Show that x

^{3}- 2x

^{2}+ 4x - 6 = 0 has a root between 1 and 2. Find the root correct to two decimal places.

Posted 26 April 2011 - 11:53 AM

quite easy to show it has a root between 1 and 2

x^{3} - 2x^{2} + 4x - 6

let x=1

1^{3} - (2)(1^{2}) + 4(1) - 6 = -3

let x=2

2^{3} - (2)(2^{2}) + 4(2) - 6 =2

since polynomials are continuous functions (they don't have jumps or breaks), since at 1 it is below the x-axis and at 2 it is above the x-axis at some point in between it must have crossed it.

As for finding the root to 2 dp, the only way to do this in higher is trial and error, start at a number say 1.5 substitute it in if negative increase the number, if positive decrease it.

[Answer should be 1.71 as 1.71 is negative but 1.715 is positive so root has to be between them, which rounded to 2 dp, is always 1.71]

Hope this helps

x

let x=1

1

let x=2

2

since polynomials are continuous functions (they don't have jumps or breaks), since at 1 it is below the x-axis and at 2 it is above the x-axis at some point in between it must have crossed it.

As for finding the root to 2 dp, the only way to do this in higher is trial and error, start at a number say 1.5 substitute it in if negative increase the number, if positive decrease it.

[Answer should be 1.71 as 1.71 is negative but 1.715 is positive so root has to be between them, which rounded to 2 dp, is always 1.71]

Hope this helps