document.write( "Question 192330: 3+(sqroot5x-1)=x \n" ); document.write( "

Algebra.Com's Answer #144389 by RAY100(1637) $\"\"$ $\"About$

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3+sq rt (5x-1) = x\r
\n" ); document.write( "\n" ); document.write( "subt 3 each side\r
\n" ); document.write( "\n" ); document.write( "sq rt(5x-1) = x-3\r
\n" ); document.write( "\n" ); document.write( "sq both sides\r
\n" ); document.write( "\n" ); document.write( "+-(5x-1) = (X-3)^2 \r
\n" ); document.write( "\n" ); document.write( "USING ONLY + answer ( - answer gives imaginary answers)\r
\n" ); document.write( "\n" ); document.write( "+(5x-1) = (x-3)(x-3)= x^2-6x+9\r
\n" ); document.write( "\n" ); document.write( "subt 5x each side\r
\n" ); document.write( "\n" ); document.write( "-1 = x^2 -6x +9 -5x\r
\n" ); document.write( "\n" ); document.write( "add 1 to each side\r
\n" ); document.write( "\n" ); document.write( "0= X^2 -6x +9 -5x +1
\n" ); document.write( "simplifying\r
\n" ); document.write( "\n" ); document.write( "0=x^2 -11x +10\r
\n" ); document.write( "\n" ); document.write( "factoring
\n" ); document.write( "0=(x-10)(x-1)
\n" ); document.write( "x=10 and x=1
\n" ); document.write( "checking
\n" ); document.write( "3 +sq rt (5x-1) = x
\n" ); document.write( "3 + sq rt (5(1)-1) = (1)
\n" ); document.write( "3 + sqrt 4=1
\n" ); document.write( "3+-2 =1 ok with -2\r
\n" ); document.write( "\n" ); document.write( "for x=10
\n" ); document.write( "3+sqrt (5(10)-1) = (10)
\n" ); document.write( "3+ sqrt 49 =10
\n" ); document.write( "3+-7=10 ok with +7 \n" ); document.write( "

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