If x=1/(2-√3),then the value of (x^3-2x^2-7x+5)

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Solution

dear student -4+2√3 is NOT simplified as -2(2+√3) -2*√3 ...but from -4+2√3 we took -2 outside..then we will get -2(2-√3)
this value -2(2-√3) is substituted in the expression (2 + √3) [-4 + 2√3] + 5...instaed of [-4 + 2√3]

= -2(2 + √3)(2 - √3) + 5

then we know that (a+b)(a-b)=a²-b²

similarly (2 + √3)(2 - √3) is in the form of (a+b)(a-b)..so we get (2 + √3)(2 - √3)=2²-√3²=4-3=1

then we get

-2(2 + √3)(2 - √3) + 5= -2(1) + 5

= 3

you can also solve this question in another way

$x=\frac{1}{2-\sqrt{3}}$
Multiply the numerator and denominator by the conjugate of the denominator
$x=\frac{1}{2-\sqrt{3}}\cdot\frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}}{2^2-\sqrt{3}^2}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}$
$\implies\,x^2=(2+\sqrt{3})^2=4+4\sqrt{3}+3=7+4\sqrt{3}$

Now consider the given expression and put the values of x and 2²
$x^3-2x^2-7x+5\\=x^3-7x-2x^2+5\\=x(x^2-7)-2x^2+5\\=(2+\sqrt{3})(7+4\sqrt{3}-7)-2(7+4\sqrt{3})+5\\=(2+\sqrt{3})(4\sqrt{3})-14-8\sqrt{3}+5\\=8\sqrt{3}+12-14-8\sqrt{3}+5\\=\boxed{3}$

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Similar questions

If $x = \frac{1}{(2 - \sqrt{3}),}$ then find the value of $(x^{3} - 2 x^{2} - 7 x + 5)$ .

If x=(/(2-3^1/2), then the value of (x^3-2x^2-7x+5

x = \frac{1}{2 - \sqrt{3}}

x^{3} - 2 x^{2} - 7 x + 5

x = \frac{1}{2 - \sqrt{3}}

x^{3} + 2 x^{2} - 7 x + 5

x = \frac{1}{2 - \sqrt{3}}

x^{3} - 2 x^{2} - 7 x + 5 = m

m

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