Question

Hey guys


Question:


If x = 1/3-✓8, find the value of x^3-2x^2-7x+5.


Please answer this question and no spams

It's very urgent please solve it correctly.

Answer 1

Answer:

49+16✓8

Step-by-step explanation:

x = 1/3-✓8

=(3+✓8)/(3-✓8)(3+✓8)

=3+✓8/9-8

x=3+✓8

=================

x²=(3+✓8)^2=9+8+6✓8=17+6✓8

x³= (17+6✓8) (3+✓8)

=51+17✓8+18✓8+6✓8*✓8

=51+35✓8+48

x³=99+35✓8

====================

Now x³-2x²-7x+5

=99+35✓8-2*(17+6✓8)-7(3+✓8)+5

=99+35✓8-34-12✓8-21-7✓8+5

=99-34-21+5+35✓8-19✓8

=49+16✓8

Answer 2

$x = \frac{1}{3 - \sqrt{8} } = \frac{1}{3 - \sqrt{8} } \times \frac{(3 + \sqrt{8}) }{(3 + \sqrt{8} )} = \frac{(3 + \sqrt{8} }{(3)^{2} - ( \sqrt{8} )^{2} } \\ = \frac{(3 + \sqrt{8} }{9 - 8} = (3 + \sqrt{8} )$

$x = 3 + \sqrt{8} =>x - 3 = \sqrt{8} \\ =>(x - 3)^{2} = \sqrt{( {8})^{2} } = 8 \\ => {x}^{2} + 9 - 6x = 8=> {x}^{2} - 6x + 1 = 0$

${x}^{3} - {2x}^{2} - 7x + 5 = x( {x}^{2} - 6x + 1) + 4( {x}^{2} 6x + 1) + 16x + 1 \\ = x \times 0 \times + 4 \times 0 + 16(3 + \sqrt{8} ) + 1 \\ = 48 + 16 \sqrt{8} + 1 = 49 + 32 \sqrt{2}$