document.write( "Question 202322: approximate to the nearest tenth, the real root of the equation f(x)=x^3-4=0 \n" ); document.write( "

Algebra.Com's Answer #152583 by solver91311(24713) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "Since\r
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\n" ); document.write( "\n" ); document.write( "we can say\r
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\n" ); document.write( "\n" ); document.write( "And then simply take the cube root of both sides to find the real number root. There are actually two other complex number roots, but those are not what we are looking for here.\r
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\n" ); document.write( "\n" ); document.write( "The simple way is to punch in 4 inverse x^3 into your calculator and then round off to the nearest tenth. But it is also pretty easy to calculate on paper since you don't have to go beyond the 1st decimal place.\r
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\n" ); document.write( "\n" ); document.write( "First of all, we know that

and

so we know that 2 is too big and 1 is too small. So let's try 1.5.

. Closer, but still a little small, so let's try 1.6.

, a touch large. So now we have it bracketed.

. But we can also see that

is closer to 4 than

. Hence the answer, to the nearest tenth, is 1.6.\r
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