document.write( "Question 1088659: Please help to solve that there is a root of the equation $\"x%5E4%2B3x%5E2-7x%2B1=0\"$ , using intermediate value theorem. \n" ); document.write( "

Algebra.Com's Answer #702969 by natolino_2017(77) $\"\"$ $\"About$

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if p(x) = x^4+3x^2-7x+1, we need to find p(a) = 0.\r
\n" ); document.write( "\n" ); document.write( "according to the Fundamental algebra theorem, there must be 4 roots on the complex number.\r
\n" ); document.write( "\n" ); document.write( "+-1 could be a rational root.\r
\n" ); document.write( "\n" ); document.write( "p(1) = -2 and p(-1) = 12, so the polynomial does not have any rational root.\r
\n" ); document.write( "\n" ); document.write( "but using the intermediate value theorem p(1)*p(-1) = -24<0 so there's at least a root between (-1,1), beacuse the images has different sign, so must be a root between the two pre-images. That solve the question.\r
\n" ); document.write( "\n" ); document.write( "****bonus: p(2) = 15, and as before p(1)*(p2) = -30<0 so there's at least a root between (1,2).\r
\n" ); document.write( "\n" ); document.write( "Using derivative and a software, we can see that there's 2 real roots:\r
\n" ); document.write( "\n" ); document.write( "x1= 0,15296291586997 and x2= 1,31770465653627 which are irrational numbers and are consistent with intermediate value theorem intervals, the other two root are complex and cannot be calculated with my software.*****\r
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