Question

If x=1/(2-√3),find the value of x3-2x2-7x+5.

Answer 1

The value of $x^{3}-2x^{2}-7x+5$ is $3$.

Step-by-step explanation:

Given,

$x=\frac{1}{2-\sqrt{3} }$ then $x^{3}-2x^{2}-7x+5=?$

∴ $x=\frac{1}{2-\sqrt{3} }$

⇒$x=\frac{2+\sqrt{3} }{(2-\sqrt{3})(2+\sqrt{3} )}$

⇒$x=\frac{2+\sqrt{3} }{(4-3)}$

⇒$x=(2+\sqrt{3})$

And $[tex]x^{2} =(2+\sqrt{3})^{2} =4+4\sqrt{3}+3$[/tex]

⇒$x^{2}=7+4\sqrt{3}$

∴ $x^{3}-2x^{2}-7x+5$

$=x^{2}(x-2)-7x+5$

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

$=(7+4\sqrt{3})(\sqrt{3})-14-7\sqrt{3})+5$

$=7\sqrt{3}+12-14-7\sqrt{3})+5$

$=3$

So, The value of $x^{3}-2x^{2}-7x+5$ is $3$.

Answer 2

Given : $x=\frac{1}{2-\sqrt{3}}$

To Find :$x^3-2x^2-7x+5$

Solution:

Given : $x=\frac{1}{2-\sqrt{3}}$

Rationalizing

$x=\frac{1}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}}\\x=\frac{2+\sqrt{3}}{2^2-(\sqrt{3})^2}\\x=\frac{2+\sqrt{3}}{4-3}\\x=2+\sqrt{3}\\x=3.732$

Given expression :$x^3-2x^2-7x+5$

Substitute the value of x in expression

$(3.732)^3-2(3.732)^2-7(3.732)+5$

2.998

Hence the value of $x^3-2x^2-7x+5$is 2.998