document.write( "Question 203898: 2 times the sq root of 90 + 3 times the sq root of 40 - 4 times the sq root of 10 = ?? \n" ); document.write( "
Algebra.Com's Answer #153836 by lazi(15)\"\" \"About 
You can put this solution on YOUR website!
x^(1/2) = square root of x\r
\n" ); document.write( "\n" ); document.write( "2(90)^(1/2) + 3(40)^(1/2) - 4(10)^(1/2) =\r
\n" ); document.write( "\n" ); document.write( "to add or subtract them we need to make sure they have the same square root; its like having the same terms. (i.e. 2(x) + 3(x) = 5(x)...)\r
\n" ); document.write( "\n" ); document.write( "like finding the common denominator we look for what number they have in common.\r
\n" ); document.write( "\n" ); document.write( "90 = 9 * 10
\n" ); document.write( "40 = 4 * 10
\n" ); document.write( "10 = 10\r
\n" ); document.write( "\n" ); document.write( "9 = 3^2
\n" ); document.write( "4 = 2^2\r
\n" ); document.write( "\n" ); document.write( "2(3^2 * 10)^(1/2) + 3(2^2 *10)^(1/2) - 4(10)^(1/2) =
\n" ); document.write( "[2 * 3 * (10)^(1/2)] + [3 * 2 * (10)^(1/2)] - [4 * (10)^(1/2)] =
\n" ); document.write( "6(10)^(1/2) + 6(10)^(1/2) - 4(10)^(1/2) = \r
\n" ); document.write( "\n" ); document.write( "now it's just simply adding and subtracting the number before (10)^(1/2)\r
\n" ); document.write( "\n" ); document.write( "6+6-4= 12-4=8\r
\n" ); document.write( "\n" ); document.write( "the answer is 8(10)^(1/2) ; 8 times the square root of 10 \n" ); document.write( "
\n" );