Question

if x is equals to 1 upon 2 minus root 3 then show that the value of x cube minus 2 x square - 7 x + 5 is equals to minus 3

Answer 1

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

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}

Answer 2

Answer:

Step-by-step explanation: answer will be 3 instead of minus 3