document.write( "Question 1037665: The equation x^3-5x=34 has a solution between 3 and 4. Give
\n" ); document.write( "your answer correct to 1 d.p. \r
\n" ); document.write( "\n" ); document.write( "So I did 3.5 and 3.6 and there too high and then I did 3.7 which gave me 32.153 and I did 3.8 which gave me 35.872. Then I found the difference between 34 and 32.153 and it was 1.847 and the difference between 35.872 and 34 which gave me 1.872. Does that mean that the solution 3.7 is closer to 34 and therefore the most accurate solution? The question is worth 4 marks.Please help this was in my GCSE. \n" ); document.write( "
Algebra.Com's Answer #652422 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
The equation x³-5x = 34 and it has a solution between 3 and 4.
\n" ); document.write( "
\r\n" );
document.write( "\r\n" );
document.write( "You should first write the equation with a 0 on the right:\r\n" );
document.write( "x³-5x-34 = 0, for it's easier to tell which is closer to 0\r\n" );
document.write( "than it is to tell which is closer to 34.\r\n" );
document.write( "\r\n" );
document.write( "When you get a POSITIVE answer, then the next time choose\r\n" );
document.write( "a value between what you tried to get that POSITIVE answer,\r\n" );
document.write( "and what you tried the last time you got a NEGATIVE answer.\r\n" );
document.write( "And vice-versa. \r\n" );
document.write( "\r\n" );
document.write( "We substitute 3: 3³-5(3)-34 = -22\r\n" );
document.write( "We substitute 4: 4³-5(4)-34 = 10\r\n" );
document.write( "\r\n" );
document.write( "We substitute a number between 3 and 4\r\n" );
document.write( "with 1 decimal place that is close as possible\r\n" );
document.write( "to halfway between them.\r\n" );
document.write( "\r\n" );
document.write( "We substitute 3.5: 3.5³-5(3.5)-34 = -8.625\r\n" );
document.write( "That's a NEGATIVE number. \r\n" );
document.write( "\r\n" );
document.write( "So now we know that the solution is between 3.5 and \r\n" );
document.write( "what we substituted the last time we got a POSITIVE \r\n" );
document.write( "number, which was when we substituted 4 and got 10.\r\n" );
document.write( "\r\n" );
document.write( "We substitute a number between 3.5 and 4\r\n" );
document.write( "with 1 decimal place that is close as possible\r\n" );
document.write( "to halfway between them.\r\n" );
document.write( "\r\n" );
document.write( "So we substitute 3.7  (we could substitute 3.8, if we like,\r\n" );
document.write( "since it's just as close to halfway between 3.5 and 4 as 3.7)\r\n" );
document.write( "\r\n" );
document.write( "We substitute 3.7: 3.7³-5(3.7)-34 = -1.847\r\n" );
document.write( "\r\n" );
document.write( "That's another NEGATIVE number.\r\n" );
document.write( "\r\n" );
document.write( "So now we know that the solution is between 3.7 and \r\n" );
document.write( "what we substituted the last time we got a POSITIVE \r\n" );
document.write( "number, which was when we substituted 4 and got 10.\r\n" );
document.write( "\r\n" );
document.write( "We substitute a number between 3.7 and 4\r\n" );
document.write( "with 1 decimal place that is close as possible\r\n" );
document.write( "to halfway between them.\r\n" );
document.write( "\r\n" );
document.write( "So we substitute 3.8  (we could substitute 3.9, if we like)\r\n" );
document.write( "\r\n" );
document.write( "We substitute 3.8: 3.8³-5(3.8)-34 = 1.842\r\n" );
document.write( "\r\n" );
document.write( "That's a POSITIVE number.\r\n" );
document.write( "\r\n" );
document.write( "So now we know that the solution is between 3.8 and \r\n" );
document.write( "what we substituted the last time we got a NEGATIVE \r\n" );
document.write( "number, which was when we substituted 3.7 and got -1.847.\r\n" );
document.write( "\r\n" );
document.write( "So the solution is between 3.7 and 3.8, and since\r\n" );
document.write( "1.842 is closer to 0 than -1.847, the solution\r\n" );
document.write( "correct to 1 decimal place is 3.8.\r\n" );
document.write( "\r\n" );
document.write( "Edwin
\n" ); document.write( "
\n" );