the square root of (2x+5)\r \n" ); document.write( "\n" ); document.write( "again squaring both sides we get\r \n" ); document.write( "\n" ); document.write( "(x + 4)^2 = [2(the square root of (2x+5)]^2\r \n" ); document.write( "\n" ); document.write( "now we apply the earlier fromula on LHS i.e \n" ); document.write( "(a+b)^2 = (a)^2 + (b)^2 + 2(a)(b) \n" ); document.write( "here a = x and b = 4 \n" ); document.write( "on applying formula we get...\r \n" ); document.write( "\n" ); document.write( "(x)^2 + (4)^2 + 2(x)(4) = [2(the square root of (2x+5)]^2\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "x^2 + 16 + 8x = 4(2x+5)\r \n" ); document.write( "\n" ); document.write( "x^2 + 16 + 8x = 8x+ 20\r \n" ); document.write( "\n" ); document.write( "subtract 8x from both sides ,,we get\r \n" ); document.write( "\n" ); document.write( "x^2 + 16 = 20\r \n" ); document.write( "\n" ); document.write( "subtract 16 from both sides ,,we get\r \n" ); document.write( "\n" ); document.write( "x^2 = 4 \r \n" ); document.write( "\n" ); document.write( "taking square root of both sides we get..\r \n" ); document.write( "\n" ); document.write( "either x = 2 0r x = -2\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "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..\r \n" ); document.write( "\n" ); document.write( "hope this will help you \r \n" ); document.write( "\n" ); document.write( "Please feel free to revert back for any further queries.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |