document.write( "Question 459718: Solve these radical equations.
\n" ); document.write( "2x= sqrt(10x+6)
\n" ); document.write( "sqrt(x+5)= 2x
\n" ); document.write( "sqrt(2x-15)= x/4
\n" ); document.write( "sqrt(2x-1)= x+8/2
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Algebra.Com's Answer #315302 by Gogonati(855)\"\" \"About 
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To solve the equation \"2x=sqrt%2810x%2B6%29\", we squared both sides of equation\r
\n" ); document.write( "\n" ); document.write( "\"%282x%29%5E2=10x%2B6\", set the equation equal to zero \"4x%5E2-10x-6=0\", using the \r
\n" ); document.write( "\n" ); document.write( "quadratic formula we find the roots of the equation.\r
\n" ); document.write( "\n" ); document.write( "\"x=%2810%2B-sqrt%2810%5E2-4%2A4%2A%28-6%29%29%29%2F2%2A4=%2810%2B-sqrt%28196%29%29%2F8\", and the roots are:\r
\n" ); document.write( "\n" ); document.write( "\"x=%2810%2B14%29%2F8=3\", and \"x=%2810-14%29%2F8=-1%2F2\".\r
\n" ); document.write( "\n" ); document.write( "After checking the roots we reject the root x=-1/2 because it doesn't satisfy the original equation.\r
\n" ); document.write( "\n" ); document.write( "Answer: The solution is x=3.\r
\n" ); document.write( "\n" ); document.write( "In the same way you can solve the other equations.
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