Question

51. If x 1 (2-√3) show that the value of (x2 - 2x2 - 7x+5) is 3.​

Answer 1

Step-by-step explanation:

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Answer 2

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

Multiply the numerator and denominator by the conjugate of the denominator

$x=\frac{1}{2-\sqrt{3}}×\frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}}{2^2-\sqrt{3}^2}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}$

$\implies\,x^2=(2+\sqrt{3})^2=4+4\sqrt{3}+3=7+4\sqrt{3}$

Now consider the given expression

$\begin{gathered}x^3-2x^2-7x+5\\=x^3-7x-2x^2+5\\=x(x^2-7)-2x^2+5\\=(2+\sqrt{3})(7+4\sqrt{3}-7)-2(7+4\sqrt{3})+5\\=(2+\sqrt{3})(4\sqrt{3})-14-8\sqrt{3}+5\\=8\sqrt{3}+12-14-8\sqrt{3}+5\\=\boxed{3}\end{gathered}$