Question

Prove that x³-2x²-7x+5=3 when x=1/(2-√3)

Answer 1

1/(2-√3)= 2+√3/1 = (2+√3)^3-( 2+√3)^2 - 7(2+√3) + 5=( 26+15√3) - (14 - 8√3)-(14 - 7√3) + 5 = 3 HENCE , PROVED.

Answer 2

$\frac{1}{2- \sqrt{3} } = \frac{1}{2- \sqrt{3} } * \frac{2+ \sqrt{3}}{2+ \sqrt{3} } = \frac{ 2+ \sqrt{3}}{2^2-3}=\frac{ 2+ \sqrt{3}}{1} = 2+ \sqrt{3}$

x³ - 2x² - 7x + 5

$(2+ \sqrt{3})^3 - 2(2+ \sqrt{3})^2 - 7(2+ \sqrt{3}) + 5$

$8 + 3 \sqrt{3} + 6 \sqrt{3} (2+ \sqrt{3}) - 2(4+3+4 \sqrt{3}) - 7(2+ \sqrt{3}) + 5$

$8 + 3 \sqrt{3} + 12 \sqrt{3} + 18 - 14-8 \sqrt{3} - 14-7 \sqrt{3} + 5$

$31 + 15 \sqrt{3} - 28-15 \sqrt{3}$

3