Question

if x=3+root 5 /2 then 7x^4-3x^3-58x^2 +51x+29=?​

Answer 1

Answer:

$\frac{8371}{16}+\frac{1917}{8}\sqrt{5}$

Step-by-step explanation:

Given, $x=3+\frac{\sqrt{5} }{2}$ .... (1)

We have to find the value of,

$7x^{4}-3x^{3}-58x^{2}+51x+29$

Squaring both sides of equation (1), we get

$x^{2} =9+\frac{5}{4}+3\sqrt{5}$=$\frac{41}{4}+3\sqrt{5}$ ...... (2)

Now, $x^{3}= x^{2}.x=(3+\frac{\sqrt{5} }{2})(\frac{41}{4}+3\sqrt{5})$

[From equations (1) and (2)]

=$\frac{123}{4}+\frac{15}{2}+9\sqrt{5}+\frac{41\sqrt{5} }{8}$

=$\frac{123+30}{4}+(9+\frac{41}{8})\sqrt{5}$

⇒$x^{3}=\frac{153}{4}+\frac{113\sqrt{5}}{8}$ .......(3)

Now squaring both sides of (2), we get

$x^{4}=\frac{1681}{16}+45+2*\frac{41}{4}*3\sqrt{5}$

⇒$x^{4}=\frac{2401}{16}+\frac{123\sqrt{5} }{2}$ ......(4)

Hence,

$7x^{4}-3x^{3}-58x^{2}+51x+29$

Using equations (1),(2),(3), and (4)

=$(\frac{16807}{16}+\frac{861\sqrt{5}}{2})-(\frac{459}{4}+\frac{339\sqrt{5}}{8})-(\frac{2378}{4}+174\sqrt{5})+(153+\frac{51\sqrt{5} }{2})+29$

=$\frac{16807-1836-9512+2912}{16}+\frac{3444-339-1392+204}{8}\sqrt{5}$

=$\frac{8371}{16}+\frac{1917}{8}\sqrt{5}$

(Answer)