Question

If x= 7/8-5√2 then the value of (2x³-24x² + 71x + 47) is​

Answer 1

$\textsf{After substituting the value of x :}$

$\sf{\displaystyle 2\left(\frac{7}{8-5\sqrt{2}}\right)^3-24\left(\frac{7}{8-5\sqrt{2}}\right)^2+71\left(\frac{7}{8-5\sqrt{2}}\right)+47}$

$\sf{\displaystyle=2\left(\frac{7}{8-5\sqrt{2}}\right)^3-24\left(\frac{7}{8-5\sqrt{2}}\right)^2+71\cdot \frac{7}{8-5\sqrt{2}}+47}$

$\sf{\displaystyle=\frac{343}{856-605\sqrt{2}}-\frac{588}{57-40\sqrt{2}}+\frac{497}{8-5\sqrt{2}}+47}$

$\sf{=\frac{343\left(856-605\sqrt{2}\right)}{\left(8-5\sqrt{2}\right)\left(57-40\sqrt{2}\right)\left(856-605\sqrt{2}\right)}-\frac{588\left(12898-9120\sqrt{2}\right)}{\left(8-5\sqrt{2}\right)\left(57-40\sqrt{2}\right)\left(856-605\sqrt{2}\right)}+\frac{497\left(97192-68725\sqrt{2}\right)}{\left(8-5\sqrt{2}\right)\left(57-40\sqrt{2}\right)\left(856-605\sqrt{2}\right)}}$

$\sf{=\dfrac{343\left(856-605\sqrt{2}\right)-588\left(12898-9120\sqrt{2}\right)+497\left(97192-68725\sqrt{2}\right)}{\left(8-5\sqrt{2}\right)\left(57-40\sqrt{2}\right)\left(856-605\sqrt{2}\right)}+47}$

$\sf{=\dfrac{343\left(856-605\sqrt{2}\right)-588\left(12898-9120\sqrt{2}\right)+497\left(97192-68725\sqrt{2}\right)}{1464786-1035760\sqrt{2}}+47}$

$\sf{=\dfrac{41014008-29001280\sqrt{2}}{1464786-1035760\sqrt{2}}+47}$

$\sf{=\dfrac{56\left(732393-517880\sqrt{2}\right)}{1464786-1035760\sqrt{2}}+47}$

$\sf{=28+47}$

$\bf{=75}$